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Example 1 Find the positive integer root of the equation
$$
\frac{2}{n}+\frac{3}{n+1}+\frac{4}{n+2}=\frac{133}{60}
$$
(1990, Shanghai Junior High School Mathematics Competition)
|
Solution: Since $n$ is a natural number, by the properties of natural numbers and
$$
\begin{array}{l}
\frac{1}{n}>\frac{1}{n+1}>\frac{1}{n+2} \\
\therefore \frac{2+3+4}{n+2}<\frac{2}{n}+\frac{3}{n+1}+\frac{4}{n+2} \\
<\frac{2+3+4}{n} .
\end{array}
$$
That is, $\frac{9}{n+2}<\frac{133}{60}<\frac{9}{n}$.
Solving this, we get $2 \frac{8}{133}<n<4 \frac{8}{133}$.
Since $n$ is a natural number, $n$ can only be 3 or 4. Upon verification, $n=3$ is the positive integer solution of the equation.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, when arranging student dormitories, it is stipulated that two students live in one room. The school's preference order for various combinations of students from grades $A, B, C, D, E$ is
$$
A A, A B, A C, A D, A E, B B, B C, B D,
$$
$B E, C C, C D, C E, D D, D E, E E$.
That is, two students from grade $A$ living together is better than one student from grade $A$ and one student from grade $B$ living together, and so on. Assign values $a, b, c, d, e$ to each grade, such that $a a, a b, a c, \cdots$, follow the same order as (1). For integers $a, b, c, d, e$, find the minimum value of $a+b+c+d+e$ and the values of $a, b, c, d, e$.
|
Three, Solution: According to the problem, we have
$$
a a>a b>a c>a d>a e>b b>b c>b d>b e>c c>c d
$$
$>c e>d d>d e>e e$, which means
$$
\left\{\begin{array}{l}
a>b>c>d>e, \\
a>\frac{b^{2}}{e}, \\
b>\frac{c^{2}}{e}, \\
c>\frac{d^{2}}{e} .
\end{array}\right.
$$
From (2), (3), and (4), we know that for a given $t$, to make $a+b+c+d+e$ the smallest, $d, c, b, a$ must take the corresponding minimum values, i.e., $d=e+1, c=\left[\frac{d^{2}}{e}\right]+1, b=\left[\frac{c^{2}}{e}\right]+1, a=\left[\frac{b^{2}}{e}\right]+1$ when $a+b+c+d+e$ takes the minimum value.
$$
\begin{array}{l}
c=\left[\frac{d^{2}}{e}\right]+1=\left[\frac{(e+1)^{2}}{e}\right]+1=\left[e+2+\frac{1}{e}\right]+1 \\
=\left\{\begin{array}{l}
e+4=5, e=1 \text { when } \\
e+3 . e>1 \text { when }
\end{array}\right. \\
b=\left[\frac{c^{2}}{e}\right]+1=\left\{\begin{array}{l}
26, \quad e=1, \\
e+7+\left[\frac{9}{e}\right], e>1
\end{array}\right. \\
=\left\{\begin{array}{l}
26, e=1 \\
13,2 \leqslant e \leqslant 5 \\
e+8,6 \leqslant e \leqslant 9 \\
e+7, e \geqslant 10
\end{array}\right. \\
a=\left[\frac{b^{2}}{e}\right]+1=\left\{\begin{array}{l}
677, e=1 \\
{\left[\frac{169}{e}\right]+1,2 \leqslant e \leqslant 5} \\
{\left[\frac{(e+8)^{2}}{e}\right]+1,6 \leqslant e \leqslant 9} \\
{\left[\frac{(e+7)^{2}}{e}\right]+1 . e \geqslant 10}
\end{array}\right. \\
=\left\{\begin{array}{l}
677, e=1 \\
85, e=2 \\
57, e=3 \\
43, e=4 \\
37, e=5 \\
33,6 \leqslant e \leqslant 9 \\
e+9,10 \leqslant e \leqslant 12 \\
\geqslant e+15, e \geqslant 13
\end{array}\right. \\
\end{array}
$$
Therefore, we have $a+b+c+d+e$
$$
=\left\{\begin{array}{l}
711, e=1 \\
108, e=2 \\
83, e=3 \\
72, e=4 \\
69, e=5 \\
69, e=6 \\
4 e+45 \geqslant 73,7 \leqslant e \leqslant 9 \\
\geqslant e+(e+1)+(e+3)+(e+7)+(e+15) \geqslant 76, \\
e \geqslant 10
\end{array}\right.
$$
Therefore, when $a=37, b=13, c=8, d=6, e=5$ or $a=33, b=14, c=9, d=7, e=6$, $a+b+c+d+e$ takes the minimum value of 69.
|
69
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
I. (25 points) The Xiguang Factory glasses workshop has received a batch of tasks, requiring the processing of 6000 Type A parts and 2000 Type B parts. This workshop has 214 workers, and each of them can process 3 Type B parts in the time it takes to process 5 Type A parts. These workers are to be divided into two groups, with both groups working simultaneously, each group processing one type of part. To complete this batch of tasks in the shortest time, how should the workers be divided?
|
One, Solution: Let the number of people in the group that includes those with the surname $A$ be $x$, and in a unit of time, the number of $\mathrm{C} . \mathrm{A}$ parts processed by one person is $5 k$, then the number of the other parts is $3 k$.
The time required to process type $A$ parts is $t_{A}(x)=\frac{6000}{5 k x}$;
The time required to process type $B$ parts is $t_{B}(x)=$ $\frac{2000}{3 k(214-x)}$.
$\therefore$ The total time to complete the entire task is
$$
t(x)=\max \left\{t_{A}(x), t_{B}(x)\right\}=\frac{2000}{k} \cdot f(x),
$$
where, $f(x)=\max \left\{\frac{3}{5 x}, \frac{1}{3(214-x)}\right\}$.
Thus, the problem is converted to finding the natural number $x(1 \leqslant x \leqslant 213)$, such that the function $f(x)$ takes the minimum value.
In the interval $[1,213]$, $\frac{3}{5 x}$ is a decreasing function, and $\frac{1}{3(214-x)}$ is an increasing function, so the minimum value of $f(x)$ is obtained at $x_{0}=137 \frac{4}{7}$, where $x_{0}$ satisfies the equation $\frac{3}{5 x}=\frac{1}{3(214-x)}$.
$\because x_{0}=137 \frac{4}{7}$ is not an integer, and $f(137)<f(138)$,
$\therefore$ The number of people in the group processing type $A$ parts is 137, and the number of people in the other group is 77.
|
137
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (35 points) The real number sequence $a_{1}, a_{2} \cdots, a_{1997}$ satisfies:
$$
\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+\left|a_{1996}-a_{1997}\right|=
$$
1997. If the sequence $\left\{b_{n}\right\}$ satisfies:
$$
b_{k}=\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}(k=1,2, \cdots, 1997),
$$
find the maximum possible value of $\left|b_{1}-b_{2}\right|+\left|b_{2}-b_{3}\right|+\cdots+\left|b_{1996}-b_{1997}\right|$.
|
$$
\begin{array}{l}
\text { III. Solution: From } \left.\left|b_{k}-b_{k+1}\right|=\frac{1}{k(k+1)} \right\rvert\,(k+1)\left(a_{1}+a_{2}\right. \\
\left.+\cdots+a_{k}\right)-k\left(a_{1}+a_{2}+\cdots+a_{k+1}\right) \mid \\
=\frac{1}{k(k+1)}\left|a_{1}+a_{2}+\cdots+a_{k}-k a_{k+1}\right| \\
\left.=\frac{1}{k(k+1)} \right\rvert\,\left(a_{1}-a_{2}\right)+2\left(a_{2}-a_{3}\right)+3\left(a_{3}-a_{4}\right)+ \\
\cdots+=k\left(a_{k}-a_{k+1}\right) \mid .
