problem
stringlengths 2
5.64k
| solution
stringlengths 2
13.5k
| answer
stringlengths 2
43
| problem_type
stringclasses 8
values | question_type
stringclasses 4
values | problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
value | source
stringclasses 6
values | synthetic
bool 1
class |
|---|---|---|---|---|---|---|---|---|
8. There are 9 students participating in a math competition in the same classroom, with seats arranged in 3 rows and 3 columns, represented by a $3 \times 3$ grid, where each cell represents a seat. To prevent cheating, three types of exams, $A$, $B$, and $C$, are used, and it is required that any two adjacent seats (cells sharing a common edge) receive different types of exams. The number of ways to distribute the exams that meet the conditions is $\qquad$ kinds.
|
8. 246.
Let $a_{ij}$ denote the square at the $i$-th row and $j$-th column. First, consider the number of ways to distribute type $A$ papers at $a_{22}$.
$$
\begin{array}{l}
\text { Let } M=\left\{a_{12}, a_{21}, a_{32}, a_{23}\right\}, \\
N=\left\{a_{11}, a_{31}, a_{33}, a_{13}\right\} .
\end{array}
$$
Consider the types of papers on the squares in $M$. There are several scenarios:
(1) All are type $B$ papers. In this case, there is a unique way to distribute the papers in $M$, and each square in $N$ has 2 ways to distribute the papers, resulting in $2^4 = 16$ ways.
(2) All are type $C$ papers. Similarly to (1), there are $2^4 = 16$ ways.
(3) One square is a type $B$ paper, and the other three are type $C$ papers (as shown in Figure 2). In this case, there are 4 ways to choose one square in $M$ to place the type $B$ paper, and the other squares in $M$ have a unique distribution method. Each square in $N$ has $1 \times 1 \times 2 \times 2 = 4$ ways to distribute the papers, resulting in $4 \times 4 = 16$ ways.
(4) One square is a type $C$ paper, and the other three are type $B$ papers. Similarly to (3), there are $4 \times 4 = 16$ ways.
(5) Two squares are type $B$ papers, and the other two are type $C$ papers. In this case, if the two squares with type $B$ papers are in the same row (or column) (as shown in Figure 3(a)), there are 2 ways to choose the squares in $M$ to place the type $B$ papers, and the other squares in $M$ have a unique distribution method. Each square in $N$ has a unique distribution method, resulting in 2 ways. If the two squares with type $B$ papers are in different rows and columns (as shown in Figure 3(b)), there are 4 ways to choose the squares in $M$ to place the type $B$ papers, and the other squares in $M$ have a unique distribution method. Each square in $N$ has $2 \times 2 \times 1 \times 1 = 4$ ways to distribute the papers, resulting in $4 \times 4 = 16$ ways. Thus, in this scenario, there are $2 + 16 = 18$ ways.
By symmetry, the total number of ways to distribute the papers that meet the requirements is $3(4 \times 16 + 18) = 246$ ways.
|
246
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Find the smallest positive integer $k$, such that $625^{k} \equiv 1(\bmod 343)$
|
$$
\left(5^{4}\right)^{k} \equiv 1\left(\bmod 7^{3}\right),
$$
which can be transformed into $(89 \times 7+2)^{k}-1 \equiv 0\left(\bmod 7^{3}\right)$.
Expanding it, we get
$$
\begin{array}{l}
\mathrm{C}_{k}^{2}(89 \times 7)^{2} \times 2^{k-2}+\mathrm{C}_{k}^{1}(89 \times 7) \times 2^{k-1}+2^{k}-1 \\
\equiv 0\left(\bmod 7^{3}\right) .
\end{array}
$$
Thus, $7 \mid\left(2^{k}-1\right)$, i.e., $2^{k} \equiv 1(\bmod 7)$.
Given $2^{1} \equiv 2(\bmod 7), 2^{2} \equiv 4(\bmod 7), 2^{3} \equiv$ $1(\bmod 7)$, the order of 2 modulo 7 is 3, hence $3 \mid k$.
Let $k=3 s$. Substituting into equation (1), we get
$$
\begin{array}{l}
\mathrm{C}_{38}^{2}(89 \times 7)^{2} \times 2^{34-2}+\mathrm{C}_{35}^{1}(89 \times 7) \times 2^{34-1}+(7+1)^{5}-1 \\
\equiv 0\left(\bmod 7^{3}\right) .
\end{array}
$$
Then $\mathrm{C}_{3}^{2}(89 \times 7)^{2} \times 2^{3-2}+\mathrm{C}_{3}^{1}(89 \times 7) \times 2^{3-1}+$ $\mathrm{C}_{s}^{2} 7^{2}+\mathrm{C}_{s}^{1} 7 \equiv 0\left(\bmod 7^{3}\right)$.
Thus, $\mathrm{C}_{38}^{2} 89^{2} \times 7 \times 2^{3 s-2}+\mathrm{C}_{38}^{1} 89 \times 2^{34-1}+$ $\mathrm{C}_{3}^{2} 7^{2}+\mathrm{C}_{3}^{1}=0\left(\bmod 7^{2}\right)$.
Therefore, $3 s \times 89 \times 2^{3 s-1}+s \equiv 0(\bmod 7)$.
Also, $2^{3-1} \equiv 2^{3-3} \times 2^{2} \equiv\left(2^{3}\right)^{s-1} \times 2^{2} \equiv 2^{2}(\bmod 7)$, so $3 s \times 40 \times 2^{2}+s \equiv 0(\bmod 7)$, i.e., $481 s \equiv 0(\bmod 7)$.
Since $(481,7)=1$, it follows that $7 \mid s$.
Let $s=7 t$. Substituting into equation (2), we get
$$
\begin{array}{l}
\frac{21 t(21 t-1)}{2} \times 89^{2} \times 7 \times 2^{21 t-2}+21 t \times 89 \times 2^{21 t-1}+ \\
\frac{7 t(7 t-1)}{2} \times 7+7 t=0\left(\bmod 7^{2}\right) . \\
\text { Then } \frac{21 t(21 t-1)}{2} \times 89^{2} \times 2^{21 t-2}+3 t \times 89 \times 2^{21 t-1}+ \\
\frac{7 t(7 t-1)}{2}+t=0(\bmod 7) .
\end{array}
$$
Therefore, $3 t \times 89 \times 2^{21 t-1}+t \equiv 0(\bmod 7)$.
Also, $2^{21 t-1} \equiv 2^{21 t-3} \times 2^{2} \equiv\left(2^{3}\right)^{7 t-1} \times 2^{2}$ $=2^{2}(\bmod 7)$,
so $3 t \times 40 \times 2^{2}+t \equiv 0(\bmod 7)$, i.e., $481 t \equiv 0(\bmod 7)$.
Since $(481,7)=1$, it follows that $7 \mid t$.
Let $t=7 r$. Then $k=3 s=21 t=147 r \geqslant 147$. Therefore, the minimum value of $k$ is 147.
|
147
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
The 277th National Junior High School Mathematics Competition consists of 14 questions (5 multiple-choice questions, 5 fill-in-the-blank questions, 4 problem-solving questions), with a full score of 150 points. Among them, each correct answer for multiple-choice and fill-in-the-blank questions earns 7 points, and a wrong answer earns 0 points, with no other point values; each problem-solving question is worth 20 points, and the step scores can only be $0, 5, 10, 15, 20$ points, with no other point values. How many different possible scores are there?
|
The score can be selected from 16 5s and 10 7s, let the total number of 5s and 7s selected be $k$.
Since selection and non-selection are relative, i.e., the score obtained by selecting $m$ 5s and $n$ 7s is the same as the score obtained by selecting $16-m$ 5s and $10-n$ 7s, which sums up to 150, we only need to consider half of the scores. Let the number of different scores that can be obtained between 0 and 74 be $a$. Since 75 can be obtained, the total number of different scores that can be obtained is $2a+1$. Therefore, we only need to discuss up to $k=14$.
(1) When $k=0$, the score is 0;
(2) When $k=2,4,6,8,10$, the scores are all even numbers between $10 \sim 14, 20 \sim 28, 30 \sim 42, 40 \sim 56, 50 \sim 70$ respectively;
(3) When $k=12,14$, the scores are all even numbers between $60 \sim 80, 70 \sim 90$ respectively;
(4) When $k=1$, the scores are all odd numbers between $5 \sim 7$;
(5) When $k=3,5,7,9$, the scores are all odd numbers between $15 \sim 21, 25 \sim 35, 35 \sim 49, 45 \sim 63$ respectively;
(6) When $k=11,13$, the scores are all odd numbers between $55 \sim 75, 65 \sim 85$ respectively.
In summary, the odd numbers that can be obtained between $0 \sim 74$ are 31, and the even numbers are 32. Therefore, $a=63$.
Thus, $2a+1=127$.
|
127
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. There are 25 cards numbered $1, 3, \cdots, 49$. If a card with number $a$ is drawn, the next card to be drawn is the card with the largest odd divisor of $99-a$. This process is repeated until no more cards can be drawn. Xiao Hua drew the card numbered 1. When the operation ends, how many cards are left?
(A) 9
(B) 10
(C) 16
(D) 17
|
4. C.
The card numbers drawn are
$$
1 \rightarrow 49 \rightarrow 25 \rightarrow 37 \rightarrow 31 \rightarrow 17 \rightarrow 41 \rightarrow 29 \rightarrow 35 \rightarrow 1 \text {. }
$$
A total of 9 cards were drawn, leaving 16 cards.
|
16
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
11. Observe the sequence of arrays $(1),(3,5),(7,9,11)$,
$(13,15,17,19), \cdots$. Then 2009 is in the $\qquad$ group.
|
$$
\begin{array}{l}
\text { Notice that 2009 is the 1005th positive odd number. } \\
1+2+\cdots+44<1005 \\
<1+2+\cdots+45,
\end{array}
$$
Then 2009 is in the 45th group.
|
45
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. As shown in Figure 1, given $A(-2,0)$, $B(0,-4)$, and $P$ as any point on the hyperbola $y=\frac{8}{x}$ $(x>0)$. A perpendicular line from $P$ to the x-axis meets at point $C$, and a perpendicular line from $P$ to the y-axis meets at point $D$. The minimum value of the area of quadrilateral $A B C D$ is $\qquad$
|
5.16.
Let point \( P\left(x_{0}, y_{0}\right) \). Then \( y_{0}=\frac{8}{x_{0}}\left(x_{0}>0\right) \). Hence \( C\left(x_{0}, 0\right) \) and \( D\left(0, \frac{8}{x_{0}}\right) \).
From the given conditions,
\[
\begin{array}{l}
|C A|=x_{0}+2,|D B|=\frac{8}{x_{0}}+4 . \\
\text { Therefore, } S=\frac{1}{2}\left(x_{0}+2\right)\left(\frac{8}{x_{0}}+4\right) \\
=2\left(x_{0}+\frac{4}{x_{0}}\right)+8 \geqslant 16 .
\end{array}
\]
Thus, the minimum value of the area of quadrilateral \( A B C D \) is 16.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In a Cartesian coordinate system, draw all rectangles that simultaneously satisfy the following conditions:
(1) The sides of these rectangles are parallel or coincide with the coordinate axes;
(2) All vertices of these rectangles (repeated vertices are counted only once) are exactly 100 integer points (points with both coordinates as integers are called integer points).
Question: What is the maximum number of such rectangles that can be drawn? Explain your reasoning.
|
4. First, prove that the number of such rectangles does not exceed 2025.
Take any 100 integer points. Let $O$ be one of the 100 integer points we have chosen. We call a rectangle "good" if $O$ is one of its vertices, the other three vertices are also chosen from these 100 integer points, and the sides are parallel or coincide with the coordinate axes.
We will prove: There are at most 81 good rectangles.
In fact, draw lines $l_{1}$ and $l_{2}$ through $O$ parallel to the coordinate axes, and let $l_{1} \backslash\{O\}$ contain $m$ points chosen from the 100 integer points, and $l_{2} \backslash\{O\}$ contain $n$ points chosen from the 100 integer points. Let $P$ be one of the 100 integer points chosen, and not on $l_{1}$ and $l_{2}$. Then, at most one good rectangle can have $P$ as one of its vertices. There are at most $99-m-n$ such points, and each good rectangle must have one vertex as such a point. Therefore,
(i) If $m+n \geqslant 18$, then the number of good rectangles is at most $99-m-n \leqslant 81$;
(ii) If $m+n \leqslant 18$, consider the point pairs $(R, Q)$, where $R \in l_{1} \backslash\{O\}$ and $Q \in l_{2} \backslash\{O\}$. Each pair $(R, Q)$ can form at most one good rectangle, so the number of good rectangles is less than or equal to
$$
m n \leqslant m(18-m) \leqslant 9 \times 9=81 \text { . }
$$
In summary, for any point $O$ among the 100 integer points chosen, the number of good rectangles with $O$ as one of its vertices is at most 81. Therefore, the number of rectangles that meet the conditions is less than or equal to $\frac{81 \times 100}{4}=2025$ (here we divide by 4 because each rectangle has 4 vertices).
Below is an example of 2025 such rectangles.
Consider the point set
$$
A=\{(x, y) \mid 1 \leqslant x \leqslant 10,1 \leqslant y \leqslant 10, x y \in \mathbf{N}\} .
$$
By choosing 100 points from the point set $A$, we can exactly draw 2025 rectangles that meet the problem's conditions.
Therefore, the maximum number of such rectangles that can be drawn is 2025.
|
2025
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9.6. In a day, 1000 dwarfs wearing red or blue hats meet each other in pairs. Dwarfs wearing red hats tell lies, while those wearing blue hats tell the truth. Each dwarf may change the color of their hat several times (i.e., red to blue, blue to red). It is known that when any two dwarfs meet, they both say that the other is wearing a red hat. Find the minimum total number of hat changes in a day.
|
9.6. Clearly, when two dwarfs meet, they say that the other is wearing a red hat if and only if their hat colors are different.
Thus, in a day, if three dwarfs never change the color of their hats, then two of them must have the same hat color, and they cannot both say that the other is wearing a red hat when they meet. Therefore, at least 998 dwarfs must have changed the color of their hats, and the total number of hat color changes is greater than or equal to 998.
Below is an explanation that 998 times is achievable.
Let the dwarfs be $1, 2, \cdots, 1000$. Initially, 1 wears a blue hat, and $2, 3, \cdots, 1000$ wear red hats. After 1 meets $2, 3, \cdots, 1000$, 2 changes the color of his hat to blue and then meets 3, $4, \cdots, 1000$, $\cdots \cdots$ 999 changes the color of his hat to blue and then meets 1000, for a total of 998 hat color changes.
|
998
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. As shown in Figure 6, in isosceles $\triangle ABC$, it is known that $AB = AC = k BC$ ($k$ is a natural number greater than 1), points $D$ and $E$ are on sides $AB$ and $AC$ respectively, and $DB = BC = CE$, $CD$ intersects $BE$ at point $O$. Find the smallest positive integer $k$ such that $\frac{OC}{BC}$ is a rational number.
(1991, Shanghai Junior High School Mathematics Competition)
|
Connect $D E$. It is easy to see that quadrilateral $B C E D$ is an isosceles trapezoid.
By the given condition $\angle B D C=\angle B C D=\angle E B C$, hence
$$
O C \cdot C D=B C^{2} \text {. }
$$
On the other hand,
$$
\frac{O C}{O D}=\frac{B C}{D E}=\frac{A B}{A D}=\frac{A B}{A B-D B}=\frac{k}{k-1} .
$$
Therefore, $\frac{O C}{C D}=\frac{k}{2 k-1}$.
(1) $\times$ (2) gives $\frac{O C}{B C}=\sqrt{\frac{k}{2 k-1}}$ is a rational number, and $(k, 2 k-1)=1$.
Thus, both $k$ and $2 k-1$ are perfect squares.
When $k=4,9,16$, $2 k-1=7,17,31$;
When $k=25$, $2 k-1=49$.
Therefore, the smallest positive integer $k$ that makes $\frac{O C}{B C}$ a rational number is $k=25$.
|
25
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given that the area of quadrilateral $ABCD$ is 32, the lengths of $AB$, $CD$, and $AC$ are all integers, and their sum is 16.
(1) How many such quadrilaterals are there?
(2) Find the minimum value of the sum of the squares of the side lengths of such quadrilaterals.
(2003, National Junior High School Mathematics League)
|
(1) As shown in Figure 3, let $A B=a, C D=b, A C=l$, and let the height from $A B$ in $\triangle A B C$ be $h_{1}$, and the height from $D C$ in $\triangle A D C$ be $h_{2}$.
Then
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D} \\
=S_{\triangle A B C}+S_{\triangle A D C} \\
=\frac{1}{2}\left(h_{1} a+h_{2} b\right) \\
\leqslant \frac{1}{2} l(a+b) .
\end{array}
$$
Equality holds if and only if $h_{1}=h_{2}=l$, i.e., $A C \perp A B$ and $A C \perp A D$.
From the given, we have $32 \leqslant \frac{1}{2} l(a+b)$.
Also, by the problem's condition, $a+b=16-l$, so we get
$$
64 \leqslant l(16-l)=64-(l-8)^{2} \leqslant 64 \text {. }
$$
Thus, it must be that $l=8, a+b=8$, and at this time, $A C \perp A B$ and $A C \perp C D$.
Therefore, such quadrilaterals are as follows:
$$
\begin{array}{l}
a=1, b=7, l=8 ; a=2, b=6, l=8 ; \\
a=3, b=5, l=8 ; a=b=4, l=8 .
\end{array}
$$
These are all trapezoids or parallelograms with $A C$ as the height.
(2) From $A B=a, C D=8-a$, we get
$$
B C^{2}=8^{2}+a^{2}, A D^{2}=8^{2}+(8-a)^{2} \text {. }
$$
Therefore, the sum of the squares of the side lengths of such quadrilaterals is
$$
\begin{array}{l}
2 a^{2}+2(8-a)^{2}+2 \times 8^{2} \\
=4(a-4)^{2}+192 .
\end{array}
$$
Hence, when $a=b=4$, the above sum of squares is minimized, and the minimum value is 192.
|
192
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A scientist invented a time machine, which looks like a circular subway track. Now (2010) is the first platform, the 2nd, 3rd, ..., 2009th platforms are 2011, 2012, ..., 4018, and the 2010th platform returns to the present (the departure platform). Later, the machine had a programming error, changing its operation rules to: the passenger specifies a time (i.e., platform number), the machine first arrives at the specified platform, then stops every 4 stations, at the 5th station. If the platform number is a positive integer power of 2, it moves back 2 stations to stop (e.g., $17 \rightarrow 22 \rightarrow 27 \rightarrow 32 \rightarrow 30 \rightarrow 35 \rightarrow \cdots$); if it stops at the first platform, it stops working. Try to answer:
(1) Can this machine get lost in the time track and fail to return to the present (i.e., not stop at the first platform)?