\end{array}
$$
We have $\left|b_{1}-b_{2}\right|+\left|b_{2}-b_{3}\right|+\cdots+\left|b_{1996}-b_{1997}\right|$
$$
\begin{aligned}
& =\frac{1}{1 \times 2}\left|a_{1}-a_{2}\right|+\frac{1}{2 \times 3}\left|\left(a_{1}-a_{2}\right)+2\left(a_{2}-a_{3}\right)\right| \\
& \quad+\frac{1}{3 \times 4}\left|\left(a_{1}-a_{2}\right)+2\left(a_{2}-a_{3}\right)+3\left(a_{3}-a_{4}\right)\right| \\
& \left.\quad+\cdots+\frac{1}{1996 \times 1997} \right\rvert\,\left(a_{1}-a_{2}\right)+2\left(a_{2}-a_{3}\right) \\
& +\cdots+1996\left(a_{1996}-a_{1997}\right) \mid \\
& \leqslant\left(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\cdots+\frac{1}{1996 \times 1997}\right)\left|a_{1}-a_{2}\right| \\
& +2\left(\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{1996 \times 1997}\right)\left|a_{2}-a_{3}\right| \\
& \quad+\cdots+1996 \frac{1}{1996 \times 1997}\left|a_{1996}-a_{1997}\right| \\
= & \left(1-\frac{1}{1997}\left|a_{1}-a_{2}\right|+\left(1-\frac{2}{1997}\right)\left|a_{2}-a_{3}\right|+\cdots\right. \\
+ & \left(1-\frac{1996}{1997}\right)\left|a_{1996}-a_{1997}\right| \\
= & \left(\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+\left|a_{1996}-a_{1997}\right|\right) \\
& -\frac{1}{1997}\left(\left|a_{1}-a_{2}\right|+2\left|a_{2}-a_{3}\right|+\cdots\right. \\
& \left.+1996\left|a_{1996}-a_{1997}\right|\right)
\end{aligned}
$$
$\leqslant 1997-\frac{1}{1997}\left(\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+\mid a_{1996}\right.$
$-a_{1997} \mid$ )
$$
=1997-1=1996 \text {. }
$$
When $a_{1}=1997, a_{2}=a_{3}=\cdots=a_{1997}=0$, the equality holds in the above expression.
Therefore, the maximum value is 1996.
$$
|
1996
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 The first 100 natural numbers are arranged in a certain order, then the sum of every three consecutive numbers is calculated, resulting in 98 sums, of which the maximum number of sums that can be odd is how many?
(21st Russian Mathematical Olympiad)
|
Proof: First, we prove that it is impossible for all 98 sums to be odd. We use proof by contradiction.
If all the sums are odd, then the arrangement of these 100 natural numbers can only be one of the following four cases:
(1) odd odd odd odd odd odd ...; (2) odd even even odd even even ...;
(3) even odd even even odd even ...; (4) even even odd even even odd ....
(1) indicates that the first 100 natural numbers are all odd, (2) and (4) indicate that the number of odd numbers among the first 100 natural numbers is more than the number of even numbers, which contradicts the fact.
Therefore, it is impossible for all 98 sums to be odd.
Next, we show that 97 of the 98 sums can be odd. We arrange the numbers from 1 to 100 in the following order:
Thus, only the 75th sum is obtained by adding even odd odd, which should be even, while the other 97 sums are all odd.
Therefore, at most 97 of the sums can be odd.
|
97
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Solve the equation
\[
\begin{array}{l}
\sqrt{x^{2}+5 x-14}+\sqrt{x+7}+\sqrt{2-x}+x-5 \\
=0 .
\end{array}
\]
|
Solution: By the properties of quadratic radicals, we get
$$
\left\{\begin{array}{l}
x^{2}+5 x-14 \geqslant 0, \\
x+7 \geqslant 0, \\
2-x \geqslant 0 .
\end{array}\right.
$$
Solving, we get $-7 \leqslant x \leqslant-7$ or $2 \leqslant x \leqslant 2$.
That is, $x=-7$ or $x=2$.
Upon verification, $x=2$ is a root of the original equation.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 For a finite set $A$, function $f: N \rightarrow A$ has only the following property: if $i, j \in N, |H| i-j |$ is a prime number, then $f(i) \neq f(j)$. How many elements does set $A$ have at least?
|
Solution: Let $|A|$ denote the number of elements in the finite set $A$, and estimate the lower bound of $|A|$.
Since the absolute value of the difference between any two numbers among $1, 3, 6, 8$ is a prime number, by the problem's condition: $f(1)$, $f(3)$, $f(6)$, $f(8)$ are four distinct elements in $A$. Therefore, $|A| \geqslant 4$.
Construct the set $A=\{0,1,2,3\}$, then the function $f: N \rightarrow A$ has the following correspondence:
. If $x \in N, x=4 k+r$, then $f(x)=r$. That is, $k \in N \bigcup\{0\}, r=0,1,2,3$.
Next, we prove that $f$ satisfies the conditions of the problem:
For any $x, y \in N$, if $|x-y|$ is a prime number, then $f(x) \neq f(y)$. Otherwise, if $f(x)=f(y)$, then
$x \equiv y(\bmod 4)$.
Thus, $4 \mid |x-y|$, which contradicts the fact that $|x-y|$ is a prime number.
From the above, we conclude that $|A|_{\text {min }}=4$.
The proof of the inequality involves various techniques, such as constructing counterexamples, constructing sequences, constructing propositions, etc. Please list them.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, let $A=\{1,2,3, \cdots, 17\}$. For any function $f: A \rightarrow A$, denote
$$
f^{[1]}(x)=f(x), f^{[k+1]}(x)=f\left(f^{[k]}(x)\right)
$$
$(k \in \mathbb{N})$. Find the natural number $M$ such that:
$$
\begin{array}{l}
\text { (1) When } m<M, 1 \leqslant i \leqslant 16, \text { we have } \\
f^{[m]}(i+1)-f^{[m]}(i) \neq \pm 1(\bmod 17), \\
f^{[m]}(1)-f^{[m]}(17) \neq \pm 1(\bmod 17) ;
\end{array}
$$
(2) When $1 \leqslant i \leqslant 16$, we have
$$
\begin{array}{l}
f^{[M]}(i+1)-f^{[M]}(i) \\
\quad \equiv 1 \text { or }-1(\bmod 17), \\
\quad f^{[M]}(1)-f^{[M]}(17) \equiv 1 \text { or }-1(\bmod 17) .
\end{array}
$$
Determine the maximum possible value of $M$ for all functions $f$ that satisfy the above conditions, and prove your conclusion.
|
Five, the required $M_{0}=8$.
First, prove $M_{0} \geqslant 8$.
In fact, we can define the mapping $f(i) \equiv 3 i-2(\bmod 17)$, where $i \in A, f(i) \in A$.
If $f(i) \equiv f(j)(\bmod 17)$,
then $3 i-2 \equiv 3 j-2(\bmod 17)$,
we have $i \equiv j(\bmod 17)$,
$$
\therefore i=j \text {. }
$$
The mapping $f$ is a mapping from $A$ to $A$.
From the definition of the mapping $f$, we know
$$
\begin{array}{l}
f^{[n]}(i) \equiv 3^{n} \cdot i-3^{n}+1(\bmod 17) \text {. } \\
\text { If }\left\{\begin{array}{l}
f^{[M]}(i+1)-f^{[M]}(i) \equiv 1 \text { or }-1(\bmod 17), \\
f^{[M]}(1)-f^{[M]}(17) \equiv 1 \text { or }-1(\bmod 17),
\end{array}\right. \\
\text { then }\left\{\begin{array}{l}
{\left[3^{M}(i+1)-3^{M}+1\right]-\left[3^{M} \cdot i-3^{M}+1\right]} \\
\equiv+1 \text { or }-1(\bmod 17), \\
1 \quad\left[3^{M} \times 17-3^{M}+1\right]=+1 \text { or }-1(\bmod 17) .
\end{array}\right. \\
\end{array}
$$
Then $3^{M}=+1$ or $-1(\bmod 17)$
$$
\begin{aligned}
\text { Since } 3^{1} \equiv 3,3^{2} \equiv 9,3^{3} \equiv 10,3^{4}=13,3^{5}=5,3^{6} \equiv 15, \\
\left.3^{7} \equiv 11,3^{4} \equiv-1 \text { (mod } 17\right),
\end{aligned}
$$
Thus $M_{0} \geqslant 8$.
Proof: $M_{0} \leqslant 8$.
Consider a 17-sided polygon $A_{1} A_{2} \cdots A_{17}$, denoted as $G$. We first define:
When $i+1=18$, take $i+1$ as 1; when $i-1=0$, take $i-1$ as 17. Then, according to the rule, connect the segments: if $i \leqslant mp>q>0$, such that
$$
\begin{array}{l}
f_{[f]}^{[f]}(i)=f^{[\varphi]}(j), \\
f_{[f]}^{[f]}(i+1)=f^{[q]}(j+1) \\
f^{[p]}(i+1)=f^{[q]}(j-1) .
\end{array}
$$
Thus $f^{[\rho]}(i+1)=f^{[q]}(j-1)$.
From this, we have
$$
\left\{\begin{array}{l}
f^{[p-q]}(i)=j, \\
f^{[p-q]}(i+1)=j+1 \text { or } f^{[p-q]}(i+1)=j-1,
\end{array}\right.
$$
Therefore, the connected segments do not overlap.
Since $G$ has $17 \times 7$ diagonals.
Thus, $17 \times\left(M_{0}, 1\right) \leqslant 17 \times 7$.
Thus $M=8$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7. Solve the equation
$$
\begin{array}{l}
\sqrt{x-1}+\sqrt{2 x-3}+\sqrt{3 x-5} \\
+\sqrt{4 x-7}=5 x-6 .
\end{array}
$$
(First Yangtze Cup Correspondence Competition for Junior High School Students)
|
Solution: By applying the method of completing the square, the original equation can be transformed into
$$
\begin{array}{l}
(\sqrt{x-1}-1)^{2}+(\sqrt{2 x-3}-1)^{2} \\
+(\sqrt{3 x-5}-1)^{2}+(\sqrt{4 x-7}-1)^{2}=0 . \\
\therefore \sqrt{x-1}-1=\sqrt{2 x-3}-1 \\
\quad=\sqrt{3 x-5}-1=\sqrt{4 x-7}-1=0 .
\end{array}
$$
Solving this, we get $x=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
24. (15 points) Given the parabola $y^{2}=\operatorname{tar}(0<a<1)$ with focus $\vec{F}$, a semicircle is constructed above the $x$-axis with center $A(a+4,0)$ and radius $|A F|$, intersecting the parabola at two distinct points $M$ and $N$. Let $P$ be the midpoint of segment $M N$.
(1) Find the value of $|M F|+|N F|$;
(2) Does there exist a value of $a$ such that $|M F|$, $|P F|$, $|N F|$ form an arithmetic sequence? If so, find the value of $a$; if not, explain the reason.
|
24. (1) From the given, we have $F(a, 0)$, and the semicircle is
$$
[x-(a+4)]^{2}+y^{2}=16(y \geqslant 0) \text {. }
$$
Substituting $y^{2}=4 a x$ into the equation, we get
$$
x^{2}-2(4-a) x+a^{2}+8 a=0 \text {. }
$$
Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$. Then, by the definition of the parabola, we have
$$
\begin{aligned}
|M F|+|N F| & =x_{1}+x_{2}+2 a \\
& =2(4-a)+2 a=8 .
\end{aligned}
$$
(2) If $|M F|,|P F|,|N F|$ form an arithmetic sequence, then
$$
2|P F|=|M F|+|N F| \text {. }
$$
On the other hand, let the projections of $M, P, N$ on the directrix of the parabola be $M^{\prime}, P^{\prime}, N^{\prime}$. Then, in the right trapezoid $M^{\prime} M N N^{\prime}$, $P^{\prime} P$ is the midline, and we have
$$
\begin{aligned}
2\left|P^{\prime} P\right| & =\left|M^{\prime} M\right|+\left|N^{\prime} N\right| \\
& =|F M|+|F N|,
\end{aligned}
$$
thus $|P F|=\left|P^{\prime} P\right|$. This indicates that point $P$ should be on the parabola.
However, it is known that $P$ is the midpoint of segment $M N$, i.e., $P$ is not on the parabola.
$\therefore$ There does not exist a value of $a$ such that $|M F|,|P F|,|N F|$ form an arithmetic sequence.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (15 points) The function $f(k)$ is defined on $N$ and takes values in $N$, being a strictly increasing function (if for any $x_{1}, x_{2} \in A$, when $x_{1}<x_{2}$, we have $f\left(x_{1}\right)<f\left(x_{2}\right)$, then $f(x)$ is called a strictly increasing function on $A$), and satisfies the condition $f(f(k))=3k$. Try to find the value of $f(1)+f(9)+f(96)$.
|
For $k \in N, f(f(k))=3_{i}^{2}$.
(1)
$\therefore f[f(f(\hat{i})]:=\hat{j}(3 k)$
Also, $f[f(f(i))]=3 f(k)$,
$\therefore f(3 k)=3 f(k)$.
If $f(1)=1$, substituting into (1) gives $f(1)=3$, a contradiction.
$\therefore f(1)=a>1$, but $f(f(1))=f(a)=3$.
By the strict monotonicity of $f(k)$, i.e., $1<a \Rightarrow f(1)<f(a)=3$,
$\therefore$ it can only be that $f(1)=2, f(2)=3$.
At this point, $f(3)=3 f(1)=6$,
$$
\begin{array}{l}
f(6)=f(3 \times 2)=3 f(2)=9, \\
f(9)=3 f(3)=18, \\
f(18)=3 f(6)=27, \\
f(27)=3 f(9)=54, \\
f(54)=3 f(18)=81 .