(2) If it can eventually return to the present, what is the maximum number of platforms the machine can stop at?
|
8. (1) The time track has 2009 stations, and the positive integer powers of 2 not exceeding 2009 are
$2,4,8,16,32,64,128,256,512,1024$.
According to the rules,
(i) (1) $5 k+1(k=1,2,3)$
$\xrightarrow{3-k \text { times }} 16 \rightarrow 14=5 \times 2+4$;
(2) $5 k+1(k=4,5, \cdots, 51)$
$\xrightarrow{51-k \text { times }} 256 \rightarrow 254=5 \times 50+4$;
(3) $5 k+1(k=52,53, \cdots, 401)$
$\xrightarrow{401-k \text { times }} 2006 \rightarrow 2$.
(ii) (1) $2 \rightarrow 2009 \rightarrow 5$;
$$
\begin{array}{l}
\text { (2) } 5 k+2(k=1,2, \cdots, 6) \\
\xrightarrow{6-k \text { times }} 32 \rightarrow 30=5 \times 6 \text {; }
\end{array}
$$
$$
\begin{array}{l}
\text { (3) } 5 k+2(k=7,8, \cdots, 102) \\
\xrightarrow{102-k \text { times }} 512 \rightarrow 510=5 \times 102 \text {; } \\
\text { (4) } 5 k+2(k=103,104, \cdots, 401) \\
\xrightarrow{401-k \text { times }} 2007 \rightarrow 3 \text {. } \\
\end{array}
$$
$$
\begin{array}{l}
\text { (iii) (1) } 3 \rightarrow 8 \rightarrow 6=5+1 \text {; } \\
\text { (2) } 5 k+3(k=1,2, \cdots, 25) \\
\xrightarrow{25-k \text { times }} 128 \rightarrow 126=5 \times 25+1 \text {; } \\
\text { (3) } 5 k+3(k=26,27, \cdots, 401) \\
\xrightarrow{401-k \text { times }} 2008 \rightarrow 4 \text {. } \\
\end{array}
$$
(iv) (1) $4 \rightarrow 2 \rightarrow 2009 \rightarrow 5$;
(2) $5 k+4(k=1,2, \cdots, 12)$
$\xrightarrow{12-k \text { times }} 64 \rightarrow 62=5 \times 12+2$;
(3) $5 k+4(k=13,14, \cdots, 204)$
$\xrightarrow{204-k \text { times }} 1024 \rightarrow 1022=5 \times 204+2$;
(4) $5 k+4(k=205,206, \cdots, 401)$
$\xrightarrow{401-k \text { times }} 2009 \rightarrow 5$.
(v) $5 k(k=1,2, \cdots, 401) \xrightarrow{401-k \text { times }} 2005 \rightarrow 1$.
Let the passenger's designated station number be $a(1<a \leqslant 2009)$.
From (i) to (v), we know that regardless of the value of $a$, the machine can always return to the present.
(2) Let the track from station 2009 to station 1 be denoted as $A$.
Let the number of times the machine passes through $A$ be $s$, and the total number of stations it stops at be $t$, where it stops at stations with numbers that are powers of 2 $v$ times, and at other stations $u$ times. Then
$2009 s+1=a+5(u-1)-2 v=a+5 t-5-7 v$.
Thus, $t=\frac{2009 s+7 v-a+6}{5}$
We now prove that the machine will not stop at the same station twice.
Otherwise, the sequence of stations the machine stops at would form a loop. The result would be that it would either never stop at the first station (contradicting (1)), or it would have already stopped at the first station in the first loop, thus stopping and not forming a loop, which is a contradiction.
Therefore, $v \leqslant 10$.
Consider the station number the machine stops at first after each pass through $A$, which is contained in $\{1,2,3,4,5\}$.
Notice
$3 \rightarrow 8 \rightarrow 6 \xrightarrow{2 \text { times }} 16 \rightarrow 14 \xrightarrow{10 \text { times }} 64 \rightarrow 62$
$\xrightarrow{90 \text { times }} 512 \rightarrow 510 \xrightarrow{299 \text { times }} 2005 \rightarrow 1$,
$$
4 \rightarrow 2 \rightarrow 2009 \rightarrow 5 \xrightarrow{400 \text { times }} 2005 \rightarrow 1 \text {. }
$$
Thus, the machine either stops at station 3, or at stations 4, 2, 5, but it cannot stop at both.
Therefore, $s \leqslant 2$.
If $s=1$, then $t \leqslant \frac{2009+70-0+6}{5}=417$.
If $s=2$, then the second loop is
$3 \xrightarrow{407 \text { times }} 1$, or $2 \xrightarrow{403 \text { times }} 1$, or $4 \xrightarrow{404 \text { times }} 1$.
We now use (1) (i) to (v) to trace back the first loop from 3 or 2, 4:
$$
\begin{array}{l}
254 \leftarrow 256 \stackrel{26}{\leftarrow} 126 \leftarrow 128 \stackrel{23}{\longleftarrow} 13 \text {, } \\
2 \leftarrow 2006 \stackrel{349}{\leftarrow} 261 \text {, } \\
4 \leftarrow 2008 \stackrel{375 \text { times }}{\longleftarrow} 133 \text {. } \\
\text { In summary, if } a=13 \text {, then } v=7, t=812 \text {; } \\
\text { if } a=261 \text {, then } v=1, t=754 \text {; } \\
\text { if } a=133 \text {, then } v=2, t=781 \text {. } \\
\end{array}
$$
Therefore, the machine can stop at a maximum of 812 stations.
|
812
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.4. In a $100 \times 100$ grid, each cell contains a positive integer. If the sum of the numbers in the cells of a rectangle (composed of several cells) is a multiple of 17, then the rectangle is called "good". We can color the cells of some good rectangles in the grid, with each cell being colored at most once. It is known that for any such number grid, at least $d$ cells can be colored according to the above rule. Find the maximum value of $d$.
|
10.4. $d_{\max }=9744=100^{2}-16^{2}$.
First, we prove a lemma.
Lemma: In a $1 \times n$ grid, each cell contains a positive integer, then at least $n-16$ cells can be colored. Proof by induction on $n$.
When $n \leqslant 16$, the conclusion is obviously true.
Assume $n=k \geqslant 17$, and for all $n < k$, the conclusion holds.
Suppose the numbers in the leftmost 17 cells are $a_{1}$, $a_{2}, \cdots, a_{17}$. Then among
$$
0, a_{1}, a_{1}+a_{2}, \cdots, a_{1}+a_{2}+\cdots+a_{17}
$$
there exist two numbers such that their difference is divisible by 17, i.e.,
$$
17 \mid\left(a_{i}+a_{i+1}+\cdots+a_{j}\right) .
$$
Thus, the cells from the $i$-th to the $j$-th form a good rectangle A. Removing A from the grid, we get a new $1 \times (k-(j-i+1))$ grid. By the induction hypothesis, the new grid can be expressed as the union of pairwise disjoint good rectangles, except for at most 16 cells. If none of these good rectangles contain both the $(i-1)$-th and $(j+1)$-th cells of the original grid, then they are also good rectangles in the original grid; if one good rectangle contains both the $i$-th and $j$-th cells of the original grid, then its union with A forms a good rectangle in the original grid. In either case, the conclusion holds for $n=k$.
Next, we prove: at least 9744 cells in the grid can be colored.
Consider each column as a large cell, with the number being the sum of the numbers in that column. By the lemma, except for at most 16 columns, the grid can be expressed as the union of several good rectangles of height 100. For the remaining columns, by the lemma, all but at most 16 cells in each column can be colored. Thus, the entire grid has at most $16^{2}=256$ cells that are not colored.
Below is an example where at least 256 cells are not colored.
In a $16 \times 16$ subgrid B of the grid, each cell is filled with 1, and the rest of the cells are filled with 17, then no cell in B can be colored.
In fact, no good rectangle can contain such a cell.
Let A be any good rectangle. If A contains 1, let its intersection with B be an $a \times b$ rectangle, then
$$
a b \equiv 0(\bmod 17),
$$
which contradicts $a, b \leqslant 16$.
|
9744
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given three positive integers $a$, $b$, and $c$ whose squares sum to 2011, and the sum of their greatest common divisor and least common multiple is 388. Then the sum of the numbers $a$, $b$, and $c$ is $\qquad$ .
|
Let the greatest common divisor of $a$, $b$, and $c$ be $d$.
Then $a = d a_{1}$, $b = d b_{1}$, $c = d c_{1}$.
Assume without loss of generality that $a_{1} \geqslant b_{1} \geqslant c_{1}$, and the least common multiple of $a_{1}$, $b_{1}$, and $c_{1}$ is $m$.
From the problem, we have
$$
a_{1}^{2} + b_{1}^{2} + c_{1}^{2} = \frac{2011}{d^{2}}, \quad 1 + m = \frac{388}{d}.
$$
Thus, $d^{2} \mid 2011$, $d \mid 388$, and $d < 45$, $388 = 2^{2} \times 97$, and $d$ is odd.
Therefore, $d = 1$.
Hence, $m = 387 = 9 \times 43$, and $a_{1}$, $b_{1}$, and $c_{1}$ are all less than 45, and they are divisors of $m$, which are $1, 3, 9, 43$.
Upon inspection, $a_{1} = 43$, $b_{1} = c_{1} = 9$.
Therefore, $a + b + c = 61$.
|
61
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Among the triangles with side lengths being consecutive natural numbers and a perimeter not exceeding 100, the number of acute triangles is $\qquad$
$(1987$, National Junior High School Mathematics League)
|
提示: Set the three sides of the triangle to be $n-1$, $n$, and $n+1$. When $n>24$, $(n-1)^{2}+n^{2}>(n+1)^{2}$, so the triangle is an acute triangle. Therefore, the number of acute triangles that meet the requirements is 29.
|
29
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. A company invested in a project in 2009, with both cash inputs and cash revenues every year. It is known that
(1) In 2009, the company invested 10 million yuan, and the investment will decrease by $20\%$ each subsequent year;
(2) In 2009, the company earned 5 million yuan, and the revenue will increase by $25\%$ each subsequent year.
Based on this, the company will recover all its investments by $\qquad$ year.
|
7.2013.
Let the total investment of the project over $n$ years from 2009 be $A_{n}$ million yuan, and the total revenue be $B_{n}$ million yuan.
From (1) we know
$$
\begin{array}{l}
A_{n}=\sum_{k=1}^{n} 1000(1-20 \%)^{k-1} \\
=\frac{1000\left[1-\left(\frac{4}{5}\right)^{n}\right]}{1-\frac{4}{5}}=5000\left[1-\left(\frac{4}{5}\right)^{n}\right] .
\end{array}
$$
Similarly, from (2) we know
$$
B_{n}=\frac{500\left[\left(\frac{5}{4}\right)^{n}-1\right]}{\frac{5}{4}-1}=2000\left[\left(\frac{5}{4}\right)^{n}-1\right] \text {. }
$$
To recover the entire investment, we should have $B_{n} \geqslant A_{n}$, which means
$$
\begin{array}{l}
2000\left[\left(\frac{5}{4}\right)^{n}-1\right] \geqslant 5000\left[1-\left(\frac{4}{5}\right)^{n}\right] \\
\Leftrightarrow\left[5\left(\frac{4}{5}\right)^{n}-2\right]\left[\left(\frac{4}{5}\right)^{n}-1\right] \geqslant 0 \\
\Leftrightarrow 5\left(\frac{4}{5}\right)^{n}-2 \leqslant 0 \\
\Leftrightarrow\left(\frac{4}{5}\right)^{n} \leqslant \frac{2}{5} \Leftrightarrow 0.8^{n} \leqslant 0.4 .
\end{array}
$$
Given $0.8^{5}=0.32768<0.4<0.4096=0.8^{4}$, we get $n \geqslant 5$.
Therefore, the company will recover the entire investment by 2013.
|
2013
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. For a positive integer $n$, let $S(n)$ denote the sum of the digits of $n$, and let $\varphi(n)$ denote the number of positive integers less than $n$ that are coprime to $n$. If $n$ is a three-digit number and satisfies $n=34 S(n)$, then the maximum value of $\varphi(n)$ is $\qquad$
|
8. 128 .
Let the three-digit number be
$$
n=100 a+10 b+c,
$$
where $a, b, c \in\{0,1, \cdots, 9\}, a>0$.
Then $100 a+10 b+c=34(a+b+c)$,
$$
8 b=22 a-11 c=11(2 a-c) .
$$
But $(8,11)=1$, so $11 \mid b \Rightarrow b=0$.
Thus, $2 a-c=0$.
When $a=1$, $c=2, n=102$;
When $a=2$, $c=4, n=204$;
When $a=3$, $c=6, n=306$;
When $a=4$, $c=8, n=408$.
Therefore, the maximum value of $\varphi(n)$ is
$$
\begin{array}{l}
\varphi(408)=\varphi\left(2^{3} \times 3 \times 17\right) \\
=408\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{17}\right)=128 .
\end{array}
$$
|
128
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Given
$$
\begin{array}{l}
\frac{1}{1 \times \sqrt{2}+2 \sqrt{1}}+\frac{1}{2 \sqrt{3}+3 \sqrt{2}}+\cdots+ \\
\frac{1}{n \sqrt{n+1}+(n+1) \sqrt{n}}
\end{array}
$$
is greater than $\frac{19}{20}$ and less than $\frac{20}{21}$. Then the difference between the maximum and minimum values of the positive integer $n$ is $\qquad$
(2009, I Love Mathematics Junior High School Summer Camp Mathematics Competition)
|
Consider the general term. The $k$-th term is
$$
\begin{array}{l}
\frac{1}{k \sqrt{k+1}+(k+1) \sqrt{k}} \\
=\frac{1}{\sqrt{k} \sqrt{k+1}(\sqrt{k}+\sqrt{k+1})} \\
=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k} \sqrt{k+1}} \\
=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}(k=1,2, \cdots, n) .
\end{array}
$$
Thus, the original expression is
$$
\begin{array}{l}
=\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\cdots+\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right) \\
=1-\frac{1}{\sqrt{n+1}} .
\end{array}
$$
From the given information,
$$
\begin{array}{l}
\frac{19}{20}\frac{1}{\sqrt{n+1}}>\frac{1}{21} \\
\Rightarrow 20<\sqrt{n+1}<21 \\
\Rightarrow 20^{2}-1<n<21^{2}-1 .
\end{array}
$$
Since $n$ is a positive integer,
$$
n_{\max }=21^{2}-2, n_{\text {min }}=20^{2} \text {. }
$$
Therefore, $21^{2}-2-20^{2}=39$ is the answer.
|
39
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Initial 281 Given that there exist $k(k \in \mathbf{N}, k \geqslant 2)$ consecutive positive integers, the mean of their squares is a perfect square. Try to find the minimum value of $k$.
|
Let the $k$ positive integers be $n+1, n+2, \cdots, n+k (n \in \mathbf{N})$. Then the mean of their squares is
$$
\begin{array}{l}
f(n, k)=\frac{1}{k} \sum_{i=1}^{k}(n+i)^{2} \\
=n^{2}+n(k+1)+\frac{(k+1)(2 k+1)}{6} .
\end{array}
$$
Since $f(n, k) \in \mathbf{Z}$, then $\frac{(k+1)(2 k+1)}{6} \in \mathbf{Z}$.
Thus, $2 \times k, 3 \times k$.
When $2<k \leqslant 13$, we have
$$
\left(n+\frac{k+1}{2}\right)^{2}<f(n, k)<\left(n+\frac{k+1}{2}+1\right)^{2} \text {. }
$$
In fact,
$$
\begin{array}{l}
\text { Equation (1) } \Leftrightarrow n^{2}+n(k+1)+\frac{k^{2}+2 k+1}{4} \\
<n^{2}+n(k+1)+\frac{(k+1)(2 k+1)}{6} \\
<n^{2}+n(k+3)+\frac{k^{2}+6 k+9}{4} \\
\Leftrightarrow \frac{k^{2}+2 k+1}{4}<\frac{2 k^{2}+3 k+1}{6}<\frac{k^{2}+6 k+9}{4}+2 n \\
\Leftrightarrow 3\left(k^{2}+2 k+1\right)<2\left(2 k^{2}+3 k+1\right) \\
<3\left(k^{2}+6 k+9\right)+24 n \\
\Leftrightarrow 1<k^{2}<12 k+25+24 n .
\end{array}
$$
The above inequality is clearly true.
At this point, $\sqrt{f(n, k)} \notin \mathbf{Z}$.
When $14<k<26$, we have
$$
\left(n+\frac{k+1}{2}\right)^{2}<f(n, k)<\left(n+\frac{k+1}{2}+2\right)^{2} \text {. }
$$
In fact,
$$
\begin{array}{l}
\text { Equation (2) } \Leftrightarrow \frac{k^{2}+2 k+1}{4}<\frac{2 k^{2}+3 k+1}{6} \\
<\frac{k^{2}+10 k+25}{4}+4 n \\
\Leftrightarrow 3\left(k^{2}+2 k+1\right)<2\left(2 k^{2}+3 k+1\right) \\
<3\left(k^{2}+10 k+25\right)+48 n \\
\Leftrightarrow 1<k^{2}<24 k+73+48 n .
\end{array}
$$
The above inequality is clearly true.
At this point, $f(n, k)=\left(n+\frac{k+1}{2}+1\right)^{2}$, i.e., $k^{2}=24 n+12 k+25$.
Then $(k-6)^{2}=24 n+61=8(3 n+7)+5$.
But the remainder of a perfect square when divided by 8 is 0 or 1, which is a contradiction.
When $k=29$,
$$
f(n, 29)=n^{2}+30 n+295=(n+15)^{2}+70 \text {, }
$$
its remainder when divided by 8 is 6 or 7, which is a contradiction.
When $k=31$,
$$
\begin{array}{c}
f(n, 31)=n^{2}+32 n+336 \\
=(n+18)^{2}-4 n+12 .
\end{array}
$$
Taking $n=3$, we get $f(3,31)=21^{2}$.
Therefore, $k_{\text {min }}=31$.
(Zhai Xiaojun, Donghai Senior High School, Jiangsu Province, 222300)
|
31
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $x$ and $y$ be positive integers such that
$$
\sqrt{x-116}+\sqrt{x+100}=y \text {. }
$$
Find the maximum value of $y$.
|
Prompt: By analogy with Example 6, we can prove that $\sqrt{x-116}$ and $\sqrt{x+100}$ are both natural numbers.
Let $\sqrt{x-116}=a, \sqrt{x+100}=b$. Then $b^{2}-a^{2}=216 \Rightarrow (b+a)(b-a)=216$.
Also, $b+a \equiv (b-a) \pmod{2}$, and since 216 is even, both $b-a$ and $b+a$ are even and positive.
Thus, $b-a \geqslant 2$.