\end{array}
$$
Note, as the variable changes from 27 to 54, it increases by 27 numbers, and the function value correspondingly increases from 54 to 81, also by 27 numbers. By the strict monotonicity of $f$,
$$
\begin{array}{l}
f(28)=55, f(29)=56, f(30)=57, \\
f(31)=58, f(32)=59, \cdots .
\end{array}
$$
And $f(96)=f(3 \times 32)=3 f(32)=3 \times 59=177$.
$$
\therefore f(1)+f(9)+f(96)=2+18+177=197 \text {. }
$$
|
197
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Does there exist a six-digit number $A$, such that the last six digits of any number in $A, 2A, 3A, \cdots, 500000A$ are not all the same?
|
5. Let $A$ be a six-digit number. If $A$ is even, then $500000A$ ends with six zeros; if $A$ is divisible by 5, then $200000A$ ends with six zeros. Therefore, we can assume that the last digit of $A$ is $1, 3, 5, 7,$ or $9$. For any odd digit $d$, there exists a one-digit number $b$ such that the last digit of $Ab$ is $d$.
Choose $b_{0}$ such that the last digit of $Ab_{0}$ is 1. Let the last two digits of $Ab_{0}$ be $c_{1}1$. Choose a one-digit number $d_{1}$ such that $d_{1} + c_{1} = 1$ or 11. There exists $b_{1}$ such that the last digit of $Ab_{1}$ is $d_{1}$, so the last two digits of $A(10b_{1} + b_{0})$ are 11. Let the last three digits of $A(10b_{1} + b_{0})$ be $c_{2}11$. Choose a one-digit number $d_{2}$ such that $d_{2} + c_{2} = 1$ or 11. There exists $b_{2}$ such that the last digit of $Ab_{2}$ is $d_{2}$, so the last three digits of $A \times (100b_{2} + 10b_{1} + b_{0})$ are 111. By continuing this process, there exists $B_{1} = 10^{5}b_{5} + 10^{4}b_{4} + 10^{3}b_{3} + 10^{2}b_{2} + 10b_{1} + b_{0}$ such that the last six digits of $AB_{1}$ are all 1.
For $2 \leqslant k \leqslant 9$, if $B_{k}$ is the last six digits of $kB_{1}$, then the last six digits of $AB_{k}$ are all $k$. If $A = 999999$, then $B_{1} = 888889$. Let $A = 888889$, then for $1 \leqslant k \leqslant 9, B_{k} = 1000000 - k > 500000$, so 888889 satisfies the problem's requirements.
|
888889
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 Find $A^{2}$, where $A$ is the sum of the absolute values of all roots of the following equation:
$$
x=\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}
$$
(9th American Invitational Mathematics Examination)
|
Solution: According to the structural characteristics of the original equation, set
$$
\begin{array}{ll}
x=\sqrt{19}+\frac{91}{y}, & y=\sqrt{19}+\frac{91}{z}, \\
z=\sqrt{19}+\frac{91}{u}, & u=\sqrt{19}+\frac{91}{v}, \\
v=\sqrt{19}+\frac{91}{x} . &
\end{array}
$$
Assume $x>y$ and $x<y$ respectively, then
$$
\begin{array}{l}
x>y \Rightarrow y>z \Rightarrow z>u \Rightarrow u>v \Rightarrow v>x, \\
x<y \Rightarrow y<z \Rightarrow z<u \Rightarrow u<v \Rightarrow v<y .
\end{array}
$$
Both conclusions contradict the assumptions. Therefore, only $x=y$. Substituting
$$
x=y \text { into } x=\sqrt{19}+\frac{91}{y} \text {, we get } x=\sqrt{19}+\frac{91}{x} \text {. }
$$
Solving, we get $x_{1}=\frac{\sqrt{19}+\sqrt{383}}{2}, x_{2}=\frac{\sqrt{19}-\sqrt{383}}{2}$.
$$
\begin{aligned}
\therefore A^{2}= & \left(\left|\frac{\sqrt{19}+\sqrt{383}}{2}\right|\right. \\
& \left.+\left|\frac{\sqrt{19}-\sqrt{383}}{2}\right|\right)^{2} \\
= & (\sqrt{383})^{2}=383 .
\end{aligned}
$$
|
383
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 If the real numbers $x, y, m$ satisfy the relation
$$
\begin{array}{l}
\sqrt{3 x+5 y-2-m}+\sqrt{2 x+3 y-m} \\
=\sqrt{x-199+y} \cdot \sqrt{199-x-y},
\end{array}
$$
try to determine the value of $m$.
(1994, Beijing Junior High School Mathematics Competition)
|
Analysis: An equation with one unknown cannot be solved by brute force, but only by intelligent methods. Observing the characteristics of the equation, all are square roots. By definition, the right side of the equation has
$$
\left\{\begin{array}{l}
x-199+y \geqslant 0, \\
199-x-y \geqslant 0,
\end{array}\right.
$$
which means
$$
\left\{\begin{array}{l}
x+y \geqslant 199, \\
x+y \leqslant 199 .
\end{array}\right.
$$
$$
\therefore x+y=199 \text {. }
$$
At this point, the equation simplifies to
$$
\sqrt{595-m+2 y}+\sqrt{398-m+y}=0 .
$$
Observing this equation again, the left side is the sum of two non-negative numbers, so we have
$$
\left\{\begin{array}{l}
595-m+2 y=0, \\
398-m+y=0 .
\end{array}\right.
$$
Eliminating $y$, we get $m=201$. Then we find $x=2$,
$$
y=197 . \quad \text {. }
$$
|
201
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 As shown in Figure 2, the areas of $\triangle A_{1} B_{2} B_{3}$, $\triangle P B_{3} A_{2}$, and $\triangle P A_{3} B_{2}$ are $5, 8, 3$ respectively. Find the area of $\triangle A_{1} A_{2} A_{3}$. (Letters have been changed, the same below)
(1994, Japan Mathematical Olympiad Preliminary Problem)
|
$$
\begin{array}{c}
\frac{\lambda_{1}}{\mu_{1}}=\frac{\lambda_{0}}{v_{1}}=\frac{S_{\triangle A_{1} A_{2} A_{3}}}{S_{\triangle A_{1} B_{2} B_{3}}} \\
=\frac{\lambda_{1}+\mu_{1}+5+8+3}{5} . \\
\text { Also } \frac{8}{\lambda_{1}}=\frac{\mu_{1}}{3}, \\
\therefore \lambda_{1}=12, \mu_{1}=2, \\
S_{\triangle A_{1} A_{2} A_{3}}=30 .
\end{array}
$$
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 In the border desert area, patrol vehicles travel 200 kilometers per day, and each vehicle can carry enough gasoline to travel for 14 days. There are 5 patrol vehicles that set out from base $A$ simultaneously, complete their tasks, and then return along the same route to the base. To allow 3 of them to patrol as far as possible and then return together, vehicles Jia and Yi travel to point $B$ along the way, leave behind the gasoline necessary for their own return to the base, and give the remaining gasoline to the other 3 vehicles to use. How far can these 3 vehicles travel in kilometers?
(1995, Hebei Province Junior High School Mathematics Competition)
|
Analysis: The key is to list the relationship formula for the distance traveled by the car, and then solve it by analyzing the characteristics of the formula.