Then $y=b+a \leqslant 108$.
When $\left\{\begin{array}{l}b+a=108, \\ b-a=2\end{array}\right.$, i.e., $\left\{\begin{array}{l}b=55, \\ a=53,\end{array}\right.$, i.e., $x=53^{2}+116$, $y$ reaches its maximum value.
|
108
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Let real numbers $x, y, z, w$ satisfy
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{x^{2}}{8^{2}-1^{2}}+\frac{y^{2}}{8^{2}-3^{2}}+\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1 .
\end{array}\right.
$$
Find the value of $x^{2}+y^{2}+z^{2}+w^{2}$.
|
Consider the function
$$
f(x)=\prod_{i=1}^{4}\left[x-(2 i-1)^{2}\right]-\prod_{i=1}^{4}\left[x-(2 i)^{2}\right] \text {. }
$$
Then $f\left(k^{2}\right)=\prod_{i=1}^{4}\left[k^{2}-(2 i-1)^{2}\right](k=2,4,6,8)$.
By the Lagrange interpolation formula, we have
$$
f(x)=\sum_{i=1}^{4} f\left((2 i-1)^{2}\right) \prod_{\substack{j \neq i<i<4 \\ 1 \leq j<4}} \frac{x-(2 i-1)^{2}}{(2 i-1)^{2}-(2 j-1)^{2}} .
$$
Comparing the coefficients of $x^{3}$ in equations (1) and (3), we get
$$
\begin{array}{l}
\sum_{i=1}^{4}\left[(2 i)^{2}-(2 i-1)^{2}\right] \\
=\sum_{i=1}^{4} f\left((2 i-1)^{2}\right) \prod_{\substack{j \in i \\
1 \leqslant i<4}}\left[(2 i-1)^{2}-(2 j-1)^{2}\right]^{-1} .
\end{array}
$$
In equation (3), let $x=k^{2}(k=2,4,6,8)$, we get
$$
\begin{array}{l}
f\left(k^{2}\right) \\
= \sum_{i=1}^{4} f\left((2 i-1)^{2}\right) \prod_{\substack{j \in i<4 \\
1 \neq j \leqslant 4}} \frac{k^{2}-(2 j-1)^{2}}{(2 i-1)^{2}-(2 j-1)^{2}} \\
= \sum_{i=1}^{4} \frac{f\left((2 i-1)^{2}\right)}{k^{2}-(2 i-1)^{2}} \prod_{\substack{j+i<i \\
1<j \leqslant 4}}\left[(2 i-1)^{2}-(2 j-1)^{2}\right]^{-1} . \\
\prod_{t=1}^{4}\left[k^{2}-(2 t-1)^{2}\right] .
\end{array}
$$
From equations (2) and (4), we have
$$
\begin{array}{l}
\sum_{i=1}^{4} \frac{f\left((2 i-1)^{2}\right)}{k^{2}-(2 i-1)^{2}} \prod_{\substack{j \neq i<i \\
1 \leqslant \leqslant<4}}\left[(2 i-1)^{2}-(2 j-1)^{2}\right]^{-1} \\
\quad=1(k=2,4,6,8) .
\end{array}
$$
By the uniqueness of the solution of the system of equations, we have
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}+w^{2} \\
=\sum_{i=1}^{4} f\left((2 i-1)^{2}\right) \prod_{\substack{1 \neq i<\\
1<j<4}}\left[(2 i-1)^{2}-(2 j-1)^{2}\right]^{-1} \\
=\sum_{i=1}^{4}\left[(2 i)^{2}-(2 i-1)^{2}\right]=36 .
\end{array}
$$
**Summary**: The construction of the function is not unique. For example, let
$$
\begin{aligned}
f(m)= & \left(\frac{x^{2}}{m-1^{2}}+\frac{y^{2}}{m-3^{2}}+\frac{z^{2}}{m-5^{2}}+\frac{w^{2}}{m-7^{2}}\right) . \\
& \prod_{i=1}^{4}\left[m-(2 i-1)^{2}\right] .
\end{aligned}
$$
Then by the Lagrange interpolation formula, we have
$$
f(m)=\sum_{i=1}^{4} f\left((2 i)^{2}\right) \prod_{\substack{j \neq i \\ 1 \leqslant j \leqslant 4}} \frac{m-(2 j)^{2}}{(2 i)^{2}-(2 j)^{2}} \text {. }
$$
Comparing the coefficients of $m^{3}$, we get the value of $x^{2}+y^{2}+z^{2}+w^{2}$.
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Find the smallest positive integer $n$, such that for any sequence of $n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $\sum_{i=1}^{n} a_{i}=2007$, there must be a sum of some consecutive terms equal to 30.
(Fourth China Southeast Mathematical Olympiad)
|
First, construct an integer sequence $a_{1}, a_{2}, \cdots, a_{1017}$ with 1017 terms, such that no consecutive terms sum to 30. For this, take
$$
a_{1}=a_{2}=\cdots=a_{29}=1, a_{30}=31,
$$
and $a_{30 m+i}=a_{i}(i=1,2, \cdots, 30, m \in \mathbf{N})$.
Thus, $\left\{a_{k}\right\}$ is:
$$
\begin{array}{l}
1,1, \cdots, 1,31 ; 1,1, \cdots, 1,31 ; \cdots ; \\
1,1, \cdots, 1,31 ; 1,1, \cdots, 1
\end{array}
$$
(There are 34 segments, the first 33 segments each have 30 terms, and the last segment has 27 terms, totaling 1017 terms).
Second, when the number of terms is less than 1017, it is sufficient to merge some consecutive numbers in certain segments into larger numbers.
For any positive integer sequence $a_{1}, a_{2}, \cdots, a_{1018}$ with 1018 terms that satisfies the condition $\sum_{i=1}^{1018} a_{i}=2007$, we will prove that there must be consecutive terms whose sum equals 30.
In fact, let $S_{k}=\sum_{i=1}^{k} a_{i}$. Then
$$
1 \leqslant S_{1}<S_{2}<\cdots<S_{1018}=2007 \text {. }
$$
Now consider the grouping of elements in the set $\{1,2, \cdots, 2007\}$: $\square$
$$
\begin{array}{c}
(60 k+i, 60 k+30+i), \\
(k=0,1, \cdots, 32 ; i=1,2, \cdots, 30), \\
1981,1982, \cdots, 2007 .
\end{array}
$$
There are $33 \times 30=990$ pairs and 27 numbers not in pairs. From these, any 1018 numbers taken as the values of $S_{k}$ must include two numbers from the same pair, say $\left(S_{k}, S_{k+m}\right)$. Then $S_{k+m}-S_{k}=30$, meaning that in the sequence,
$$
a_{k+1}+a_{k+2}+\cdots+a_{k+m}=30 .
$$
Therefore, the minimum value of $n$ is 1018.
|
1018
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For a natural number $n$, let the sum of its digits be denoted as $a_{n}$, for example,
$$
\begin{array}{l}
a_{2009}=2+0+0+9=11, \\
a_{2010}=2+0+1+0=3 .
\end{array}
$$
Then $a_{1}+a_{2}+\cdots+a_{2010}=$ ( ).
(A) 28062
(B) 28065
(C) 28067
(D) 28068
|
6. D.
Consider all natural numbers from 1 to 2010 as four-digit numbers (if $n$ is less than four digits, add 0s at the beginning to make it four digits, which does not change the value of $a_{n}$).
Notice that, the number of times 1 appears in the thousands, hundreds, tens, and units place are $10^{3}$, $2 \times 10^{2}$, $2 \times 10^{2}+1$, and $2 \times 10^{2}+1$ respectively. Therefore, the total number of times 1 appears is
$$
10^{3}+2 \times 10^{2} \times 3+2=1602 \text {. }
$$
The number of times 2 appears in the thousands, hundreds, tens, and units place are $11$, $2 \times 10^{2}$, $2 \times 10^{2}$, and $2 \times 10^{2}+1$ respectively. Therefore, the total number of times 2 appears is
$$
11+2 \times 10^{2} \times 3+1=612 \text {. }
$$
Similarly, the total number of times $k(k=3,4, \cdots, 9)$ appears is $2 \times 10^{2} \times 3+1=601$.
$$
\begin{array}{l}
\text { Hence } a_{1}+a_{2}+\cdots+a_{2010} \\
=1602 \times 1+612 \times 2+601(3+4+\cdots+9) \\
=28068 \text {. }
\end{array}
$$
|
28068
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 The number of positive integer values of $n$ that satisfy $\left|\sqrt{\frac{n}{n+2009}}-1\right|>\frac{1}{1005}$ is $\qquad$
(2009, I Love Mathematics Junior High School Summer Camp Mathematics Competition)
|
Given that $n$ is a positive integer, we have
$$
\begin{array}{l}
0\frac{1}{1005} \\
\quad \Rightarrow \sqrt{\frac{n}{n+2009}}<\frac{1004}{1005} .
\end{array}
$$
Let $1004=a$. Then
$$
\begin{array}{l}
\frac{n}{n+2 a+1}<\frac{a^{2}}{(a+1)^{2}} \\
\Rightarrow\left[(a+1)^{2}-a^{2}\right] n<a^{2}(2 a+1) \\
\Rightarrow(2 a+1) n<a^{2}(2 a+1) \\
\Rightarrow n<a^{2} \Rightarrow n<1004^{2} .
\end{array}
$$
Since $n$ is a positive integer, therefore, $n \leqslant 1004^{2}-1$.
Thus, the number of positive integers $n$ that satisfy the condition is $1004^{2}-1$, which is 1008015.
|
1008015
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{3}+y^{3}=19, \\
x+y=1 .
\end{array}\right.
$$
then $x^{2}+y^{2}=$ $\qquad$
|
II, 1.13.
From $x^{3}+y^{3}=19$, we get $(x+y)\left[(x+y)^{2}-3 x y\right]=19$. Substituting $x+y=1$ yields $x y=-6$. Therefore, $x^{2}+y^{2}=(x+y)^{2}-2 x y=13$.
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Arrange several balls of two colors, red and black, in a row, requiring that both colors of balls must appear, and any two balls separated by 5 or 10 balls must be of the same color. Arrange according to this requirement, the maximum number of balls that can be placed is.
|
4. 15.
Label the positions of these balls in order as $1,2, \cdots$.
According to the problem, when $|i-j|=6$ or 11, the $i$-th ball and the $j$-th ball are the same color, denoted as $i \sim j$.
Thus, $6 \sim 12 \sim 1 \sim 7 \sim 13 \sim 2 \sim 8 \sim 14 \sim 3 \sim 9$ $\sim 15 \sim 4 \sim 10,5 \sim 11$.
Therefore, 15 balls can be placed according to the requirement.
If the number of balls is more than 15, then $10 \sim 16 \sim 5$. Thus, the colors of the balls numbered 1 to 16 are all the same. Further, it can be known that all the balls have the same color, which does not meet the requirement.
Therefore, according to this requirement, the maximum number of balls that can be placed is 15.
|
15
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. The number of integer solutions to the inequality $\log _{6}(1+\sqrt{x})>\log _{25} x$ is $\qquad$ .
|
10. 24 .
Let $\log _{25} x=t$. Then $x=25^{t}$.
Thus the original inequality $\Leftrightarrow \log _{6}\left(1+5^{t}\right)>t$
$$
\Leftrightarrow 1+5^{t}>6^{t} \Leftrightarrow\left(\frac{1}{6}\right)^{t}+\left(\frac{5}{6}\right)^{t}>1 \text {. }
$$
Define $f(t)=\left(\frac{1}{6}\right)^{t}+\left(\frac{5}{6}\right)^{t}$. Then $f(t)$ is a decreasing function, $f(t)>1=f(1)$.
Hence $t=\log _{25} x<1$.
Therefore, $0<x<25$, a total of 24.
|
24
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Given real numbers $x, y, z$ satisfy
$$
\sqrt[5]{x-y}+\sqrt[5]{y-z}=3 \text{, and } x-z=33 \text{. }
$$
Find the value of the algebraic expression $x-2 y+z$.
|
Let $\sqrt[5]{x-y}=x_{1}, \sqrt[5]{y-z}=x_{2}$. Then $x_{1}+x_{2}=3, x_{1}^{5}+x_{2}^{5}=33$.
Let $x_{1} x_{2}=\lambda$. Then $x_{1} 、 x_{2}$ are the two roots of $x^{2}-3 x+\lambda=0$.
Let $x_{1}^{n}+x_{2}^{n}=a_{n}$ ( $n$ is a positive integer $)$. Then
$$
a_{n}-3 a_{n-1}+\lambda a_{n-2}=0 \text {, }
$$
i.e.,
$$
a_{n}=3 a_{n-1}-\lambda a_{n-2}(n \geqslant 3) .
$$
From $x_{1}+x_{2}=3, x_{1} x_{2}=\lambda$
$$
\begin{array}{l}
\Rightarrow a_{1}=3, a_{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=9-2 \lambda \\
\Rightarrow a_{3}=3(9-2 \lambda)-3 \lambda=27-9 \lambda \\
\Rightarrow a_{4}=3(27-9 \lambda)-\lambda(9-2 \lambda) \\
\quad=81-36 \lambda+2 \lambda^{2} \\
\Rightarrow a_{5}=3\left(81-36 \lambda+2 \lambda^{2}\right)-\lambda(27-9 \lambda)=33 \\
\Rightarrow 81-36 \lambda+2 \lambda^{2}-\lambda(9-3 \lambda)=11 \\
\Rightarrow \lambda^{2}-9 \lambda+14=0
\end{array}
$$
$$
\Rightarrow \lambda=2 \text { or } 7 \text {. }
$$
From $x^{2}-3 x+\lambda=0$ having two real roots, we get
$$
\Delta=(-3)^{2}-4 \lambda \geqslant 0 \Rightarrow \lambda \leqslant \frac{9}{4}
$$
$\Rightarrow \lambda=2 \Rightarrow x_{1} 、 x_{2}$ are the two roots of $x^{2}-3 x+2=0$
$$
\Rightarrow\left\{\begin{array} { l }
{ x _ { 1 } = 1 , } \\
{ x _ { 2 } = 2 }
\end{array} \text { or } \left\{\begin{array}{l}
x_{1}=2, \\
x_{2}=1 .
\end{array}\right.\right.
$$
Since $x-2 y+z=x_{1}^{5}-x_{2}^{5}$, we have
$$
1^{5}-2^{5}=-31 \text { or } 2^{5}-1^{5}=31
$$
This is the solution.
|
31
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. $[x]$ represents the greatest integer not exceeding the real number $x$ (for example, $[\pi]=3,[-\pi]=-4,[-4]=-4$). Let $M=[x]+[2x]+[3x]$. Positive integers that cannot be expressed in the form of $M$ are called "invisible numbers". If the invisible numbers are arranged in ascending order, the 2009th invisible number is $\qquad$
|
3.6028 .
Let $n$ be a natural number.
When $n \leqslant x < n+\frac{1}{3}$,
$2 n \leqslant 2 x < 2 n+\frac{2}{3}$,
$3 n \leqslant 3 x < 3 n+1$.
Thus, $M=n+2 n+3 n=6 n$.
When $n+\frac{1}{3} \leqslant x < n+\frac{1}{2}$,
$2 n+\frac{2}{3} \leqslant 2 x < 2 n+1$,
$3 n+1 \leqslant 3 x < 3 n+\frac{3}{2}$.
Thus, $M=n+2 n+(3 n+1)=6 n+1$.
When $n+\frac{1}{2} \leqslant x < n+\frac{2}{3}$,
$2 n+1 \leqslant 2 x < 2 n+\frac{4}{3}$,
$3 n+\frac{3}{2} \leqslant 3 x < 3 n+2$.
Thus, $M=n+(2 n+1)+(3 n+1)=6 n+2$.
When $n+\frac{2}{3} \leqslant x < n+1$,
$2 n+\frac{4}{3} \leqslant 2 x < 2 n+2$,
$3 n+2 \leqslant 3 x < 3 n+3$.
Thus, $M=n+(2 n+1)+(3 n+2)=6 n+3$.
Therefore, the hidden numbers are $6 n+4$ or $6 n+5 (n \in \mathbf{N})$.
The 2009th hidden number is
$$
6 \times \frac{2009-1}{2}+4=6028
$$
|
6028
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given a positive integer $M$ has $k$ positive divisors, among which, only two divisors are prime, and the sum of the reciprocals of these $k$ positive divisors is $\frac{342}{287}$. Find all values of $M$.
---
The text has been translated while preserving the original formatting and line breaks.
|
$$
\begin{array}{l}
\text{Because } 287=7 \times 41, \text{ so, 7 and 41 are exactly the two prime factors of } M. \\
\text{Let } M=7^{m} \times 41^{n}\left(m, n \in \mathbf{N}_{+}\right), d_{1}, d_{2}, \cdots, d_{k} \text{ be all the positive divisors of } M \text{ arranged in ascending order. Thus, } d_{1}=1, d_{k}=M. \text{ Then,} \\
\frac{342}{287}=\frac{1}{d_{1}}+\frac{1}{d_{2}}+\frac{1}{d_{3}}+\cdots+\frac{1}{d_{k}} \\
=\frac{d_{k}}{d_{k}}+\frac{d_{k-1}}{d_{k}}+\frac{d_{k-2}}{d_{k}}+\cdots+\frac{d_{1}}{d_{k}} \\
=\frac{d_{1}+d_{2}+d_{3}+\cdots+d_{k}}{d_{k}} \\
=\frac{\left(1+7+7^{2}+\cdots+7^{m}\right)\left(1+41+41^{2}+\cdots+41^{n}\right)}{7^{m} \times 41^{n}} . \\
\text{Hence, } \frac{7^{m+1}-1}{6} \cdot \frac{41^{n+1}-1}{40}=342 \times 7^{m-1} \times 41^{n-1} . \\
\text{By } \left(41^{n-1}, \frac{41^{n+1}-1}{40}\right)=1, \text{ we know that } \\
41^{n-1} \mid \frac{7^{m+1}-1}{6} . \\
\text{Let } 7^{m+1}-1=41^{n-1} \times 6 t\left(t \in \mathbf{N}_{+}\right) .
\end{array}
$$
Obviously, $(t, 7)=1$. Then,
$$
\frac{41^{n+1}-1}{40}=\frac{342}{t} \times 7^{m-1} \text{. }
$$
Therefore, $t \mid 342 \Rightarrow t \leqslant 342$.
Substituting equation (1) into equation (2) and rearranging, we get
$$
\begin{array}{l}
7^{m-1}\left(41^{2} \times 7^{2}-6 \times 13680\right)=41^{2}+6 t \\
\Rightarrow 289 \times 7^{m-1}=1681+6 t \\
\Rightarrow 1681<289 \times 7^{m-1} \leqslant 1681+6 \times 342 \\
\Rightarrow 5<\frac{1681}{289}<7^{m-1} \leqslant \frac{3733}{289}<13 \\
\Rightarrow m=2 .
\end{array}
$$
Substituting into equation (3), we get $t=57$.
Substituting into equation (1), we get $n=1$.
Thus, $M=7^{m} \times 41^{n}=2009$.
(Xie Wenxiao, Huanggang High School, Hubei, 438000)
$$
|
2009
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Using the vertices of a regular dodecagon as the vertices of triangles, the total number of acute and obtuse triangles that can be formed is $\qquad$.
untranslated part: $\qquad$ (This part is typically left as a blank space for the answer to be filled in.)
|
-1.160 .
The circumference of a circle with 12 points can form $\mathrm{C}_{12}^{3}=220$ triangles; among the chords of a circle equally divided into 12 parts, there are 6 diameters, and each diameter corresponds to 10 right-angled triangles, thus yielding 60 right-angled triangles. Therefore, acute and obtuse triangles amount to $220-60=160$ (triangles).
|
160
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $S=\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+$ $\cdots+\sqrt{1+\frac{1}{1999^{2}}+\frac{1}{2000^{2}}}$. Find the greatest integer not exceeding $S$.
(2000, Taiyuan Junior High School Mathematics Competition)
|
Prompt: Following Example 2, the largest integer not exceeding $S$ is
$$
1999 .
$$
|
1999
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (40 points) Prove: The set $E$ of real numbers $x$ that satisfy the inequality
$$
\frac{1}{x-1}+\frac{2}{x-2}+\cdots+\frac{200}{x-200}>10
$$
can be expressed as the union of some mutually disjoint open intervals. Find the total length of these intervals.
|
Consider the function
$$
f(x)=\frac{1}{x-1}+\frac{2}{x-2}+\cdots+\frac{200}{x-200}-10 \text {. }
$$
For any real number $x$.
For any $k$ in the set $\{1,2, \cdots, 200\}$, when $x \rightarrow k-0$, $f(x) \rightarrow-\infty$, and when $x \rightarrow k+0$, $f(x) \rightarrow+\infty$. Also, when $x \rightarrow+\infty$, $x \rightarrow-10$. Therefore, the equation $f(x)=0$ has one solution in each of the intervals
$$
(1,2),(2,3), \cdots,(199,200),(200,+\infty).
$$
Let these 200 solutions be denoted as $x_{1}, x_{2}, \cdots, x_{200}$.
Construct the polynomial
$$
P(x)=(x-1)(x-2) \cdots(x-200) f(x) \text {. }
$$
Since $P(x)$ is a 200th-degree polynomial, the equation $P(x)=0$ has at most 200 distinct roots. Clearly, each $x_{i}$ that satisfies $f(x)=0$ is also a root of $P(x)=0$.
Therefore, $x_{1}, x_{2}, \cdots, x_{200}$ are all the roots of $P(x)=0$.
This indicates that each $x_{k}$ is the unique root in its respective interval $(k, k+1)$ $(k=1,2, \cdots, 199)$ and $(200,+\infty)$.
Thus, the solution set of the inequality $f(x)>0$ is
$$
E=\left(1, x_{1}\right) \cup\left(2, x_{2}\right) \cup \cdots \cup\left(200, x_{200}\right) \text {. }
$$
Hence, the total sum of the lengths of all intervals is
$$
\begin{aligned}
S & =\left(x_{1}-1\right)+\left(x_{2}-2\right)+\cdots+\left(x_{200}-200\right) \\
& =\left(x_{1}+x_{2}+\cdots+x_{200}\right)-(1+2+\cdots+200) \\
& =\sum_{i=1}^{200} x_{i}-10 \times 2010 .
\end{aligned}
$$
Notice that
$$
P(x)=\left[\prod_{i=1}^{200}(x-i)\right]\left(\sum_{i=1}^{200} \frac{i}{x-i}-10\right),
$$
If we expand $P(x)$, the coefficient of the highest degree term is -10.
Thus, the coefficient of $x^{199}$ in $P(x)$ is
$$
(10+1) \sum_{k=1}^{200} k=11 \times 20100 \text {. }
$$
Therefore, $\sum_{i=1}^{200} x_{i}=11 \times 2010$.
From equation (1), we get
$$
S=\sum_{i=1}^{200} x_{i}-10 \times 2010=2010 .
$$
|
2010
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) A positive integer is called "simple" if it does not have any square factors greater than 1. Determine how many numbers in $1,2, \cdots, 2010$ are simple.
|
Three, notice that $2010 < 45^2$.
Thus, if a number $n$ in the set $M = \{1, 2, \cdots, 2010\}$ is not simple, it must contain one or more square factors from the set of primes
$$
N = \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43\}.
$$
Let
$$
\left[\frac{2010}{p^2}\right] = s(p), \left[\frac{2010}{p^2 q^2}\right] = s(p, q),
$$
and so on.
Then $s(2) = 502, s(3) = 223, s(5) = 80$,
$$
\begin{array}{l}
s(7) = 41, s(11) = 16, s(13) = 11, \\
s(17) = 6, s(19) = 5, s(23) = 3, \\
s(29) = s(31) = 2, \\
s(37) = s(41) = s(43) = 1.
\end{array}
$$
Using $\left[\frac{\left[\frac{a}{n}\right]}{m}\right] = \left[\frac{a}{mn}\right]$, we get
$$
\begin{array}{c}
s(2,3) = 55, s(2,5) = 20, s(2,7) = 10, \\
s(2,11) = 4, s(2,13) = 2, s(2,17) = 1, \\
s(2,19) = 1, s(2,23) = 0, s(3,5) = 8, \\
s(3,7) = 4, s(3,11) = s(3,13) = 1, \\
s(3,17) = 0, s(5,7) = 1, s(5,11) = 0; \\
s(2,3,5) = 2, s(2,3,7) = 1, \\
s(2,3,11) = s(3,5,7) = 0.
\end{array}
$$
Thus, the desired result is
$$
\begin{array}{l}
2010 - \sum_{p \in N} s(p) + \sum_{p, q \in N} s(p, q) - \sum_{p, q, r \in N} s(p, q, r) \\
= 2010 - (502 + 223 + 80 + 41 + 16 + \\
11 + 6 + 5 + 3 + 2 + 2 + 1 + 1 + 1) + \\
(55 + 20 + 10 + 4 + 2 + 1 + 1 + 8 + 4 + \\
1 + 1 + 1) - (2 + 1) \\
= 2010 - 894 + 108 - 3 = 1221.
\end{array}
$$
|
1221
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $\alpha=\frac{\sqrt{5}+1}{2}$. Then $\left[\alpha^{16}\right]=$ $\qquad$ .
(2008, "Five Sheep Cup" Mathematics Competition (Junior High School))
[Analysis] Note that $\alpha=\frac{\sqrt{5}+1}{2}$ and $\beta=\frac{\sqrt{5}-1}{2}$ $(0<\beta<1)$ can make $\alpha+\beta=\sqrt{5}$ and $\alpha \beta=1$. Therefore, the value of $\alpha^{16}+\beta^{16}$ can be calculated to make a judgment.
|
Let $\beta=\frac{\sqrt{5}-1}{2}$. Then
$$
\begin{array}{l}
\alpha+\beta=\sqrt{5}, \alpha \beta=1 . \\
\text { Therefore, } \alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=3, \\
\alpha^{4}+\beta^{4}=\left(\alpha^{2}+\beta^{2}\right)^{2}-2(\alpha \beta)^{2}=7, \\
\alpha^{8}+\beta^{8}=\left(\alpha^{4}+\beta^{4}\right)^{2}-2(\alpha \beta)^{4}=47, \\
\alpha^{16}+\beta^{16}=\left(\alpha^{8}+\beta^{8}\right)^{2}-2(\alpha \beta)^{8}=2207 .
\end{array}
$$
Since $0<\beta<1$, we know $0<\beta^{16}<1$.
Therefore, $\left[\alpha^{16}\right]=2206$.
|
2206
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Calculate $[\sqrt{2008+\sqrt{2008+\cdots+\sqrt{2008}}}]$ (2008 appears 2008 times).
(2008, International Youth Math Invitational Competition)
【Analysis】Although there are 2008 square root operations, as long as you patiently estimate from the inside out, the pattern will naturally become apparent.
|
Let $a_{n}=\sqrt{2008+\sqrt{2008+\cdots+\sqrt{2008}}}$ (with $n$ 2008s, $n=1,2, \cdots$ ).
Given $44<\sqrt{2008}<45$, we have
$$
\begin{array}{l}
45^{2}<2008+44<2008+\sqrt{2008} \\
<2008+45<46^{2} .
\end{array}
$$
Thus, $45<\sqrt{2008+\sqrt{2008}}<46$, which means $45<a_{2}<46$.
Therefore, $45^{2}<2008+45<2008+a_{2}$
$$
<2008+46<46^{2} \text {. }
$$
Hence, $45<\sqrt{2008+a_{2}}<46$, which means $45<a_{3}<46$.
Similarly, $45<a_{4}<46, \cdots \cdots 45<a_{2008}<46$.
Therefore, $\left[a_{2008}\right]=45$.
|
45
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12 Given a positive integer $n$ less than 2006, and $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\frac{n}{2}$. Then the number of such $n$ is $\qquad$.
(2006, National Junior High School Mathematics Competition)
|
From the properties of the Gaussian function, we know that $\left[\frac{n}{3}\right] \leqslant \frac{n}{3}$, with equality holding if and only if $\frac{n}{3}$ is an integer; $\left[\frac{n}{6}\right] \leqslant \frac{n}{6}$, with equality holding if and only if $\frac{n}{6}$ is an integer.
Therefore, $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right] \leqslant \frac{n}{3}+\frac{n}{6}=\frac{n}{2}$.
Equality holds in the above inequality if and only if $\frac{n}{6}$ is an integer.
Thus, $\frac{n}{6}$ is an integer.
Since the positive integer $n$ is less than 2006, the number of $n$ that satisfy the condition is $\left[\frac{2005}{6}\right]=334$.
|
334
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The algebraic expression $\left[(\sqrt{6}+\sqrt{5})^{6}\right]=$ $\qquad$
(2004, "Five Sheep Cup" Mathematics Competition (Grade 9))
|
Prompt: Example 1. Answer: 10581.
|
10581
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The sum of all real numbers $x$ that satisfy $25\{x\}+[x]=125$ is $\qquad$
(2007, International Invitational Competition for Young Mathematicians in Cities)
|
Hint: $25\{x\}+[x]=125$ can be transformed into $25 x-24[x]=125$.
Example 6. Answer: 2837.
|
2837
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Algebraic Expression
$[\sqrt[3]{1 \times 2 \times 3}]+[\sqrt[3]{2 \times 3 \times 4}]+\cdots+$ $[\sqrt[3]{2000 \times 2001 \times 2002}]=(\quad)$.
(A) 2000000
(B) 2001000
(C) 2002000
(D) 2003001
(2000, "Five Sheep Cup" Mathematics Competition (Grade 9))
[Analysis] Notice that each term in the sum can be expressed as $[\sqrt[3]{k(k+1)(k+2)}]$ (where $k$ is a positive integer). We need to estimate $\sqrt[3]{k(k+1)(k+2)}$.
|
Since $k^{3}<k(k+1)(k+2)<(k+1)^{3}$,
thus, $k<\sqrt[3]{k(k+1)(k+2)}<k+1$.
Therefore, $[\sqrt[3]{k(k+1)(k+2)}]=k$.
Hence the original expression $=1+2+\cdots+2000=2001000$.
|
2001000
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Calculate $\left[\frac{23 \times 1}{101}\right]+\left[\frac{23 \times 2}{101}\right]+\cdots+$ $\left[\frac{23 \times 100}{101}\right]$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Since 23 and 101 are coprime, for $k=1$, $2, \cdots, 100$, $\frac{23 k}{101}$ is not an integer. Therefore,
$$
\begin{array}{l}
\frac{23 k}{101}-1<\left[\frac{23 k}{101}\right]<\frac{23 k}{101}, \\
\frac{23(101-k)}{101}-1<\left[\frac{23(101-k)}{101}\right] \\
<\frac{23(101-k)}{101}. \\
\text{ Hence }\left(\frac{23 k}{101}-1\right)+\left[\frac{23(101-k)}{101}-1\right] \\
<\left[\frac{23 k}{101}\right]+\left[\frac{23(101-k)}{101}\right] \\
<\frac{23 k}{101}+\frac{23(101-k)}{101}, \\
21<\left[\frac{23 k}{101}\right]+\left[\frac{23(101-k)}{101}\right]<23. \\
\end{array}
$$
Therefore, $\left[\frac{23 k}{101}\right]+\left[\frac{23(101-k)}{101}\right]=22$.
Thus, we can pair $\left[\frac{23 \times 1}{101}\right],\left[\frac{23 \times 2}{101}\right], \cdots$, $\left[\frac{23 \times 100}{101}\right]$ from the beginning to the end, forming 50 pairs, with each pair summing to 22.
$$
\begin{array}{l}
\text{ Hence }\left[\frac{23 \times 1}{101}\right]+\left[\frac{23 \times 2}{101}\right]+\cdots+\left[\frac{23 \times 100}{101}\right] \\
=22 \times 50=1100. \\
\end{array}
$$
|
1100
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $a^{2}(b+c)=b^{2}(a+c)=2010$, and $a \neq b$. Then $c^{2}(a+b)=$
|
1. 2010 .
Notice
$$
\begin{array}{l}
a^{2}(b+c)-b^{2}(a+c) \\
=(a-b)(a c+b c+a b)=0 .
\end{array}
$$
Since $a \neq b$, it follows that $a c+b c+a b=0$.
$$
\begin{array}{l}
\text { Then } c^{2}(a+b)-b^{2}(a+c) \\
=(c-b)(a c+b c+a b)=0 .
\end{array}
$$
Therefore, $c^{2}(a+b)=b^{2}(a+c)=2010$.
|
2010
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Wang Qiang has four colors of small cylindrical rods, Table 1 lists the lengths of the rods of different colors.
Table 1
\begin{tabular}{|c|c|c|c|c|}
\hline Color & Green & Yellow-Red & Purple & Red \\
\hline Length & $3 \mathrm{~cm}$ & $4 \mathrm{~cm}$ & $8 \mathrm{~cm}$ & $9 \mathrm{~cm}$ \\
\hline
\end{tabular}
Now, several small rods are to be taken and joined together to form a long rod of length $2010 \mathrm{~cm}$, and each of the four colors of small rods must be used at least 81 times. Then the number of different ways to do this is.
|
9. 91.
From the problem, we have
$$
\begin{array}{l}
3(81+a)+4(81+b)+8(81+c)+9(81+d) \\
=2010,
\end{array}
$$
where $a, b, c, d$ are non-negative integers.
$$
\begin{array}{l}
\text { Simplifying, we get } 3(a+3 d)+4(b+2 c)=66 . \\
\text { Therefore, }(a+3 d, b+2 c) \\
=(4 i+2,15-3 i)(i=0,1, \cdots, 5) . \\
\text { When } a+3 d=4 i+2 \text {, }(a, d) \text { has }\left[\frac{4 i+2}{3}\right]+1
\end{array}
$$
possible values;
$$
\text { When } b+2 c=15-3 i \text {, }(b, c) \text { has }\left[\frac{15-3 i}{2}\right]+1
$$
possible values.
Therefore, the total number of different combinations is
$$
1 \times 8+3 \times 7+4 \times 5+5 \times 4+7 \times 2+8 \times 1=91 \text {. }
$$
|
91
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. In a competition with 20 singers, 9 judges respectively assign them ranks from 1 to 20. It is known that for each singer, the difference between any two ranks does not exceed 3. If the sum of the ranks each singer receives is arranged in an increasing sequence: $C_{1} \leqslant C_{2} \leqslant \cdots \leqslant C_{20}$, then the maximum value of $C_{1}$ is . $\qquad$
|
10. 24.
If 9 judges all give a singer the first place, then $C_{1}=9$.
If two singers both get the first place, then one of them gets no less than 5 first places, while the other 4 ranks are no higher than fourth place, so $C_{1} \leqslant 5 \times 1+4 \times 4=21$.
If three singers all get the first place, then their other ranks are no higher than fourth place, and their total ranks do not exceed $1 \times 9+3 \times 9+4 \times 9=72$, so $C_{1} \leqslant 24$.
If four singers all get the first place, then their total ranks do not exceed $1 \times 9+2 \times 9+3 \times 9+4 \times 9=90$, so $C_{1} \leqslant 24$.
The situation where five or more singers all get the first place is impossible, so $C_{1} \leqslant 24$.
Here is an example where $C_{1}=24$:
The judges give the three singers with rank sums $C_{1} 、 C_{2} 、 C_{3}$ the ranks $1,1,1,3,3,3,4,4,4$;
The judges give the three singers with rank sums $C_{4} 、 C_{5} 、 C_{6}$ the ranks $2,2,2,5,5,5,6,6,6$;
While the judges give the rest of the singers ranks between 7 and 20.
At this point, $C_{1}=24$.
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Find the smallest odd number $a$ greater than 5 that satisfies the following conditions: there exist positive integers $m_{1}, n_{1}, m_{2}, n_{2}$, such that
$$
a=m_{1}^{2}+n_{1}^{2}, a^{2}=m_{2}^{2}+n_{2}^{2} \text {, }
$$
and $m_{1}-n_{1}=m_{2}-n_{2}$.
|
8. From $261=15^{2}+6^{2}, 261^{2}=189^{2}+180^{2}$,
$15-6=189-180$,
we know that 261 has the property described in the problem.
Next, we prove that 261 is the smallest odd number greater than 5 with the property described in the problem, i.e., there is no such odd number between 5 and 261.
If not, then there exists an odd number $a$ between 5 and 261 and positive integers $m_{1}, n_{1}, m_{2}, n_{2}$, such that
$a=m_{1}^{2}+n_{1}^{2}, a^{2}=m_{2}^{2}+n_{2}^{2}, m_{1}-n_{1}=m_{2}-n_{2}$.
Let $m_{1}-n_{1}=l$, by symmetry, we can assume $l \geqslant 0$.
Since $a$ is odd, $m_{1}$ and $n_{1}$ have different parities.
Therefore, $l$ is an odd number.
Also, $m_{1}500>261$, a contradiction.
If $l=5$, then $a^{2} \equiv 2 n_{2}^{2}(\bmod 5)$.
Since 2 is not a quadratic residue modulo 5, $51 a$.
Also, $0 \equiv a \equiv 2 n_{1}^{2}(\bmod 5)$, so $5 \mid n_{1}$.
If $n_{1} \geqslant 10$, then $m_{1} \geqslant 15$.
Thus, $a \geqslant 15^{2}+10^{2}=325>261$, a contradiction;
If $n_{1}=5$, then
$m_{1}=10, a=10^{2}+5^{2}=125$.