Let 5 cars travel to point $B$ in $x$ days, and 3 cars continue to travel for $y$ days, with the distance $s=200(x+y)$, which is to find the maximum value of $s$. According to the problem, we have
$$
2[3(x+y)+2 x]=14 \times 5,
$$
which simplifies to $5 x+3 y=35$, and $x>0, y>0$.
Thus, $14 \times 5-(5+2) x \leqslant 14 \times 3, x \geqslant 4$, then $y \leqslant 5$.
Therefore, the maximum value of $s$ is obtained when $x=4, y=5$, which is $s=200 \times(4+5)=1800$ kilometers.
|
1800
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given that $x, y, z$ are two non-negative rational numbers, and satisfy $3x+2y+z=5, x+y-z=2$. If $S=2x+y-z$, then what is the sum of the maximum and minimum values of $S$?
(1996, Tianjin City Junior High School Mathematics Competition)
|
Analysis: By representing the algebraic expression $S$ with one letter and determining the range of this letter, we can find the maximum or minimum value of $S$.
From the given, we solve $y=\frac{7-4 x}{3}, z=\frac{1-x}{3}$, and.
$$
\left\{\begin{array}{l}
\frac{7-4 x}{3} \geqslant 0, \\
\frac{1-x}{3} \geqslant 0, \\
x \geqslant 0.
\end{array} \text { Thus, } \left\{\begin{array}{l}
x \leqslant \frac{7}{4}, \\
x \leqslant 1, \\
x \geqslant 0,
\end{array}\right.\right.
$$
That is, $0 \leqslant x \leqslant 1$.
And $S=2 x+\frac{7-4 x}{3}-\frac{1-x}{3}=x+2$, obviously 2
$\leqslant S \leqslant 3$. Therefore, the sum of the maximum and minimum values of $S$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given that $n$ is a positive integer, and $n^{2}-71$ is divisible by $7 n+55$. Try to find the value of $n$.
Translating the text into English while preserving the original formatting and line breaks, the result is as follows:
Example 7 Given that $n$ is a positive integer, and $n^{2}-71$ is divisible by $7 n+55$. Try to find the value of $n$.
|
Analysis: Let $\frac{n^{2}-71}{7 n+55}=k$ ( $k$ is an integer), then $n^{2}$ $-7 k n-(55 k+71)=0$, and $\Delta=49 k^{2}+4$ (5jk $+71)=49 k^{2}+220 k+284$ should be a perfect square.
$$
\begin{array}{l}
\text { and }(7 k+15)^{?}=49 k^{2}+210 k+225 \\
<49 k^{2}+220 k+284 \\
<49 k^{2}+238 k+289 \\
=(7 k+17)^{2}, \\
\therefore \Delta=(7 k+16)^{2}=49 k^{2}+220 k+284 .
\end{array}
$$
Thus $k=7$. Therefore, $n^{2}-49 n-456=0$.
Solving gives $n=-8$ or 57. Therefore, $n=57$.
|
57
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Given that $a$, $b$, and $c$ are all positive integers, and the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
$(1996$, National Junior High School Mathematics League)
|
$$
\begin{array}{l}
\text{Analysis: Let } x_{1} \text{ and } x_{2} \text{ be the x-coordinates of the points where the parabola intersects the x-axis, then } x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a}. \text{ It is easy to know that } -\frac{b}{a}<0, \text{ and } -116 c(a-b+c). \\
\because a-b+c>0, c \geqslant 1, \\
\therefore a^{2}>16, a>4 .
\end{array}
$$
$>20 c$. Thus, the minimum value of $b$ is $5$, the minimum value of $c$ is $1$, and the parabola is $y=5 x^{2}+5 x+1$, intersecting the x-axis at $\left(\frac{-5+\sqrt{5}}{10}, 0\right),\left(\frac{-5-\sqrt{5}}{10}, 0\right)$, which meets the requirements.
Therefore, $(a+b+c)_{\text{min}}=11$.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Arrange the natural numbers $1,2,3, \cdots, 999$. 1.5 into the number $N=1234 \cdots 998999$. Then, the sum of the digits of $N$ is $\qquad$
|
II. 1.13500.
Since $1.2, \cdots .999$ contains 999 numbers. |1. $1+998=2+997$ 999 , for sure, hence the sum of each pair is 27. There are 500 pairs. Therefore, the sum of the digits of $N$ is $500 \times 27=13500$.
|
13500
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Natural numbers $m, n$ satisfy $8 m+10 n>9 m n$. Then
$$
m^{2}+n^{2}-m^{2} n^{2}+m^{4}+n^{4}-m^{4} n^{4}=
$$
$\qquad$
|
3. 2 .
It is known that $\frac{8}{n}+\frac{10}{m}>9$. It is easy to see that one of the two natural numbers $m, n$ must be 1 (otherwise $\frac{8}{m}+\frac{10}{m} \leqslant \frac{8}{2}+\frac{10}{2}=9$ which is a contradiction). Without loss of generality, let $m=1$, then
$$
\begin{array}{l}
m^{2}+n^{2}-m^{2} n^{2}+m^{4}+n^{4}-m^{4} n^{4} \\
=1+n^{2}-n^{2}+1+n^{4}-n^{4}=2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $(-1)^{n} 2^{n} \equiv b_{n}(\bmod 9), 0 \leq b_{n} \leq 8$. Then the range of values for $b_{n}$ is \{ $\qquad$ \}.
|
$\begin{array}{l}\text { II.1.1. } \\ \text { \| }(-1)^{n} 2^{3 n}=(-8)^{n}=(1-9)^{n} \\ \quad=C_{n}^{n}-C_{n}^{1} 9+C_{n}^{2} 9^{2}+\cdots+(-1)^{n} C_{n}^{n} 9^{n}, \\ \text { know }(-1)^{n} 2^{3 n} \equiv 1(\bmod 9) .\end{array}$
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
The struggle of the land, by 1996, the forest coverage rate of the whole county had reached 30% (becoming an oasis). From 1997 onwards, each year will see such changes: 16% of the original desert area will be converted, and 4% of it will be eroded, turning back into desert.
(1) Assuming the total area of the county is 1, in 1996 the oasis area is $a_{1}=\frac{3}{10}$, after $n$ years the oasis area becomes $a_{n+1}$. Prove:
$$
a_{n+1}=\frac{4}{5} a_{n}+\frac{4}{25} \text {. }
$$
(2) How many years at least are needed to make the forest coverage rate of the whole county exceed 60% (round to the nearest whole number)?
|
Let the accumulated area be $b_{n+1}$. Then
$$
a_{1}+b_{1}=1, \quad a_{n}+b_{n}=1 .
$$
According to the problem, $a_{n+1}$ consists of two parts: one part is the remaining area of the original oasis $a_{n}$ after being eroded by $\frac{4}{100} \cdot a_{n}$, which is
$$
a_{n}-\frac{4}{100} a_{n}=\frac{96}{100} a_{n} .
$$
The other part is the newly greened area $\frac{16}{100} b_{n}$. Therefore,
$$
\begin{aligned}
a_{n+1} & =\frac{96}{100} a_{n}+\frac{16}{100} b_{n}=\frac{96}{100} a_{n}+\frac{16}{100}\left(1-a_{n}\right) \\
& =\frac{4}{5} a_{n}+\frac{4}{25} .