But from $125^{2}=\left(n_{2}+5\right)^{2}+n_{2}^{2}$, we get
$$
\left(2 n_{2}+5\right)^{2}=5^{2} \times 1249,
$$
a contradiction.
If $l=7$, then by $a<261<16^{2}+9^{2}$, we know $n_{1} \leqslant 8$.
By enumerating $n_{1}$, we find that $a$ can only be
$65,85,109,137,169,205,245$
one of these.
From $a^{2}=\left(n_{2}+7\right)^{2}+n_{2}^{2}$, we get
$\left(2 n_{2}+7\right)^{2}=2 a^{2}-49$.
Therefore, $2 a^{2}-49$ must be a perfect square.
But by verifying each of $a=65,85,109,137,169,205,245$, we find that none of these values of $a$ satisfy the condition.
In conclusion, the smallest odd number greater than 5 with the property described in the problem is 261.
(Provided by Zhu Huawei)
|
261
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (25 points) In a $7 \times 7$ grid, there are 64 intersection points (referred to as "nodes") where chess pieces can be placed, with at most 1 piece per point, totaling $k$ pieces. If no matter how the pieces are placed, there always exist 4 pieces such that the nodes they occupy form the four vertices of a rectangle (with sides parallel to the grid lines of the chessboard). Find the minimum value of $k$.
(Zhang Limin, problem contributor)
|
First, as shown in Figure 2, when $k=24$, it is possible that no 4 nodes with chess pieces form the four vertices of a rectangle, so $k \geqslant 25$.
Second, we will prove that when $k=25$, there must exist 4 nodes with chess pieces that form the four vertices of a rectangle.
If not, consider the row (or column) with the most chess pieces among all rows and columns. Since swapping any two rows (or two columns) does not affect the result, we can assume without loss of generality that the first row has the most chess pieces.
(1) When the first row has 8 chess pieces, because
$$
25-8=17,17=8 \times 2+1 \text {, }
$$
there must be another row with 2 chess pieces. Thus, the 4 chess pieces in these two rows form a rectangle, which is a contradiction.
(2) When the first row has 7 chess pieces, if no rectangle exists, then in the remaining 7 rows of the 7 columns with chess pieces, there can be at most 7 chess pieces, totaling at most 21 chess pieces, which is a contradiction.
(3) When the first row has 6 chess pieces, if no rectangle exists, then in the 6 columns with these 6 chess pieces, there can be at most $7+6=13$ chess pieces. In the remaining 2 columns, there are 12 chess pieces $(12=7+5)$, and there must be two rows with chess pieces in these two columns, forming a rectangle with the 4 chess pieces, which is a contradiction.
(4) When the first row has 5 chess pieces, if no rectangle exists, these 5 chess pieces in the 5 columns can have at most $5+7=12$ chess pieces. In the remaining 7 rows and 3 columns, there are 13 chess pieces $(13=3 \times 4+1)$. Thus, there must be one column with 5 chess pieces. The 5 chess pieces in this column can have at most 7 chess pieces in the other 3 columns, leaving 6 chess pieces in a $2 \times 3$ grid, which must form a rectangle with 4 points, which is a contradiction.
(5) When the first row has 4 chess pieces, if no rectangle exists, these 4 chess pieces in the 4 columns can have at most $4+7=11$ chess pieces. The remaining 14 chess pieces are placed in 7 rows and 4 columns. Since $14=4 \times 3+2$, there must be one column with 4 chess pieces. Thus, the 4 chess pieces in this column can have at most $4+3=7$ chess pieces in the other 4 columns, leaving 7 chess pieces in 3 rows and 3 columns, which still form a rectangle with 4 points, which is a contradiction.
In summary, the minimum value of $k$ is 25.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. $\left(\frac{1+\sqrt{5}}{2}\right)^{6}+\left(\frac{1-\sqrt{5}}{2}\right)^{6}=$
|
Let $x_{1}=\frac{1+\sqrt{5}}{2}, x_{2}=\frac{1-\sqrt{5}}{2}$. Then $x_{1}+x_{2}=1, x_{1} x_{2}=-1$.
Thus, $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=3$.
Therefore, $x_{1}^{6}+x_{2}^{6}=\left(x_{1}^{2}\right)^{3}+\left(x_{2}^{2}\right)^{3}$
$=\left(x_{1}^{2}+x_{2}^{2}\right)\left[\left(x_{1}^{2}\right)^{2}-x_{1}^{2} x_{2}^{2}+\left(x_{2}^{2}\right)^{2}\right]$
$=3\left[\left(x_{1}^{2}+x_{2}^{2}\right)^{2}-3 x_{1}^{2} x_{2}^{2}\right]$
$=3\left(3^{2}-3\right)=18$.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) If the sum, difference, product, and quotient of two unequal natural numbers add up to a perfect square, then such two numbers are called a "wise pair" (for example, $(8,2)$ is a wise pair, since $\left.(8+2)+(8-2)+8 \times 2+\frac{8}{2}=36=6^{2}\right)$.
If both of these natural numbers do not exceed 100, how many such wise pairs are there?
|
Three, let $(a, b)$ be a wise array, and without loss of generality, assume $a > b$.
By definition, we have
$$
(a+b)+(a-b)+a b+\frac{a}{b}=m^{2}
$$
$(a, b, m$ are natural numbers), which simplifies to
$$
2 a+a b+\frac{a}{b}=m^{2}.
$$
Since $2 a, a b, m^{2}$ are all natural numbers, $\frac{a}{b}$ must also be a natural number.
Without loss of generality, let $\frac{a}{b}=k$ (where $k$ is a natural number and $k \neq 1$). Then $a=k b$.
Substituting into equation (1) and simplifying, we get $m^{2}=(b+1)^{2} k$.
Thus, $(b+1)^{2} \mid m^{2} \Rightarrow (b+1) \mid m$.
Therefore, $k=\left(\frac{m}{b+1}\right)^{2} \geqslant 4$.
Also, $b < a \leqslant 100$, so
$1 \leqslant b \leqslant 25$, and $a=k b, 1 < k \leqslant \frac{100}{b}$.
Thus, when $b=1$, we have $k=2^{2}, 3^{2}, \cdots, 10^{2}$, so there are 9 pairs $(a, b)$ that satisfy the condition.
When $b=2$, we have $k=2^{2}, 3^{2}, \cdots, 7^{2}$, so there are 6 pairs $(a, b)$ that satisfy the condition.
Similarly, when $b=3,4$, there are 4 pairs $(a, b)$ each; when $b=5,6$, there are 3 pairs $(a, b)$ each; when $b=7,8, \cdots, 11$, there are 2 pairs $(a, b)$ each; when $b=12,13, \cdots, 25$, there is 1 pair $(a, b)$ each.
Therefore, there are a total of 53 wise arrays.
(Li Yaowen, No. 18 Middle School of Benzhuang City, Shandong Province, 277200)
|
53
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let the function $f(n)$ be defined on the set of positive integers, for any positive integer $n$, we have $f(f(n))=4 n+3$, and for any non-negative integer $k$, we have
$$
f\left(2^{k}\right)=2^{k+1}+1 \text {. }
$$
Then $f(2303)=$
|
5.4607.
Notice
$$
\begin{array}{l}
2303=3+4 \times 3+4^{2} \times 3+4^{3} \times 3+4^{4} \times 2^{3} \text {. } \\
\text { And } f(4 n+3)=f(f(f(n)))=4 f(n)+3 \text {, then } \\
f(2303) \\
=3+4 f\left(3+4 \times 3+4^{2} \times 3+4^{3} \times 2^{3}\right) \\
=\cdots \\
=3+4 \times 3+4^{2} \times 3+4^{3} \times 3+4^{4} f\left(2^{3}\right) \\
=3+4 \times 3+4^{2} \times 3+4^{3} \times 3+4^{4}\left(2^{4}+1\right) \\
=2303+4^{4}\left(2^{4}+1-2^{3}\right)=4607 .
\end{array}
$$
|
4607
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=1$, and for $n \geqslant 2$, $a_{n}=\left\{\begin{array}{ll}a_{\frac{n}{2}}+1, & n \text { is even; } \\ \frac{1}{a_{n-1}}, & n \text { is odd. }\end{array}\right.$ If $a_{n}=\frac{20}{11}$, then the positive integer $n=$ $\qquad$
|
8. 198 .
From the problem, we know that when $n$ is even, $a_{n}>1$; when $n(n>1)$ is odd, $a_{n}=\frac{1}{a_{n-1}}1$, so, $n$ is even. Thus, $a_{\frac{n}{2}}=\frac{20}{11}-1=\frac{9}{11}1, \frac{n}{2}-1$ is even;
$a_{\frac{n-2}{4}}=\frac{11}{9}-1=\frac{2}{9}1, \frac{n-6}{4}$ is even;
$a_{\frac{n-6}{8}}=\frac{9}{2}-1=\frac{7}{2}>1, \frac{n-6}{8}$ is even;
$a_{\frac{n-6}{16}}=\frac{7}{2}-1=\frac{5}{2}>1, \frac{n-6}{16}$ is even;
$a_{\frac{n-6}{32}}=\frac{5}{2}-1=\frac{3}{2}>1, \frac{n-6}{32}$ is even;
$a_{\frac{n-6}{4}}=\frac{3}{2}-1=\frac{1}{2}1, \frac{n-70}{64}$ is even;
$a_{\frac{n-70}{128}}=2-1=1$.
Therefore, $\frac{n-70}{128}=1 \Rightarrow n=198$.
|
198
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given seven points on a circle, connect each pair of points. Find the minimum number of intersection points of these lines inside the circle.
|
We prove that these lines intersect at most three sets of three lines at a point inside the circle, and it is obvious that there cannot be four lines intersecting at a point.
First, select seven vertices of a regular octagon,
it is clear that there are three sets of three lines intersecting at a point (as shown in Figure 3).
Second, there cannot be more than three sets of three lines intersecting at a point. We call a point other than the intersection of three lines a "remaining point".
If there are two adjacent remaining points, assume without loss of generality that these seven points are $A_{1}, A_{2}, \cdots, A_{7}$, and $A_{1}$ and $A_{2}$ are both remaining points. Then, when $A_{1}$ is a remaining point, the intersection of three lines is the intersection of $A_{2} A_{5}$, $A_{3} A_{6}$, and $A_{4} A_{7}$. When $A_{2}$ is a remaining point, the intersection of three lines is the intersection of $A_{1} A_{5}$, $A_{3} A_{6}$, and $A_{4} A_{7}$. Both intersections of three lines are the intersection of $A_{3} A_{6}$ and $A_{4} A_{7}$, hence they coincide, which is a contradiction.
Therefore, remaining points cannot be adjacent.
Thus, among the seven points, there can be at most three remaining points, meaning there can be at most three sets of three lines intersecting at a point.
In summary, the minimum number of intersection points inside the circle is $\mathrm{C}_{7}^{4}-3(3-1)=29$.
(Zhang Lei, Northeast Yucai School, 110179)
|
29
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 As shown in Figure $1, a / / b$, line $a$ has ten points $A_{1}, A_{2}$, $\cdots, A_{10}$, and line
$b$ has nine points $B_{1}, B_{2}, \cdots, B_{9}$
nine points. Connecting each point on
$a$ with each point on
$b$ can result in many line segments, given that no three line segments intersect at one point. How many intersection points do these line segments have in total?
(1996, Seoul International Mathematics Invitational Tournament China Training Team Question)
|
Solution: Take two points each on $a$ and $b$, the four points determine a unique intersection point. Taking two points from $a$ has $10 \times 9 \div 2=45$ methods, taking two points from $b$ has $9 \times 8 \div 2=36$ methods. In total, there are $45 \times 36=1620$ methods.
|
1620
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $a$, $b$, and $c$ are all positive integers, and $a b c=$ 2010. Then the minimum value of $a+b+c$ is ( ).
(A) 84
(B) 82
(C) 78
(D) 76
|
$$
\begin{array}{l}
\text { I.1.C. } \\
\text { From } 2010=2 \times 3 \times 5 \times 67=6 \times 5 \times 67 \\
=1 \times 30 \times 67=\cdots,
\end{array}
$$
we know that the minimum value of $a+b+c$ is 78.
|
78
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example: How many types of isosceles triangles with integer side lengths and a perimeter of 100 are there?
(8th "Hua Luogeng Cup" Junior Mathematics Invitational Final)
|
Let $a$, $b$, and $c$ be the lengths of the three sides of a triangle.
(1) $a=ba .\end{array}\right.$,
Solving, we get $25<a<33 \frac{1}{3}$.
Therefore, there are 8 possibilities.
(2) $a<b=c$.
In this case, $a=100-2 c$.
Thus, $\left\{\begin{array}{l}100-2 c<c \\ 1+2 c \leqslant 100\end{array}\right.$.
Solving, we get $33 \frac{1}{3}<c \leqslant 49 \frac{1}{2}$.
Therefore, there are 16 possibilities.
In total, there are 24 possibilities.
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. When $n$ is a positive integer, it is defined that
$$
n!=n \times(n-1) \times \cdots \times 2 \times 1 \text {, }
$$
which is called the factorial of $n$ (for example, $10!=10 \times 9 \times \cdots \times 2 \times 1$ $=3628800$). Therefore, in 2010!, the total number of zeros at the end is $\qquad$
|
4.501.
The number of trailing zeros depends on the number of factors of 10. Since $10=2 \times 5$, and it is clear that in 2010! the number of factors of 2 is greater than the number of factors of 5, we only need to find the number of prime factors of 5 in 2010!, which gives us the number of trailing zeros.
Among the integers from 1 to 2010, there are $\left[\frac{2010}{5}\right]$ multiples of 5, $\left[\frac{2010}{5^{2}}\right]$ multiples of $5^{2}$, $\qquad$
Since $5^{5}>2010$, the number of factors of 5 in 2010! is
$$
\begin{array}{l}
{\left[\frac{2010}{5}\right]+\left[\frac{2010}{5^{2}}\right]+\left[\frac{2010}{5^{3}}\right]+\left[\frac{2010}{5^{4}}\right]} \\
=402+80+16+3=501 .
\end{array}
$$
|
501
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six. (16 points) As shown in Figure 3, there is a fixed point $P$ inside $\angle M A N$. It is known that $\tan \angle M A N=3, P$ is at a distance $P D=12$ from the line $A N$, and $A D=30$. A line through $P$ intersects $A N$ and $A M$ at points $B$ and $C$ respectively. Find the minimum area of $\triangle A B C$.
---
The translation is provided as requested, maintaining the original formatting and structure.
|
Six, Solution 1
As shown in Figure 7, let $D B=$
$x$, draw $C E \perp A B$,
$P F / / C A$.
Let $A E=t$. Then
$$
\begin{array}{l}
\tan \angle M A N=3 \\
\Rightarrow C E=3 t .
\end{array}
$$
Similarly, by $P D=12 \Rightarrow F D=4$.
By $\triangle A B C \backsim \triangle F B P$, we get $\frac{A B}{C E}=\frac{F B}{P D}$, i.e.,
$$
\frac{30+x}{3 t}=\frac{4+x}{12} \text {. }
$$
Thus, $t=\frac{4(30+x)}{4+x}$.
Therefore, $S_{\triangle A B C}=\frac{1}{2} A B \cdot C E$
$=\frac{1}{2}(30+x) \times \frac{12(30+x)}{4+x}$
$=6 \times \frac{(30+x)^{2}}{4+x}$.
Let $m=\frac{(30+x)^{2}}{4+x}$.
Transforming into a quadratic equation in $x$ gives
$$
x^{2}+(60-m) x+900-4 m=0 \text {. }
$$
Since $x$ is a real number, we have
$$
\Delta=(60-m)^{2}-4(900-4 m) \geqslant 0 \text {, }
$$
i.e., $m(m-104) \geqslant 0$.
Thus, $m \leqslant 0$ or $m \geqslant 104$.
Therefore, when $m=104$, we have
$\left(S_{\triangle B B C}\right)_{\text {min }}=6 \mathrm{~m}=624$.
By $\frac{(30+x)^{2}}{4+x}=104$, solving for $x$ gives $x=22$.
Solution 2 can prove that when the line through point $P$ satisfies $P B=P C$, the area of $\triangle A B C$ is minimized.
In fact, let $B_{1} C_{1}$ be any line through point $P$, forming $\triangle A B_{1} C_{1}$.
As shown in Figure 8, draw $C F$
$/ / B_{1} B$. Then
$\triangle P C F \cong \triangle P B B_{1}$.
Thus, $S_{\triangle B B C}$
$$
\begin{array}{l}
=S_{\triangle P C F}+ \\
S_{\text {quadrilateral } 1 B_{1}, P C} \\
<S_{\triangle B_{1} C_{1}} \text {. } \\
\end{array}
$$
According to the above geometric conclusion, the calculation is as follows.
Since $P D=12$, we have $C E=24$.
Since $\tan \angle M A N=3$, we have $A E=8$.
By $E D=B D$, and $E D=A D-A E=22$, we get $B D=22$.
Thus, $A B=A D+B D=52$.
$\left(S_{\triangle A B C}\right)_{\text {min }}=\frac{1}{2} A B \cdot C E=624$.
|
624
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of lattice points (points with integer coordinates) inside the region (excluding the boundary) bounded by the right branch of the hyperbola $x^{2}-y^{2}=1$ and the line $x=100$ is $\qquad$
|
3.9800 .
By symmetry, we only need to consider the situation above the $x$-axis first.
Let the line $y=k(k=1,2, \cdots, 99)$ intersect the right branch of the hyperbola at point $A_{k}$, and intersect the line $x=100$ at point $B_{k}$. Then the number of integer points inside the segment $A_{k} B_{k}$ is $99-k$. Therefore, the number of integer points in the region above the $x$-axis is
$$
\sum_{k=1}^{9}(99-k)=\sum_{k=0}^{98} k=49 \times 99 \text {. }
$$
There are 98 integer points on the $x$-axis, so the total number of integer points is
$$
2 \times 49 \times 99+98=9800 .
$$
|
9800
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let positive integers $a, b$ satisfy $1 \leqslant a < b \leqslant 100$. If there exists a positive integer $k$, such that $a b \mid\left(a^{k}+b^{k}\right)$, then the pair $(a, b)$ is called a "good pair". Find the number of all good pairs.
（Xiong Bin's problem）
|
4. Let $(a, b)=d, a=s d, b=t d,(s, t)=1(t>s)$.
Thus, $s t d^{2} \mid d^{k}\left(s^{k}+t^{k}\right)$.
Since $\left(s t, s^{k}+t^{k}\right)=1$, it follows that $s t \mid d^{k}$.
Therefore, all prime factors of $st$ can divide $d$.
If $s$ or $t$ has a prime factor $p$ not less than 11, then $p \mid d$. Thus, $p^{2} \mid a$ or $p^{2} \mid b$.
Since $p^{2}>100$, this is a contradiction.
Therefore, the prime factors of $st$ can only be $2,3,5,7$.
If $st$ has at least three of the prime factors $2,3,5,7$, then
$d \geqslant 2 \times 3 \times 5=30$.
Thus, $\max \{a, b\} \geqslant 5 d>100$, which is a contradiction.
If the set of prime factors of $st$ is $\{3,7\}$, then
$\max \{a, b\} \geqslant 7 \times 3 \times 7>100$,
which is a contradiction;
Similarly, the set of prime factors of $st$ cannot be $\{5,7\}$.
If the set of prime factors of $st$ is $\{3,5\}$, $d$ can only be 15, at this time, $s=3, t=5$, so, $(a, b)=(45,75)$. There is 1 good pair.
If the set of prime factors of $st$ is $\{2,7\}$, $d$ can only be 14, at this time, $(s, t)=(2,7),(4,7)$, so, $(a, b)=$ $(28,98),(56,98)$. There are 2 good pairs.
If the set of prime factors of $st$ is $\{2,5\}$, $d$ can only be 10,20.
When $d=10$,
$(s, t)=(2,5),(1,10),(4,5),(5,8)$;
When $d=20$, $(s, t)=(2,5),(4,5)$.
There are 6 good pairs.
If the set of prime factors of $st$ is $\{2,3\}$, $d$ can only be 6, 12, 18, 24, 30.
When $d=6$,
\[
\begin{aligned}
(s, t)= & (1,6),(1,12),(2,3),(2,9), \\
& (3,4),(3,8),(3,16),(4,9), \\
& (8,9),(9,16) ;
\end{aligned}
\]
When $d=12$,
$(s, t)=(1,6),(2,3),(3,4),(3,8)$;
When $d=18$, $(s, t)=(2,3),(3,4)$;
When $d=24$, $(s, t)=(2,3),(3,4)$;
When $d=30$, $(s, t)=(2,3)$.
There are 19 good pairs.
If the set of prime factors of $st$ is $\{7\}$, then $(s, t)=$ $(1,7)$, $d$ can only be 7,14. There are 2 good pairs.
If the set of prime factors of $st$ is $\{5\}$, then $(s, t)=$ $(1,5)$, $d$ can only be $5,10,15,20$. There are 4 good pairs.
If the set of prime factors of $st$ is $\{3\}$, then
When $(s, t)=(1,3)$, $d$ can only be $3,6, \cdots, 33$;
When $(s, t)=(1,9)$, $d$ can only be $3,6,9$;
When $(s, t)=(1,27)$, $d$ can only be 3.
There are 15 good pairs.
If the set of prime factors of $st$ is $\{2\}$, then
When $(s, t)=(1,2)$, $d$ can only be $2,4, \cdots, 50$;
When $(s, t)=(1,4)$, $d$ can only be $2,4, \cdots, 24$;
When $(s, t)=(1,8)$, $d$ can only be $2,4, \cdots, 12$;
When $(s, t)=(1,16)$, $d$ can only be $2,4,6$;
When $(s, t)=(1,32)$, $d$ can only be 2.
There are 47 good pairs.
Therefore, the total number of good pairs is
$1+2+6+19+2+4+15+47=96$ (pairs).
|
96
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Initially 285 Test: How many positive integers $T$ are there such that between $\frac{2010}{T}$ and $\frac{2010+T}{2 T}$ (not including $\frac{2010}{T}$ and $\frac{2010+T}{2 T}$) there are exactly 5 different positive integers?
|
Solution: Obviously, when $T \geqslant 2010$,
$$
\frac{2010}{T} \leqslant 1, \frac{2010+T}{2 T} \leqslant 1,
$$
which does not meet the requirement.
Therefore, $T < 2010$ and
$$
\begin{array}{l}
\frac{2010}{T}>\frac{2010+T}{2 T} \\
\Rightarrow \frac{2010}{T}>\frac{2010+T}{2 T} .
\end{array}
$$
From this, we know
$$
\begin{array}{l}
45,
$$
meaning that there are at most four positive integers, 6, 7, 8, 9, between $\frac{2010}{T}$ and $\frac{2010+T}{2 T}$, which is a contradiction.
In summary, $T$ is a positive integer such that $168 \leqslant T \leqslant 200$, totaling 33.
(Yu Chenjie, Class 6, Grade 2, Shangbei Junior High School, Shanghai, 200070)
|
33
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { Three, (25 points) Let three distinct prime numbers } a, b, c \text { satisfy } \\
a \text { divides }(3 b-c), b \text { divides }(a-c), c \text { divides }(2 a-7 b), \\
20<c<80 .
\end{array}
$$
Find all values of $a^{b} c$.
|
When $a>b, a>c$, we have:
$$
-7 c<-7 b<2 a-7 b<2 a<2 c \text {. }
$$
Let $2 a-7 b=k c$,
where, $k=-6,-5,-4,-3,-2,-1,0,1$.
Then $2 a=7 b+k c \Rightarrow 7 b=2 a-k c$,
$$
\begin{array}{l}
a|(3 b-c) \Leftrightarrow a|(3 \times 7 b-7 c) \\
\Leftrightarrow a \mid[3(2 a-k c)-7 c] \\
\Leftrightarrow a \mid[6 a-(3 k+7) c] \\
\Leftrightarrow a|(3 k+7) c \Leftrightarrow a|(3 k+7), \\
b|(a-c) \Leftrightarrow b|(2 a-2 c) \Leftrightarrow b \mid(7 b+k c-2 c) \\
\Leftrightarrow b|(k-2) c \Leftrightarrow b|(k-2) .
\end{array}
$$
When $k=-6$, from equations (2) and (3) we get
$$
a|(-11), b|(-8) \text {. }
$$
Thus, $a=11, b=2$.
From equation (1) we get $c=-\frac{4}{3}$ (discard).
When $k=-5$, from equations (2) and (3) we get
$a|(-8), b|(-7)$.
Thus, $a=2, b=7$.
From equation (1) we get $c=9$ (discard).
Similarly, when $k=-4,-3,-2,1$, there are no values that meet the conditions.
When $k=-1$, $a=2, b=3, c=17$, which does not satisfy $20<c<80$.
When $k=0$, $2 a=7 b$, then $217 b, 7 \mid 2 a$.
Thus, $b=2, a=7$.
From $a \mid(3 b-c)$, we get $3 b-c=a t, c=6-7 t$. Since $20<c<80$, we have $20<6-7 t<80,-10 \frac{4}{7}<t<-2$.
Only when $t=-5$, $c$ is a prime number, $c=41$. Therefore, $a^{b} c=7^{2} \times 41=2009$.
(Xie Wenxiao, Huangfeng Middle School, Hubei Province, 438000)
|
2009
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given 2011 positive integers, the product of which is equal to their sum. How many 1s are there at least among these 2011 numbers?
|
Let the 2011 positive integers be
$a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{2011}$.
From $a_{1} a_{2} \cdots a_{2011}=a_{1}+a_{2}+\cdots+a_{2011}$
$\leqslant 2011 a_{2011}$,
we know $a_{1} a_{2} \cdots a_{2010} \leqslant 2011a b \Rightarrow(a-1)(b-1)a b \Rightarrow(a-1)(b-1)<1,
\end{array}
$$
Contradiction.
If $A=(2,3,3, a, b)$ or $(3,3,3, a, b)$, the same contradiction is derived.
Therefore, $a_{2004}=1$.
(5) When $a_{2005}=a_{2006}=\cdots=a_{2010}=2, a_{2011}$ $=32$,
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2011}=2004+2 \times 6+32 \\
=2048=a_{1} a_{2} \cdots a_{2011} .
\end{array}
$$
Combining (1)～(5), we know there are at least 2004 ones.
(Liu Donghua, Nankai High School, Tianjin, 300100)
|
2004
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. At a certain charity fundraising dinner, each person ate half a plate of rice, one-third of a plate of vegetables, and one-quarter of a plate of meat. The dinner provided a total of 65 plates of food. How many people attended this fundraising dinner?
|
2. 60 .
Each person ate $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}$ of food, and $65 \div \frac{13}{12}=60$.
Therefore, 60 people attended this charity dinner.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of triples of positive integers $(x, y, z)$ that satisfy $x y z=3^{2010}$ and $x \leqslant y \leqslant z<x+y$ is $\qquad$.
|
3. 336 .
Let $x=3^{a}, y=3^{b}, z=3^{c}$. Then
$$
0 \leqslant a \leqslant b \leqslant c, a+b+c=2010 \text {. }
$$
If $c \geqslant b+1$, then
$$
x+y=3^{a}+3^{b}<3^{b+1} \leqslant 3^{c}=z \text {, }
$$
which contradicts $z<x+y$. Hence, $c=b$.
Thus, $a+2 b=2010$.
Therefore, $670 \leqslant b \leqslant 1005$.
Hence, the number of such positive integer triples, which is the number of $b$, is $1005-670+1=336$.
|
336
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The smallest four-digit number that has exactly 14 divisors (including 1 and itself), and one of its prime factors has a units digit of 3 is $\qquad$ .
|
5.1458.
Let this four-digit number be $n$.
Since $14=14 \times 1=7 \times 2$, we have $n=p^{13}$ or $p^{6} q$ (where $p$ and $q$ are different prime numbers).
If $n=p^{13}$, by the given condition, the unit digit of $p$ is 3, so, $p \geqslant 3$.
Thus, $n \geqslant 3^{13}=1594323$, which is a contradiction.
Therefore, $n=p^{6} q$.
If $p \geqslant 5$, then the number of digits in $n$ will exceed 4, which is a contradiction.
If $p=3$, then $p^{6}=729$, and when $q=2$, $n=$ $p^{6} q=1458$, which satisfies the condition.
If $p=2$, then $p^{6}=64$, in this case, $q$ must be chosen from 3, 13, 23, 43, ... these prime numbers with a unit digit of 3.
Multiplying 64 by these numbers, the smallest four-digit number obtained is
$$
p^{6} q=64 \times 23=1472>1458 .
$$
Therefore, the smallest four-digit number that satisfies the condition is 1458.
|
1458
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given $n$ positive integers (not necessarily distinct), their sum is 100, and the sum of any seven of them is less than 15. Then the minimum value of $n$ is $\qquad$ .
|
9. 50.
Let these $n$ numbers be $a_{1}, a_{2}, \cdots, a_{n}$. Then
$$
a_{1}+a_{2}+\cdots+a_{49} \leqslant 14 \times 7=98 \text {. }
$$
Thus $n \geqslant 50$, and when these 50 numbers are all 2, the condition is satisfied.
|
50
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given point $P$ inside $\triangle A B C$, satisfying $\angle A B P$ $=20^{\circ}, \angle P B C=10^{\circ}, \angle A C P=20^{\circ}$ and $\angle P C B$ $=30^{\circ}$. Then $\angle C A P=$
|
10.20.
As shown in Figure 7, construct a regular $\triangle QBC$ on side $BC$ on the same side as point $A$.
Notice that $BA$ is the angle bisector of the regular $\triangle QBC$, so, $\angle AQC = \angle ACQ = 10^{\circ}$.
Also, $CP$ is the angle bisector of the regular $\triangle QBC$, hence
$$
\angle PQC = \angle PBC = 10^{\circ}.
$$
Since points $A$ and $P$ are on the same side of $QC$, point $A$ lies on segment $PQ$.
Thus, $\angle CAP = \angle AOC + \angle ACQ = 20^{\circ}$.
|
20
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. A farmer has 100 pigs and 100 chickens. He has four adjacent square yards, forming a $2 \times 2$ grid. The farmer wants to distribute the livestock among the yards according to the following requirements: the first row has 120 heads, the second row has 300 feet; the first column has 100 heads, the second column has 320 feet. Then there are $\qquad$ different ways to distribute them.
|
11.341.
As shown in Figure 8, label the four courtyards with letters $A_{i} (i=1,2,3,4)$. Let $A_{i}$ courtyard have $x_{i}$ pigs and $y_{i}$ chickens.
From the problem, we get
every 8
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}=100, \\
y_{1}+y_{2}+y_{3}+y_{4}=100, \\
x_{1}+y_{1}+x_{2}+y_{2}=120, \\
4\left(x_{3}+x_{4}\right)+2\left(y_{3}+y_{4}\right)=300, \\
x_{1}+y_{1}+x_{3}+y_{3}=100, \\
4\left(x_{2}+x_{4}\right)+2\left(y_{2}+y_{4}\right)=320 .
\end{array}\right.
$$
Solving, we get $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=(x, 30-x, 40-x, 30+x)$,
$$
\left(y_{1}, y_{2}, y_{3}, y_{4}\right)=(y, 90-y, 60-y, y-50) \text {, }
$$
where $x$ has 31 possible values from $0$ to $30$, and $y$ has 11 possible values from $50$ to $60$.
Therefore, there are $11 \times 31=341$ ways to distribute the livestock to the courtyards as required.
|
341
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In a candy store, candies are sold in three types of packages: small packs contain 6 candies, medium packs contain 9 candies, and large packs contain 20 candies. If you can only buy whole packs of candies, what is the maximum number of candies you cannot purchase?
|
2. Write non-negative integers in the following six-row number table:
\begin{tabular}{ccccccccc}
$\mathbf{0}$ & 6 & 12 & 18 & 24 & 30 & 36 & 42 & 48 \\
1 & 7 & 13 & 19 & 25 & 31 & 37 & 43 & 49 \\
2 & 8 & 14 & 20 & 26 & 32 & 38 & 44 & 50 \\
3 & 9 & 15 & 21 & 27 & 33 & 39 & 45 & 51 \\
4 & 10 & 16 & 22 & 28 & 34 & 40 & 46 & 52 \\
5 & 11 & 17 & 23 & 29 & 35 & 41 & 47 & 53
\end{tabular}
Each row of numbers increases by 6 from left to right.
If a number in a row can be obtained (i.e., you can buy this number of candies), then all the numbers in this row that come after it can also be obtained. Therefore, it is only necessary to find the first number in each row that can be obtained (which can be expressed in the form $9k + 20l$ where $k, l \in \mathbf{N}$), which are marked in bold in the table.
Clearly, the largest of these six numbers is 49.
Thus, the number 43, which is in the same row and immediately before it, is the largest number that cannot be obtained.
|
43
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. (40 points) 64 points on a plane form an $8 \times 8$ grid. The distance between any two adjacent points in the same row or column is 1. How many rectangles with an area of 12 can be formed using four of these 64 points as vertices?
|
4. Connecting points in the same row and column can form a grid.
First, consider rectangles with sides parallel to the grid lines. For a rectangle of shape $2 \times 6$, there are two orientation choices (horizontal and vertical). One pair of opposite sides has $8-6=2$ choices, and the other pair has $8-2=6$ choices. Therefore, there are $2 \times 2 \times 6=24$ such rectangles.
Similarly, for a rectangle of shape $3 \times 4$, there are $2 \times 4 \times 5=40$ such rectangles.
Next, consider rectangles with sides parallel to the diagonals of the grid, with shapes $\sqrt{2} \times 6 \sqrt{2}$ and $2 \sqrt{2} \times 3 \sqrt{2}$.
These rectangles also have two orientation choices. First, consider the direction where the longest side extends from the bottom left to the top right. For the first shape, there is only 1 such rectangle (as shown in Figure 12); for the second shape, there are 9 such rectangles, one of which is marked in Figure 12. The top-left vertices of these 9 rectangles are marked with black dots in Figure 12. Therefore, there are $2 \times (1+9)=20$ such rectangles.
Since $12=2^{2} \times 3$, and 3 cannot be expressed as the sum of two perfect squares, there are no other shapes of rectangles.
Therefore, the total number of rectangles that meet the criteria is $24+40+20=84$.
|
84
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. (40 points) Find the largest positive integer $n$, such that there exists a unique positive integer $k$ satisfying $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$.
|
5. The inequality can be written as $1+\frac{7}{8}>1+\frac{k}{n}>1+\frac{6}{7}$. Or $\frac{98}{112}>\frac{k}{n}>\frac{96}{112}$.
If $n=112$, then the unique value of $k$ is 97.
Assuming $n>112$, then $\frac{98 n}{112 n}>\frac{112 k}{112 n}>\frac{96 n}{112 n}$.
Between $96 n$ and $98 n$, there are at least two numbers that are multiples of 112, in which case, the value of $k$ is not unique. Therefore, the maximum value of $n$ is 112.
|
112
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. (40 points) In each small square of a $9 \times 9$ grid, fill in a number such that each row and each column contains at most four different numbers. What is the maximum number of different numbers that can be in this grid?
|
6. Suppose this square table contains 29 different numbers. According to the pigeonhole principle, there must be a row with four different numbers, let's assume it is the first row. The remaining 25 numbers are in rows $2 \sim 9$. Similarly, according to the pigeonhole principle, there must be a row with four, let's assume it is the second row.
Next, observe the numbers in each column of the square table. Each column already has two different numbers at the top, so each column below the second row can have at most two different numbers. This way, the maximum number of different numbers is $8+9 \times 2=26$, which is a contradiction. Therefore, the number of different numbers is less than 29.
Figure 13 constructs a $9 \times 9$ square table containing 28 different numbers, with each row and each column having exactly four different numbers. Another filling method is shown in Figure 14.
|
28
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the function $f(x)=x^{2}-1$ with domain $D$, and the range is $\{-1,0,1,3\}$. Determine the maximum number of such sets $D$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
2, 1.27.
Since $f(0)=-1, f( \pm 1)=0$,
$$
f( \pm \sqrt{2})=1, f( \pm 2)=3 \text {, }
$$
Therefore, $0 \in D$; at least one from each of the sets $\{-1,1\}$, $\{-\sqrt{2}, \sqrt{2}\}$, and $\{-2,2\}$ belongs to $D$.
Thus, there are $3 \times 3 \times 3=27$ such $D$'s.
|
27
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In a convex quadrilateral $ABCD$, $\angle BAD + \angle ADC = 240^{\circ}$, $E$ and $F$ are the midpoints of sides $AD$ and $BC$, respectively, and $EF = \sqrt{7}$. If two squares $A_1$ and $A_2$ are drawn with sides $AB$ and $CD$, respectively, and a rectangle $A_3$ is drawn with length $AB$ and width $CD$, find the sum of the areas of the three figures $A_1$, $A_2$, and $A_3$.
|
6.28 .
As shown in Figure 6, extend $BA$ and $CD$ to intersect at point $P$.
Given $\angle BAD + \angle ADC = 240^{\circ}$, we have
$$
\angle BPC = 60^{\circ}.
$$
Connect $BD$, and take the midpoint $G$ of $BD$, then connect $EG$ and $FG$.