\end{aligned}
$$
(2) From (1), we have
$$
\begin{aligned}
a_{n+1}-\frac{4}{5} & =\frac{4}{5}\left(a_{n}-\frac{4}{5}\right)=\left(\frac{4}{5}\right)^{2}\left(a_{n-1}-\frac{4}{5}\right) \\
& =\cdots=\left(\frac{4}{5}\right)^{n}\left(a_{1}-\frac{4}{5}\right) \\
& =-\frac{1}{2}\left(\frac{4}{5}\right)^{n} . \\
\therefore a_{n+1} & =\frac{4}{5}-\frac{1}{2}\left(\frac{4}{5}\right)^{n} .
\end{aligned}
$$
According to the problem, $a_{n+1} \geqslant 60 \%$, i.e.,
$$
\frac{4}{5}-\frac{1}{2}\left(\frac{4}{5}\right)^{n}>\frac{3}{5} \text {. }
$$
This simplifies to $\left(\frac{4}{5}\right)^{n}<\frac{2}{5}$,
or $(1-0.2)^{n}<0.4$.
To find the integer value, we use an upper approximation for $(1-0.2)^{n}$
$$
(1-0.2)^{n}<1-C_{n}^{1} \times 0.2+C_{n}^{2} \times 0.04 \leqslant 0.4,
$$
which simplifies to $n^{2}-11 n+30 \leqslant 0$.
Solving this, we get $5 \leqslant n \leqslant 6$.
Taking the smallest integer $n=5$, we verify that $a_{5}<0.6<a_{6}$, so after 5 years of effort, the greened area can exceed $60 \%$.
Here we used an approximation. We can also solve $a_{n} \leqslant 0.6<a_{n+1}$ to get $n=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, color $a$ points on a line red, $b$ points only yellow, and $c$ points blue, such that no two adjacent points have the same color. The number of such coloring methods is denoted as $F(a, b, c)$. Obviously, we have:
(1) Prove: For natural numbers $a, b, c$ not all equal to 1, we have
$$
\begin{array}{l}
F(a, b, c) \\
=F(a-1, b-1, c)+F(a-1, b, c-1) \\
+F(a, b-1, c-1)+2 F(a-1, b-1, \\
c-1) .
\end{array}
$$
(2) Using the above recursive relation, solve the following:
On a bookshelf, there are 4 different mathematics books, 5 different physics books, and 3 different chemistry books, all tightly arranged in a row. If it is required that books of the same type are not adjacent, how many different arrangements are there?
|
The number of ways to color the 1st point yellow and blue.
$$
F(a, b, c)=\sum_{i=1}^{3} F_{i}(a, b, c) .
$$
And
$$
\begin{aligned}
& F_{1}(a, b, c)=F_{2}(a-1, b, c)+F_{3}(a-1, b, c) \\
= & F_{1}(a-1, b-1, c)+F_{3}(a-1, b-1, c) \\
& +F_{1}(a-1, b, c-1)+F_{2}(a-1, b, c-1) \\
= & F_{1}(a-1, b-1, c)+F_{3}(a-1, b-1, c) \\
& +F_{1}(a-1, b, c-1)+F_{1}(a-1, b-1, c-1) \\
& +F_{3}(a-1, b-1, c-1) .
\end{aligned}
$$
For $F_{2}\left(a, b, c^{\prime}\right)$
$$
\begin{array}{l}
= F_{2}(a-1, b-1, c)+F_{1}(a, b-1, c-1) \\
+F_{2}(a, b-1, c-1)+F_{1}(a-1, b-1, c-1) \\
+F_{2}(a-1, b-1, c-1) . \\
F_{3}(a, b, c) \\
= F_{2}(a-1, b, c-1)+F_{3}(a-1, b, c-1) \\
+F_{3}(a, b-1, c-1)+F_{2}(a-1, b-1, c-1) \\
+F_{3}(a-1, b-1, c-1) .
\end{array}
$$
1
Substituting (2), (3), and (4) into (1) yields
$$
\begin{aligned}
F(a, b, c)= & \sum_{i=1}^{3} F_{i}(a-1, b-1, c) \\
& +\sum_{i=1}^{3} F_{i}(a-1, b, c-1) \\
& +\sum_{i=1}^{3} F_{i}(a, b-1, c-1) \\
& \quad+2 \sum_{i=1}^{3} F_{i}(a-1, b-1, c-1) \\
& F(a-1, b-1, c)+F(a-1, b, c-1) \\
& \quad+F(a, b-1, c-1)+2 F(a-1, b-1, c-1) .
\end{aligned}
$$
If the chemistry books are considered identical, their arrangement number is
$$
F(4,5,3)=588 \text { (process omitted). }
$$
Then, the total number of arrangements for all types of books is
$$
N=F(4,5,3) \cdot 4!\cdot 5!\cdot 3!=10160640 .
$$
|
10160640
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
The sum and product of the first 51.1995 integers are both equal to F $\cdot$ 1996. What is the sum of the absolute values of these 1995 integers? Please prove your conclusion.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
Solution: Let these 1995 integers be denoted as $x_{1}, x_{2}, \cdots, x_{1995}$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{1995}=1996, \\
x_{1} x_{2} \cdots \cdots x_{1995}=1996=2^{2} \times 499 .
\end{array}\right.
$$
(1) From (1), we know that among these 1995 numbers, there are an even number of odd numbers and an odd number of even numbers.
(2) From (2), we know that among these 1995 numbers, there are an even number of negative numbers and an odd number of positive numbers.
(3) From (2), we know that among these 1995 numbers, there can only be one or two even numbers. Combining this with (1), we know that there can only be one even number. This even number can be $\pm 4$ or $\pm 1996$.
Next, we prove that it is impossible to take $\pm 1996$. Otherwise, if $x_{1995} = 1996$, the other numbers can only be +1 or -1, and let their counts be $k_{1}$ and $k_{2}$, respectively. From (2), we know that $k_{1}$ and $k_{2}$ are both non-negative even numbers. From (1) and (2), we get
$$
\left\{\begin{array}{l}
k_{1}-k_{2}+1996=1996, \\
k_{1}+k_{2}=1994 .
\end{array}\right.
$$
It is easy to see that the solution to this system of equations is $k_{1}=k_{2}=997$, which are not even numbers. Therefore, the assumption is not valid.
If $x_{1995} = -1996$, since the sum of the absolute values of the other numbers is 1994, equation (1) has no solution, which is still impossible.
Since 2 and 499 are both prime numbers, these 1995 numbers can only be $\pm 1, \pm 4, \pm 499$.
If $x_{1995} = -499$ and $x_{1994} = 4a$, where $a$ is +1 or -1, and the rest are $k_{1}$ +1s and $k_{2}$ -1s, then
$$
k_{1}-k_{2}+4a-499=1996.
$$
Since $k_{1}+k_{2}=1993$, it is clear that the left side is less than 1996, so the equation has no solution. Therefore, we can only take $x_{1995} = 499$. In this case, we have
$$
k_{1}-k_{2}+4a+499=1996.
$$
That is,
$$
\left\{\begin{array}{l}
k_{1}-k_{2}+4a=1497, \\
k_{1}+k_{2}=1993 .
\end{array}\right.
$$
Solving this, we get $k_{1}=1745-2a, k_{2}=248+2a$.