By the Midline Theorem of a triangle, we know
$$
EG \perp \frac{1}{2} AB, FG \perp \frac{1}{2} CD, \angle EGF = 120^{\circ}.
$$
In $\triangle EGF$, by the Law of Cosines, we get
$$
\begin{array}{l}
7 = EF^2 = EG^2 + FG^2 + EG \cdot FG \\
= \left(\frac{AB}{2}\right)^2 + \left(\frac{CD}{2}\right)^2 + \left(\frac{AB}{2}\right)\left(\frac{CD}{2}\right),
\end{array}
$$
which simplifies to $AB^2 + CD^2 + AB \times CD = 28$.
Therefore, the sum of the areas of the three figures $A_1, A_2$, and $A_3$ is 28.
|
28
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ is a permutation of $1,2,3,4,5$, and satisfies $\left|a_{i}-a_{i+1}\right| \neq 1(i=1,2,3,4)$. Then the number of permutations $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ that meet the condition is $\qquad$.
|
8. 14 .
The permutations that satisfy the conditions are:
$$
\begin{array}{l}
1,3,5,2,4 ; 1,4,2,5,3 ; 2,4,1,3,5 ; \\
2,4,1,5,3 ; 2,5,3,1,4 ; 3,1,4,2,5 ; \\
3,1,5,2,4 ; 3,5,1,4,2 ; 3,5,2,4,1 ; \\
4,1,3,5,2 ; 4,2,5,1,3 ; 4,2,5,3,1 ; \\
5,2,4,1,3 ; 5,3,1,4,2 .
\end{array}
$$
|
14
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) A competitive event involves several teams from two cities, A and B. It is known that city B has 8 more teams than city A, and any two teams play exactly one match. The event rules state: the winner gets 1 point, the loser gets 0 points, and there are no ties. In the end, the total score of all teams from city B is 4 points more than the total score of all teams from city A. Find the minimum score of the best team from city A.
|
Let city A have $x$ teams participating in the competition. Then city B has $x+8$ teams. When the teams from cities A and B compete against each other, city A scores a total of $y$ points, and city B scores a total of $x(x+8)-y$ points.
From the problem, we have
$$
\begin{array}{l}
\frac{1}{2} x(x-1)+y+4 \\
=\frac{1}{2}[(x+8)-1](x+8)+[x(x+8)-y] \\
\Rightarrow y=\frac{1}{2} x^{2}+8 x+12 .
\end{array}
$$
Since $y$ is an integer, $x$ must be even.
$$
\begin{array}{l}
\text { From } x(x+8)-y \geqslant 0 \\
\Rightarrow x(x+8)-\left(\frac{1}{2} x^{2}+8 x+12\right) \geqslant 0 \\
\Rightarrow x \geqslant \sqrt{24}>4 .
\end{array}
$$
Since $x$ is even, the minimum value of $x$ is 6. Clearly, $y \geqslant x+8$.
Thus, the highest score of the best team in city A is
$$
(x-1)+(x+8)=2 x+7 \text { (points). }
$$
Therefore, when $x=6$, the minimum value of $2 x+7$ is 19. Hence, the minimum value of the highest score of the best team in city A is 19.
|
19
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2: With $2009^{12}$ as one of the legs, and all three sides being integers, the number of different right-angled triangles (congruent triangles are considered the same) is $\qquad$.
(2009, International Mathematics Tournament of the Cities for Young Mathematicians)
|
Let $a$, $2009^{12}$, and $c$ be the sides of a right triangle, with $c$ being the hypotenuse. Then,
$$
(c+a)(c-a)=2009^{24}=41^{24} \times 7^{48}.
$$
Since $c+a > c-a$ and they have the same parity, both must be odd.
Also, $2009^{24}=41^{24} \times 7^{48}$ has $25 \times 49$ different factors, which can be paired into $\frac{25 \times 49-1}{2}=612$ pairs, each pair forming a possible solution.
On the other hand, $2009^{12}$ clearly does not meet the requirements.
Therefore, there are 612 pairs of different integer-sided right triangles that satisfy the conditions.
|
612
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 There is a railway network between six cities, such that there is a direct railway between any two cities. On Sundays, some railways will be closed for maintenance. The railway department stipulates: after closing several sections of the railway, it must still be possible to travel by rail between any two cities (not necessarily directly). How many different railway maintenance methods meet the railway department's requirements? ${ }^{[1]}$
(2007, British Mathematical Olympiad)
|
【Analysis】The problem is equivalent to finding the number of all possible ways to connect a graph composed of six points and several edges, such that any two points are connected.
Let $f(n)$ denote the number of all possible ways to connect $n$ points such that any two points are connected. Then
$$
f(1)=1, f(2)=1 \text {. }
$$
For the case of three points, either all pairs of points are connected, or any two points are connected by selecting any two out of the three edges, so $f(3)=4$.
Since $n$ vertices can have $\mathrm{C}_{n}^{2}$ edges, there are a total of $2^{\mathrm{C}_{n}^{2}}$ different connection methods. At this point, any two vertices either have an edge or do not have an edge. Select one vertex $v$ from the $n$ vertices, and consider the set of $i(1 \leqslant i \leqslant n)$ vertices that include $v$. You can choose $i-1$ vertices from the remaining $n-1$ vertices to connect with $v$ to form this set of $i$ vertices, so there are $\mathrm{C}_{n-1}^{i-1}$ ways to select.
By the definition of $f(i)$, for the selected $i-1$ vertices and vertex $v$, there are $f(i)$ ways to ensure that any two points are connected. For the remaining $n-i$ points, there are $2^{\mathrm{C}_{n-i}^{2}}$ different connection methods.
Therefore, $2^{\mathrm{C}_{n}^{2}}=\sum_{i=1}^{n} \mathrm{C}_{n-1}^{i-1} f(i) 2^{\mathrm{C}_{n-i}^{2}}$.
Given $f(1)=1, f(2)=1, f(3)=4$, we get
$$
f(4)=38, f(5)=728, f(6)=26704 \text {. }
$$
Since among the 26704 ways, 26703 ways involve closing some sections of the railway, and 1 way involves not closing any section, the answer is 26703.
|
26703
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let $a_{n}$ be the coefficient of $x$ in the binomial expansion of $(3-\sqrt{x})^{n}(n=2,3, \cdots)$. Then
$$
\sum_{n=2}^{18} \frac{3^{n}}{a_{n}}=
$$
$\qquad$
|
11. 17.
Since $a_{n}=3^{n-2} C_{n}^{2}$, therefore,
$$
\frac{3^{n}}{a_{n}}=3^{2} \times \frac{2}{n(n-1)}=\frac{18}{n(n-1)} \text {. }
$$
Thus, $\sum_{n=2}^{18} \frac{3^{n}}{a_{n}}=18 \sum_{n=2}^{18} \frac{1}{n(n-1)}=17$.
|
17
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For each positive integer $n$, let $f(n)$ denote the last digit of $1+2+\cdots+n$ (for example, $f(1)=1$, $f(2)=3$, $f(3)=6$). Calculate the value of $f(1)+f(2)+\cdots+f(2011)$.
|
One, because the last digit of the sum of any 20 consecutive positive integers is 0, so,
$$
f(n+20)=f(n) .
$$
Thus, it is only necessary to calculate the $f(n)$ corresponding to 1 to 20, as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$f(n)$ & 1 & 3 & 6 & 0 & 5 & 1 & 8 & 6 & 5 & 5 \\
\hline$n$ & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\
\hline$f(n)$ & 6 & 8 & 1 & 5 & 0 & 6 & 3 & 1 & 0 & 0 \\
\hline
\end{tabular}
Also, $f(1)+f(2)+\cdots+f(20)=70$,
$2011=20 \times 100+11$,
then $f(1)+f(2)+\cdots+f(2011)$
$=70 \times 100+f(1)+f(2)+\cdots+f(11)$
$=7000+46=7046$.
|
7046
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies
$$
f(f(x)+f(y))=f(x)+y(\forall x, y \in \mathbf{R}) \text {. }
$$
then $f(2011)=$ $\qquad$ .
|
$-1.2011$.
Let $x=0$ in equation (1), we get
$$
f(f(0)+f(y))=f(0)+y,
$$
which shows that $f$ is surjective.
Let $x$ be the zero root $a$ of $f(x)$ in equation (1). Then $f(f(y))=y$.
Replace $x$ with $f(x)$ in equation (1) to get
$$
f(x+f(y))=x+y \text {. }
$$
Let $y=a$ in equation (2) to get $f(x)=x+a$.
Substitute into equation (1) to get
$$
f(x+a+y+a)=x+a+y \text {, }
$$
which simplifies to $x+y+3 a=x+y+a \Rightarrow a=0$.
Thus, $f(x)=x$.
|
2011
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a_{1}, a_{2}, \cdots, a_{2011}$ be positive real numbers, $S=\sum_{i=1}^{2011} a_{i}$, and
$$
\begin{array}{l}
(S+2011)\left(S+a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2011} a_{1}\right) \\
=4\left(\sqrt{a_{1} a_{2}}+\sqrt{a_{2} a_{3}}+\cdots+\sqrt{a_{2011} a_{1}}\right)^{2} .
\end{array}
$$
Then $S=$ . $\qquad$
|
3. 2011.
By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
& (S+2011)\left(S+a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2011} a_{1}\right) \\
= & \left(a_{1}+a_{2}+\cdots+a_{2011}+\frac{1+\cdots+1}{2011}\right) . \\
& \left(a_{2}+a_{3}+\cdots+a_{1}+a_{1} a_{2}+\cdots+a_{2011} a_{1}\right) \\
\geqslant & {\left[2\left(\sqrt{a_{1} a_{2}}+\sqrt{a_{2} a_{3}}+\cdots+\sqrt{a_{2011} a_{1}}\right)\right]^{2} . }
\end{aligned}
$$
Since equality holds, we have
$$
\frac{a_{1}}{a_{2}}=\frac{a_{2}}{a_{3}}=\cdots=\frac{a_{2011}}{a_{1}}=\frac{1}{a_{1} a_{2}}=\cdots=\frac{1}{a_{2011} a_{1}} \text {. }
$$
Thus, $a_{1}=a_{2}=\cdots=a_{2011}=1$.
Therefore, $S=2011$.
|
2011
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Try to find the last two non-zero digits of 2011!.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Let the last two non-zero digits of a positive integer $k$ be denoted as $f(k)$. Below, we discuss under modulo 100.
Clearly, $5 \times f(k!)$, and if $5 \times f(k), 5 \times f(l)$, then $f(k l) \equiv f(k) f(l)$. Therefore,
$$
\begin{array}{l}
f(2011!) \equiv f\left(\frac{2011!}{2000!}\right) f(2000!) \\
\equiv 68 f(2000!), \\
f(2000!) \\
\equiv f\left(\prod_{i \in A_{1}} g_{220}(i)\right) f\left(\prod_{i \in \Lambda_{2}} g_{200}(i)\right) f\left(g_{2000}(10)\right),
\end{array}
$$
where, $A_{1}=\{1,2,3,7,8,9\}, A_{2}=\{4,5,6\}$,
$$
g_{k}(i)=\prod_{j=0}^{k-1}(10 j+i) .
$$
It is easy to see that, $g_{k}(i) g_{k}(10-i) \equiv i^{k}(10-i)^{k}(i=$ $1,2,3,4)$, and when $20 \mid k$,
$$
\begin{array}{l}
g_{k}(i) \equiv i^{k} \equiv\left\{\begin{array}{ll}
1, & i=1,3,7,9 ; \\
76, & i=2,4,6,8 .
\end{array}\right. \\
\text { Then } f\left(\prod_{i \in \Lambda_{1}} g_{200}(i)\right) \equiv 76^{2} \equiv 76, \\
f\left(\prod_{i \in \lambda_{2}} g_{200}(i)\right) \\
\equiv\left(\prod_{j=0}^{10} \frac{10 j+4}{2}\right)\left(\prod_{j=0}^{100} \frac{10 j+6}{2}\right) f\left(2^{200} \prod_{j=0}^{109}(10 j+5)\right) \\
\equiv\left(\prod_{i \in A_{1} \backslash 1,91} g_{100}(i)\right) f\left(2^{200} \prod_{j=1}^{200}(2 j-1)\right) \\
\equiv 76 f\left(2^{200} \prod_{j=1}^{200}(2 j-1)\right), \\
f\left(g_{200}(10)\right)=f(200!) .
\end{array}
$$
$$
\begin{array}{l}
\quad \text { Hence } f(2000!) \\
\equiv 76 f\left(2^{200} \prod_{j=1}^{200}(2 j-1)\right) f(200!) \\
\equiv 76 f(400!) \\
\equiv 76 f\left(\prod_{i \in \Lambda_{1}} g_{40}(i)\right) f\left(\prod_{i \in \lambda_{2}} g_{40}(i)\right) f\left(g_{40}(10)\right) . \\
\text { Also } f\left(\prod_{i \in \Lambda_{1}} g_{40}(i)\right) \equiv 76, \\
f\left(\prod_{i \in \Lambda_{2}} g_{40}(i)\right) \\
\equiv\left(\prod_{i \in A_{1} \backslash 1,91} g_{20}(i)\right) f\left(2^{40} \prod_{j=1}^{40}(2 j-1)\right) \\
\equiv 76 f\left(2^{40} \prod_{j=1}^{40}(2 j-1)\right), \\
f\left(g_{40}(10)\right)=f(40!), \\
\text { Hence } f(2000!) \equiv 76 f(80!) \\
\equiv 76 f\left(\prod_{i \in \Lambda_{1}} g_{8}(i)\right) f\left(\prod_{i \in \Lambda_{2}} g_{8}(i)\right) f\left(g_{8}(10)\right) . \\
\text { And } f\left(\prod_{i \in A_{1}} g_{8}(i)\right) \equiv \prod_{i=1}^{3} g_{8}(i) g_{8}(10-i) \\
\equiv \prod_{i=1}^{3} i^{8}(10-i)^{8} \equiv 76,
\end{array}
$$
$$
\begin{array}{l}
f\left(\prod_{i \in A_{2}} g_{8}(i)\right)=\left(\prod_{\left.i \in A_{1} \backslash 1,9\right\}} g_{4}(i)\right) f\left(2^{8} \prod_{j=1}^{8}(2 j-1)\right) \\
=2^{4} \times 7^{4} \times 3^{4} \times 8^{4} f\left(2^{8} \times 15!!\right) \equiv 44, \\
f\left(g_{8}(10)\right)=f(8!)=32,
\end{array}
$$
Therefore, $f(2000!) \equiv 76^{2} \times 44 \times 32 \equiv 8$.
Thus, $f(2011!) \equiv 68 \times 8 \equiv 44$.
|
44
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In $\triangle A B C$, it is known that $A B=A C, \angle C$'s bisector $C D$ intersects $A B$ at point $D, B D, B C, C D$ are three consecutive integers. Find the perimeter of $\triangle A B C$.
|
Let $A B=A C=b, B C=a$.
By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
B D=\frac{a b}{a+b}, \\
B C-B D=a-\frac{a b}{a+b}=\frac{a^{2}}{a+b} \\
=k \in\{1,2\} .
\end{array}
$$
Following Example 1, we get
$$
C D^{2}=B D(B D+B C)=\frac{a^{2} b(a+2 b)}{(a+b)^{2}} \text {. }
$$
If $k=1$, then $b=a^{2}-a$.
Thus, $C D^{2}=(a-1)(2 a-1)>(a-2)^{2}$.
Therefore, $(a-1)(2 a-1)=(a+1)^{2}$.
Solving this, we get $a=5, b=20, a+2 b=45$.
If $k=2$, similarly, we get $a=3, b=\frac{3}{2}$, which is discarded.
Therefore, the perimeter of $\triangle A B C$ is 45.
|
45
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Assuming the Earth rotates once around the axis connecting the North Pole and the South Pole in 23 hours 56 minutes 4 seconds, and the Earth's equatorial radius is $6378.1 \mathrm{~km}$. Then, when you stand on the equator and rotate with the Earth, the linear velocity around the axis is $\qquad$ meters/second (rounded to meters, $\pi$ taken as 3.1416$)$.
|
1. 465
|
465
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (20 points) The numbers $1,2, \cdots, 666$ are written on a blackboard. In the first step, the first eight numbers: $1,2, \cdots, 8$, are erased, and the sum of these numbers, 36, is written after 666; in the second step, the next eight numbers: $9,10, \cdots, 16$, are erased, and the sum of these numbers, 100, is written at the end; this process continues (i.e., in each step, the first eight numbers are erased, and the sum of these eight numbers is written at the end).
(1) How many steps are required until only one number remains on the blackboard?
(2) When only one number remains on the blackboard, find the sum of all numbers that have appeared on the blackboard (if a number appears multiple times, it should be counted repeatedly).
|
(1) Since each step reduces the numbers by seven, after $\frac{666-1}{7}=95$ steps, only one number remains.
(2) From $666-512=154$, it follows that after $\frac{154}{7}=22$ steps, there are 512 numbers left.
In 22 steps, a total of $22 \times 8=176$ numbers are crossed out, and their sum is
$1+2+\cdots+176=88 \times 177$.
Let $S=1+2+\cdots+666=333 \times 667$.
Then after 22 steps, the sum of the remaining 512 numbers is still $S$.
Assume there were originally $8^{k}$ numbers, with their sum being $x$. Then after $8^{k-1}$ steps, all the original $8^{k}$ numbers are crossed out, and the sum of the remaining $8^{k-1}$ numbers on the blackboard is still $x$.
Therefore, when the blackboard is left with only one number, the sum of all the numbers is $(k+1) x$.
Thus, when only one number remains on the blackboard, the sum of all the numbers that have appeared on the blackboard is
$$
\begin{array}{l}
88 \times 177+4 S=88 \times 177+4 \times 333 \times 667 \\
=904020
\end{array}
$$
|
904020
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given the inverse function of $y=f(x+1)$ is
$$
\begin{array}{c}
y=f^{-1}(x+1) \text {, and } f(1)=4007 \text {. Then } \\
f(1998)=
\end{array}
$$
|
II, 7.2010.
From $y=f^{-1}(x+1)$, we get $x+1=f(y)$, which means $x=f(y)-1$.
Thus, the inverse function of $y=f^{-1}(x+1)$ is $y=f(x)-1$. Therefore, $f(x+1)-f(x)=-1$.
Let $x=1,2, \cdots, 1997$, add up all the equations and simplify to get $f(1998)-f(1)=-1997$.
Hence, $f(1998)=2010$.
|
2010
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given real numbers $a, b (a \neq b)$, and they satisfy
$$
\begin{array}{l}
(a+1)^{2}=3-3(a+1), \\
3(b+1)=3-(b+1)^{2} .
\end{array}
$$
Then the value of $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}$ is ( ).
(A) 23
(B) -23
(C) -2
(D) -13
|
【Analysis】Transform the known two equations into
$$
\begin{array}{l}
(a+1)^{2}+3(a+1)-3=0, \\
(b+1)^{2}+3(b+1)-3=0 .