From (2), we know that the number of negative numbers is even, so we must have $a=1$ (otherwise, there will be $k_{2}+1$ negative numbers), i.e., $k_{1}=1743, k_{2}=250$. The sum of the absolute values is $1743+250+4+499=2496$.
Note: The editor made significant modifications to the solution.
(Fu Jie, Huayuan County National Middle School, Hunan Province, 416400)
|
2496
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Initial 52. Given real numbers $a, b, x, y$ satisfy $a x^{n}+b y^{n}=1+2^{n+1}$ for any natural number $n$. Find the value of $x^{a}+y^{b}$.
|
Let $S_{n}=a x^{\prime \prime}+b y^{\prime \prime}$. Then
$$
S_{1}=a x+b y=5 \text {, }
$$
$$
\begin{array}{l}
S_{0}=a x^{2}+b y^{2}=9 . \\
S=a x^{3}+b y^{3}=17 . \\
S_{1}=a x^{4}+b y^{\prime}=33 .
\end{array}
$$
Let $x+y=A, x y=B$. Then $x, y$ are the roots of the quadratic equation $t^{2}-A t+B=0$. Therefore,
$$
\begin{array}{l}
x^{2}-A x+B=0, \\
y^{2}-A y+B=0 .
\end{array}
$$
a. $x^{n-2} \cdot$ (3) $+b y^{n-2} \cdot$ (1), we get
$$
S_{n}=A S_{n-1}-B S_{n-2} \text {. }
$$
From the problem, we have $\left\{\begin{array}{l}S_{3}=A S_{2}-B S_{1}, \\ S_{4}=A S_{3}-B S_{2},\end{array}\right.$
which gives $\left\{\begin{array}{l}17=9 A-5 B, \\ 33=17 A-9 B .\end{array}\right.$
Solving these, we get $\left\{\begin{array}{l}A=3, \\ B=2 .\end{array}\right.$.
Thus, $\left\{\begin{array}{l}x+y=3 \\ x y=2\end{array}\right.$
Substituting $x=1, y=2$ into (1) and (2), we get $a=1, b=2$.
Similarly, when $x=2, y=1$, we get $a=2, b=1$.
(1) When $x=1, y=2, a=1, b=2$,
$$
x^{4}+y^{b}=1^{1}+2^{2}=5 \text {. }
$$
(2) When $x=2, y=1, a=2, b=1$,
$$
x^{2}+y^{3}=1^{1}+2^{2}=5 \text {. }
$$
（From Yu Yuanzheng, "Zhonghua Yinghao School, Conghua, Guangzhou, 510960）
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{c}
\text { Example } 1 \text { Calculate } \frac{a}{a^{3}+a^{2} b+a b^{2}+b^{3}}+ \\
\frac{b}{a^{3}-a^{2} b+a b^{2}-b^{3}}+\frac{1}{a^{2}-b^{2}}-\frac{1}{b^{2}+a^{2}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} .
\end{array}
$$
(1995, Tianjin City, Grade 7 "Mathematics Competition)
|
$\begin{array}{l}\text { Solution: Original expression }=\frac{a}{(a+b)\left(a^{2}+b^{2}\right)} \\ +\frac{b}{(a-b)\left(a^{2}+b^{2}\right)}+\frac{2 b^{2}}{a^{4}-b^{4}}-\frac{a^{2}+3 b^{2}}{a^{4}-b^{4}} \\ =\frac{a^{2}+b^{2}}{\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right)}-\frac{a^{2}+b^{2}}{a^{4}-b^{4}} \\ =\frac{1}{a^{2}-b^{2}}-\frac{1}{a^{2}-b^{2}}=0 . \\\end{array}$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 If $a+b+c=a b c \neq 0$, find the value of $\frac{\left(1-a^{2}\right)\left(1-b^{2}\right)}{a b}+\frac{\left(1-b^{2}\right)\left(1-c^{2}\right)}{b c}+$ $\frac{\left(1-c^{2}\right)\left(1-a^{2}\right)}{a c}$.
(1990, Wuhan City Mathematics Competition)
|
$\begin{array}{l}\text { Solution: Original expression }=\frac{1-a^{2}-b^{2}+a^{2} b^{2}}{a b} \\ +\frac{1-b^{2}-c^{2}+b^{2} c^{2}}{b c}+\frac{1-a^{2}-c^{2}+a^{2} c^{2}}{a c} \\ =\frac{1}{a b}-\frac{a}{b}-\frac{b}{a}+a b+\frac{1}{b c}-\frac{b}{c}-\frac{c}{b}+b c \\ +\frac{1}{a c}-\frac{a}{c}-\frac{c}{a}+a c \\ =\left(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\right)-\frac{b+c}{a}-\frac{a+c}{b}-\frac{a+b}{c} \\ +a b+b c+c a \\ =\frac{a+b+c}{a b c}-(b c-1)-(a c-1) \\ -(a b-1)+a b+b c+c a \\ =4 \text {. } \\\end{array}$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Given that $\alpha$ is a root of the equation $x^{2}+x-\frac{1}{4}=0$. Find the value of $\frac{\alpha^{3}-1}{\alpha^{5}+\alpha^{4}-\alpha^{3}-\alpha^{2}}$.
(1995, National Junior High School Mathematics League)
|
Solution: From the given, we have $\alpha^{2}+\alpha=\frac{1}{4}$.
To form the required expression containing $\alpha^{2}+\alpha$, we have
$$
\frac{\alpha^{3}-1}{\alpha^{5}+\alpha^{4}-\alpha^{3}-\alpha^{2}}=\frac{(\alpha-1)\left(\alpha^{2}+\alpha+1\right)}{(\alpha-1)\left(\alpha^{2}+\alpha\right)^{2}}=20 .
$$
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. $1000,1001,1002, \cdots, 1996$ These 997 natural numbers appear in all the digits whose sum is
The sum of all the digits that appear in the 997 natural numbers $1000,1001,1002, \cdots, 1996$ is
|
9. 14419
|
14419
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (Full marks 20 points) Given the equation $x^{2}+m x-m+1=0$ (where $m$ is an integer) has two distinct positive integer roots. Find the value of $m$.
---
The translation maintains the original format and line breaks as requested.
|
Three, let the two distinct positive integer roots be $\alpha, \beta(\alpha<\beta)$. By Vieta's formulas, we have $\left\{\begin{array}{l}\alpha+\beta=-m, \\ \alpha \beta=-m+1 .\end{array}\right.$ Eliminating $m$, we get $\alpha \beta-\alpha-\beta=1$.
That is, $(\alpha-1)(\beta-1)=2$.
Then $\left\{\begin{array}{l}\alpha-1=1, \\ \beta-1=2\end{array} \Rightarrow \alpha=2, \beta=3\right.$.
Therefore, $m=-(\alpha+\beta)=-5$.
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (Full marks 20 points, (1) 10 points, (2) 10 points) A total of 240 students participating in a large group performance stand in a row facing the coach, and report numbers in sequence from left to right as $1,2,3,4,5, \cdots$. The coach asks all students to remember their reported numbers and perform the following actions: first, all students who reported numbers that are multiples of 3 turn around; then, all students who reported numbers that are multiples of 5 turn around; finally, all students who reported numbers that are multiples of 7 turn around. Questions:
(1) How many students are still facing the coach at this point?