\end{array}
$$
It can be seen that $a$ and $b$ are the two roots of the equation with respect to $x$
$$
(x+1)^{2}+3(x+1)-3=0
$$
Solution: From equation (1) in the analysis, simplifying and rearranging yields
$$
x^{2}+5 x+1=0 \text {. }
$$
Since $\Delta=25-4>0, a+b=-5, a b=1$, therefore, both $a$ and $b$ are negative.
Then $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}=-\frac{a^{2}+b^{2}}{\sqrt{a b}}$
$$
=-\frac{(a+b)^{2}-2 a b}{\sqrt{a b}}=-23 \text {. }
$$
Therefore, the answer is B.
|
-23
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Four, for a $7 \times 7$ grid of small squares, how many of the small squares can be shaded such that no two shaded squares are adjacent? Adjacent means they share a side.
How many of the small squares can be shaded?
|
Four, color at most 26 small squares.
The coloring in Figure 2 satisfies the conditions.
The following proves: At most 26 small squares can be colored.
First, according to the problem, for any $2 \times 2$ square grid, at most two of the small squares can be colored; for a $3 \times 3$ square grid, at most 5 of the small squares can be colored to meet the conditions.
Second, for a $5 \times 5$ square grid, it is based on a $3 \times 3$ square grid with a border of width 2 added (as shown in Figure 3). This border can be divided into 4 $2 \times 2$ square grids, with the bottom-right two squares, one being reused and the other not used. Therefore, at most 9 additional squares can be colored, and when adding 9 colored squares, the bottom-right square must be colored, while the reused square is not colored. Thus, a $5 \times 5$ square grid can have at most 14 squares colored, and when 14 squares are colored, the bottom-right square is colored.
Finally, a $7 \times 7$ square grid is based on a $5 \times 5$ square grid with a border of width 2 added (as shown in Figure 4).
Similarly, at most 13 additional squares can be colored, and when adding 13 colored squares, the bottom-right square must be colored, while the reused square is not colored. Thus, at most 27 squares can be colored. At this point, assume the square in the last row and the second-to-last column is not colored (otherwise, the square in the second-to-last row and the last column is not colored). According to the previous analysis, the two squares in the last two rows and the third-to-last column are colored, which is a contradiction.
Therefore, at most 26 squares can be colored.
In summary, at most 26 small squares can be colored.
|
26
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 If real numbers $x, y$ satisfy
$$
\begin{array}{l}
\frac{x}{3^{3}+4^{3}}+\frac{y}{3^{3}+6^{3}}=1, \\
\frac{x}{5^{3}+4^{3}}+\frac{y}{5^{3}+6^{3}}=1,
\end{array}
$$
then $x+y=$ $\qquad$
(2005, National Junior High School Mathematics Competition)
|
【Analysis】It is easy to notice that the denominators of the two given equations contain $3^{3}$ and $5^{3}$ respectively. Therefore, we can consider $3^{3}$ and $5^{3}$ as the roots of a certain equation.
Solution From the given conditions, it is easy to see that $3^{3}$ and $5^{3}$ are the roots of the equation in terms of $t$:
$$
\frac{x}{t+4^{3}}+\frac{y}{t+6^{3}}=1 \text {, }
$$
which is equivalent to the quadratic equation
$$
t^{2}+\left(4^{3}+6^{3}-x-y\right) t+\left(4^{3} \times 6^{3}-4^{3} y-6^{3} x\right)=0.
$$
By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
3^{3}+5^{3}=-\left(4^{3}+6^{3}-x-y\right) . \\
\text { Therefore, } x+y=3^{3}+4^{3}+5^{3}+6^{3}=432 .
\end{array}
$$
|
432
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. If the function $f(x)=\ln \frac{\mathrm{e} x}{\mathrm{e}-x}$, then $\sum_{k=1}^{2010} f\left(\frac{k e}{2011}\right)=$ $\qquad$ .
|
13.2010.
Notice
$$
\begin{array}{l}
f(x)+f(\mathrm{e}-x) \\
=\ln \left[\frac{\mathrm{e} x}{\mathrm{e}-x} \cdot \frac{\mathrm{e}(\mathrm{e}-x)}{\mathrm{e}-(\mathrm{e}-x)}\right]=2 . \\
\text { Therefore } \sum_{k=1}^{2010} f\left(\frac{k \mathrm{e}}{2011}\right) \\
=\sum_{k=1}^{1005}\left(f\left(\frac{k \mathrm{e}}{2011}\right)+f\left(\mathrm{e}-\frac{k \mathrm{e}}{2011}\right)\right) \\
=2 \times 1005=2010 .
\end{array}
$$
|
2010
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
19. (15 points) In a certain grade, $n$ students participate in Chinese and
Mathematics exams, with scores ranging from 0 to 100 for each subject. Suppose no two students have exactly the same scores (i.e., at least one subject score is different). Additionally, “A is better than B” means that student A’s scores in both Chinese and Mathematics are higher than student B’s scores in both subjects. Question: What is the smallest value of $n$ such that there must exist three students (let’s call them A, B, and C) where A is better than B, and B is better than C.
|
19. Establish a Cartesian coordinate system $x O y$.
If a student's Chinese score is $i$ points, and mathematics score is $j$ points, let it correspond to the integer point $(i, j)$ on the plane, called a "score point". Thus, the examination results of $n$ students are mapped to $n$ score points within the range
$$
0 \leqslant x \leqslant 100,0 \leqslant y \leqslant 100
$$
on the plane.
Consider 201 lines on the plane:
$$
y=x \pm b(b=0,1, \cdots, 100) \text {. }
$$
If a line has three score points, it indicates that there are three students, A, B, and C, such that A is better than B, and B is better than C.
Clearly, the lines $y=x+100$ and $y=x-100$ can each have at most one score point; the lines $y=x+99$ and $y=x-99$ can each have at most two score points.
Since $2 \times(201-2)+1 \times 2=400$, when $n>400$, there must be a line with three score points.
Thus, the minimum value of $n$, $n_{0} \leqslant 401$.
Let the set
$S=\{(i, j) \mid i=0,1 ; j=0,1, \cdots, 100\}$;
$T=\{(i, j) \mid i=0,1, \cdots, 100 ; j=0,1\}$.
Clearly, $|S \cup T|=400$, and there do not exist three score points on the same line in $S \cup T$.
Therefore, $n_{0} \geqslant 401$.
Thus, $n_{0}=401$.
|
401
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Given that $a$ and $b$ are integers, and satisfy $a-b$ is a prime number, $ab$ is a perfect square. If $a \geqslant 2011$, find the minimum value of $a$.
|
Three, let $a-b=m$ (where $m$ is a prime number) and $ab=n^2$ (where $n$ is a positive integer).
$$
\begin{array}{l}
\text { From }(a+b)^{2}-4ab=(a-b)^{2} \\
\Rightarrow(2a-m)^{2}-4n^{2}=m^{2} \\
\Rightarrow(2a-m+2n)(2a-m-2n)=m^{2} \times 1 .
\end{array}
$$
Since $2a-m+2n$ and $2a-m-2n$ are both positive integers, and $2a-m+2n > 2a-m-2n$ (since $m$ is a prime number), we have:
$$
2a-m+2n=m^{2}, 2a-m-2n=1 .
$$
Solving these, we get $a=\frac{(m+1)^{2}}{4}, n=\frac{m^{2}-1}{4}$.
Thus, $b=a-m=\frac{(m-1)^{2}}{4}$.
Given $a \geqslant 2011$, i.e., $\frac{(m+1)^{2}}{4} \geqslant 2011$.
Considering $m$ is a prime number, we get $m \geqslant 89$.
In this case, $a \geqslant \frac{(89+1)^{2}}{4}=2025$.
When $a=2025$,
$$
m=89, b=1936, n=1980 \text {. }
$$
Therefore, the minimum value of $a$ is 2025.
|
2025
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. There are 20 teams participating in the national league. Question: What is the minimum number of matches that must be played so that in any group of three teams, at least two teams have played against each other?
|
Let team $A$ have the minimum number of matches (which is $k$ matches). Then
(1) There are $k$ teams that have played against team $A$, and each of these teams has played at least $k$ matches;
(2) There are $19-k$ teams that have not played against team $A$, and these teams must all have played against each other, otherwise, the two teams that have not played against each other and team $A$ would not meet the requirements. Therefore, the minimum number of matches is
$$
\begin{array}{l}
N=\frac{1}{2}\left[k+k^{2}+(19-k)(18-k)\right] \\
=(k-9)^{2}+90 \geqslant 90 .
\end{array}
$$
Next, we prove that 90 matches can meet the requirements of the problem.
Divide the 20 teams into two groups, each with 10 teams, and each group conducts a single round-robin tournament.
|
90
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Among the integers $1,2, \cdots, 2011$, the number of integers that can be expressed in the form $[x[x]]$ is $\qquad$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
|
2. 1.990.
Let $x=k+a, k=[x], 0 \leqslant a<1$. Then $[x[x]]=[k(k+a)]=k^{2}+[k a]$.
Therefore, $k^{2}, k^{2}+1, \cdots, k^{2}+k-1(k=1,2$, $\cdots, 44)$ can all be expressed in the form of $[x[x]]$. Hence, the number of integers that meet the requirement is
$$
1+2+\cdots+44=990 \text { (numbers). }
$$
|
990
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $x_{1}=x_{2011}=1$, $\left|x_{n+1}\right|=\left|x_{n}+1\right|(n=1,2, \cdots, 2010)$.
Then $x_{1}+x_{2}+\cdots+x_{2010}=$ $\qquad$
|
3. -1005 .
From the given, it is easy to obtain
$$
x_{n+1}^{2}=x_{n}^{2}+2 x_{n}+1(n=1,2, \cdots, 2010) \text {. }
$$
By summing up and organizing these 2010 equations, we get
$$
\begin{array}{l}
x_{2011}^{2}=2\left(x_{1}+x_{2}+\cdots+x_{2010}\right)+2011 . \\
\text { Therefore, } x_{1}+x_{2}+\cdots+x_{2010}=-1005 .
\end{array}
$$
|
-1005
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Among the positive integers not greater than 100, the ordered integer pairs $(m, n)$ that satisfy
$$
\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n}
$$
are $\qquad$ pairs.
|
4. 170 .
Notice that $\sqrt{2} n-1<m<\sqrt{2}(n+1)$.
For each $n$, the number of $m$ is given by
$$
\begin{array}{l}
{[\sqrt{2}(n+1)]-[\sqrt{2} n-1]} \\
=[\sqrt{2}(n+1)]-[\sqrt{2} n]+1
\end{array}
$$
Since $100<71 \sqrt{2}<101<72 \sqrt{2}$, we have $n \leqslant 71$.
But when $n=71$, $m=100$.
Therefore, the number of $m$ is
$$
\begin{array}{l}
\sum_{n=1}^{70}([\sqrt{2}(n+1)]-[\sqrt{2} n]+1)+1 \\
=[71 \sqrt{2}]-[\sqrt{2}]+71 \\
=100-1+71=170 .
\end{array}
$$
|
170
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) When $1<x<5$, find the minimum value of the algebraic expression
$$
\frac{\sqrt{(7 x+1)(5-x)(x-1)(7 x+5)}}{x^{2}}
$$
|
$$
\begin{array}{l}
\frac{\sqrt{(7 x+1)(5-x)(x-1)(7 x+5)}}{x^{2}} \\
=\sqrt{\frac{\left(7 x^{2}-2 x-5\right)\left(-7 x^{2}+34 x+5\right)}{x^{4}}} \\
=\sqrt{\left(7 x-\frac{5}{x}-2\right)\left(-7 x+\frac{5}{x}+34\right)} \text {. } \\
\end{array}
$$
Let $t=7 x-\frac{5}{x}$. Then
$$
\begin{array}{l}
\text { Equation (1) }=\sqrt{(t-2)(34-t)} \\
=\sqrt{-t^{2}+36 t-68} \\
=\sqrt{-(t-18)^{2}+256} \leqslant 16 .
\end{array}
$$
When $t=7 x-\frac{5}{x}=18$, i.e., $x=\frac{9+2 \sqrt{29}}{7}$,
the expression reaches its maximum value of 16.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the equation in $x$
$$
x^{2}+(a-2010) x+a=0 \quad (a \neq 0)
$$
has two integer roots. Then the value of the real number $a$ is $\qquad$.
|
3.4024.
Let the roots of the equation be $x_{1} 、 x_{2}\left(x_{1} \leqslant x_{2}\right)$.
By Vieta's formulas, we have
$$
x_{1}+x_{2}=-(a-2010), x_{1} x_{2}=a \text {. }
$$
Then $x_{1} x_{2}+x_{1}+x_{2}=2010$, which means
$$
\left(x_{1}+1\right)\left(x_{2}+1\right)=2011 \text {. }
$$
Since 2011 is a prime number, we have
$$
\left\{\begin{array} { l }
{ x _ { 1 } = 0 , } \\
{ x _ { 2 } = 2 0 1 0 }
\end{array} \text { or } \left\{\begin{array}{l}
x_{1}=-2012, \\
x_{2}=-2 .
\end{array}\right.\right.
$$
Thus, $a=0$ (discard) or $a=4024$.
|
4024
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) In a certain drill activity, the drill leader arranged $n$ students, numbered $1 \sim n (n>3)$, in a circular formation, and they performed a $1-2-3$ cyclic counting. The drill leader recorded the numbers of the students who reported, and required students who reported 1 and 2 to leave the formation, while students who reported 3 remained in the formation, and their numbers were changed to the sum of the numbers of the three students in this cycle of counting. The counting continued in this manner. When the number of students remaining on the playground was no more than two, the counting activity ended. The drill leader recorded the numbers of the students who remained on the playground (for example, if nine students numbered $1 \sim 9$ were arranged in a circular formation, after the counting ended, only the student originally numbered 9 remained on the playground, at which point his number was 45, and the data recorded by the drill leader were $1,2,3,4,5,6,7,8,9,6,15,24,45$). It is known that 2011 students participated in the drill.
(1) What was the initial number of the student who remained on the playground?
(2) Find the sum of the numbers recorded by the drill leader.
|
Three, the numbers recorded by the leader are $a_{1}$, $a_{2}, \cdots, a_{n}, a_{n+1}, \cdots, a_{m}$, with their sum being $S_{n}$. Thus,
(1) After each 1-2-3 counting cycle, the total number of students decreases by 2, but the sum of the numbers remains unchanged.
(2) If $n=3^{k}$, after $3^{k-1}$ 1-2-3 counting cycles, $3^{k-1}$ new numbers are generated, and the sum of these numbers is $\frac{(1+n) n}{2}$; after another $3^{k-2}$ counting cycles, $3^{k-2}$ new numbers are generated, and the sum of these numbers is also $\frac{(1+n) n}{2}; \cdots \cdots$ After a total of
$$
3^{k-1}+3^{k-2}+\cdots+3^{1}+1=\frac{3^{k}-1}{2}
$$
counting cycles, only one student remains, and his number is $\frac{(1+n) n}{2}$, and his original number is $n$ (i.e., when $n=3^{k}$, the last student remains on the field).
From (1), we know $S_{n}=(k+1) \frac{(1+n) n}{2}$.
(3) If the number of students is $3 k$, and the number of the last student is $a_{p}$, then after $k$ counting cycles, $k$ new numbers are generated, and the number of students becomes $k$, with the first student's number being $a_{p+1}$.
For a general odd number $n(n>3)$, let $k$ and $r$ satisfy $n=3^{k}+2 r\left(0 \leqslant r<3^{k}\right)$.
Then $3^{k}<n<3^{k+1}, 3 r<n$. One group, the sum of their numbers is
$$
M=1+2+\cdots+3 r=\frac{3 r(3 r+1)}{2} .
$$
Let the initial sum of student numbers be $N=\frac{n(n+1)}{2}$.
From (1) and (3), we know that after $r$ 1-2-3 counting cycles, the number of students becomes $3^{k}$, with the first student's number being $a_{3 r+1}$, called the second group; after another $3^{k-1}$ counting cycles, the number of students becomes $3^{k-1}$, called the third group; repeat the counting cycles until the number of students is 1, which is the $k$ +2 group.
The original number of the last student remaining on the field is $3 r$, and
$$
\begin{array}{l}
S_{n}=M+(k+1) N \\
=\frac{3 r(3 r+1)}{2}+(k+1) \frac{(1+n) n}{2} .
\end{array}
$$
When $n=2011$, since $n=2011=3^{6}+2 \times 641$, we have $k=6, r=641$.
The original number of the last student remaining on the field is 1923. Therefore, $S_{2011}=16011388$.
|
16011388
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given $a, b, c > 0, a^{2}+b^{2}+c^{2}=14$. Prove: $a^{5}+\frac{b^{5}}{8}+\frac{c^{5}}{27} \geqslant 14$.
|
Prove the construction of the $3 \times 5$ matrix
$$
\left(\begin{array}{ccccc}
a^{5} & a^{5} & 1 & 1 & 1 \\
\frac{b^{5}}{8} & \frac{b^{5}}{8} & 4 & 4 & 4 \\
\frac{c^{5}}{27} & \frac{c^{5}}{27} & 9 & 9 & 9
\end{array}\right) .
$$
Using Carleman's inequality, we get
$$
\begin{array}{l}
{\left[\left(a^{5}+\frac{b^{5}}{8}+\frac{c^{5}}{27}\right)^{2}(1+4+9)^{3}\right]^{\frac{1}{5}}} \\
\geqslant\left(a^{5} \cdot a^{5} \cdot 1^{3}\right)^{\frac{1}{5}}+\left(\frac{b^{5}}{8} \cdot \frac{b^{5}}{8} \cdot 4^{3}\right)^{\frac{1}{5}}+\left(\frac{c^{5}}{27} \cdot \frac{c^{5}}{27} \cdot 9^{3}\right)^{\frac{1}{5}} .
\end{array}
$$
Since $a^{2}+b^{2}+c^{2}=14$, we have
$$
a^{5}+\frac{b^{5}}{8}+\frac{c^{5}}{27} \geqslant\left[\frac{\left(a^{2}+b^{2}+c^{2}\right)^{5}}{14^{3}}\right]^{\frac{1}{2}}=14 .
$$
|
14
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. If the product of the first 2011 positive integers
$$
1 \times 2 \times \cdots \times 2011
$$
can be divided by $2010^{k}$, then the maximum value of the positive integer $k$ is
|
4.30.
$$
2010=2 \times 3 \times 5 \times 67 \text {. }
$$
In $1 \times 2 \times \cdots \times 2011$, the exponent of 67 is
$$
\left[\frac{2011}{67}\right]+\left[\frac{2011}{67^{2}}\right]+\cdots=30 \text {. }
$$
Obviously, the exponents of $2,3,5$ in $1 \times 2 \times \cdots \times 2011$ are all greater than 30.
Therefore, the maximum value of $k$ is 30.
|
30
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.