(2) Among the students facing the coach, what is the number reported by the 66th student from left to right?
|
(1) Because all students who report numbers that are multiples of $15$, $21$, and $35$ but not multiples of $105$ turn twice, still facing the coach. Students who report numbers that are multiples of $3$ but not multiples of $15$ or $21$, multiples of $5$ but not multiples of $15$ or $35$, and multiples of $7$ but not multiples of $21$ or $35$ turn once, all facing away from the coach. Students who report numbers that are multiples of $105$ turn three times, all facing away from the coach. According to this, the number of students facing the coach is
$$
\begin{array}{l}
240-\left\{\left[\frac{240}{3}\right]+\left[\frac{240}{5}\right]+\left[\frac{240}{7}\right]\right\}+2\left\{\left[\frac{240}{15}\right]\right. \\
\left.+\left[\frac{240}{21}\right]+\left[\frac{240}{35}\right]\right\}-\cdots\left[\frac{240}{105}\right]=136 .
\end{array}
$$
That is, 136 students are still facing the coach. The coach has 60 students:
$$
\begin{array}{l}
105-\left\{\left[\frac{105}{3}\right]+\left[\frac{105}{5}\right]+\left[\frac{105}{7}\right]\right\} \\
+2\left\{\left[\frac{105}{15}\right]+\left[\frac{105}{21}\right]+\left[\frac{105}{35}\right]\right\}-4\left[\frac{105}{105}\right]=60 .
\end{array}
$$
The numbers reported must be greater than $105$. Now, from the sequence of integers greater than $105$, remove the multiples of $3$, $5$, and $7$ but not multiples of $15$, $21$, and $35$, leaving $106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, \ldots$ Among the remaining numbers, the 6th number is $118$, so the 66th student facing the coach reports the number $118$.
|
118
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a, b, c$ are distinct. Find the value of $\frac{2 a-b-c}{(a-b)(a-c)}$ $+\frac{2 b-c-a}{(b-c)(b-a)}+\frac{2 c-a-b}{(c-a)(c-b)}$.
|
(Answer: 0)
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Real numbers $a, b, c$ satisfy $a+b+c=0, a b c>0$. If $x=$
$$
\begin{array}{l}
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}, y=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}\right. \\
\left.+\frac{1}{b}\right) \text {. Then, } x+2 y+3 x y=
\end{array}
$$
.
|
1. 2
The above text has been translated into English, maintaining the original text's line breaks and format.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six, (Full marks 12 points) On a circle, there are
12 points, one of which is painted red, and another is painted blue, with the remaining 10 points unpainted. Among the convex polygons formed by these points, those whose vertices include both the red and blue points are called bicolored polygons; those that include only the red (blue) point are called red (blue) polygons, and those that include neither the red nor the blue point are called colorless polygons.
Try to determine, among all the convex polygons (which can range from triangles to 12-gons) formed by these 12 points, 1. whether the number of bicolored polygons is greater than the number of colorless polygons, and by how many.
|
For any bicolored $n(n \geqslant 5)$-gon, naturally, removing the red and blue vertices results in a colorless $(n-2)$-gon. Different bicolored $n$-gons, after removing the red and blue vertices, yield different colorless $(n-2)$-gons; conversely, for any colorless polygon, adding red and blue vertices can always result in a bicolored polygon. Therefore, the number of colorless polygons (from triangles to decagons) is equal to the number of bicolored polygons (from pentagons to decagons).
Thus, the number of bicolored polygons is greater.
The additional number is exactly the number of bicolored triangles and bicolored quadrilaterals. It is easy to calculate that there are 10 bicolored triangles; bicolored quadrilaterals have $\frac{1}{2} \times 10 \times 9=45$ (units), meaning bicolored polygons are 55 more.
|
55
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Anita, Kieran, and Mitchell form a team to participate in a mixed cycling and running race, which takes place on a 3-kilometer circular track. It is known that each person's running speed is 5 kilometers per hour, and their cycling speed is 20 kilometers per hour. The starting point of the track is marked by a flag, and the three start from the flag with two bicycles.
(a) Both cycling and running are done in a clockwise direction, and each person must complete one lap. The three must arrive at the flag simultaneously. How can they complete the race in 18 minutes?
(b) All rules are the same, but each person must complete two laps. Can the team complete the race in less than 36 minutes? If so, indicate how to do it; otherwise, explain why it is not possible.
|
2. (a) It is easy to see that each person takes 12 minutes to run 1 km and 3 minutes to cycle 1 km. At the start, let Anita and Kemai cycle, while Mitchell runs. When they reach the 1 km mark, Kemai gives the bike to Mitchell and starts running. When they reach the 2 km mark, Anita gives the bike to Kemai and starts running. This way, each person runs 5 km, taking 12 minutes, and cycles 2 km, taking 6 minutes, totaling 18 minutes to reach the finish line simultaneously.
(b) The team can complete the race in 27 minutes. At the start, Anita and Kemai cycle, while Mitchell runs. When the cyclists overtake Mitchell, they have already traveled 3 km more than Mitchell, as their speed is 4 times that of Mitchell, so Mitchell has just run 1 km. At this point, Kemai gives the bike to Mitchell and starts running. At the next 1 km mark, Anita leaves the bike for Kemai and starts running. This way, each person runs 1 km, taking 12 minutes, and cycles 5 km, taking 15 minutes, totaling 27 minutes to reach the finish line simultaneously.
|
18
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If three integers $a, b, c (a \neq 0)$ make the equation $a x^{2}$ $+b x+c=0$ have two roots $a$ and $b$, then $a+b+c$ equals. $\qquad$
|
3. 18 .
From the relationship between roots and coefficients, we get
$$
\left\{\begin{array}{l}
a+b=-\frac{b}{a}, \\
a b=\frac{c}{a} .
\end{array}\right.
$$
From (1), we have $a^{2}+a b+b=0, (a+1) b=-a^{2}$.
Since $a \neq 0$, then $a+1 \neq 0$, $b \neq 0$. Thus,
$$
b=-\frac{a^{2}}{a+1}=-\frac{(a+1-1)^{2}}{a+1}=-a+1-\frac{1}{a+1} \text {, }
$$
Since $b$ is an integer, it follows that $a+1$ divides 1. Since $a+1 \neq 1$ (because $a \neq 0$), then $a+1=-1, a=-2$. Therefore, $b=4, c=a^{2} b$ $=16$. Hence, $a+b+c=18$.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $X=\{-1,0,1\}, Y=\{-2,-1,0,1,2\}$, and for all elements $x$ in $X$, $x+f(x)$ is even. Then the number of mappings $f$ from $X$ to $Y$ is ( ).
(A) 7
(B) 10
(C) 12
(D) 15.
|
4. (C).
$x$ and $f(x)$ are both even, or both odd.
When $x=0$, there are three possibilities: $-2, 0, 2$;
When $x=-1, 1$, there are two possibilities each for -1 and 1, making 4 possibilities.
Therefore, the total is $3 \times 4=12$.
|
12
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. A monkey is on a ladder with $n$ steps: climbing up and down, it either ascends 16 steps or descends 9 steps each time. If it can climb from the ground to the very top step, and then return to the ground, the minimum value of $n$ is ( ).
(A) 22
(B) 23
(C) 24
(D) greater than 24.
|
6. (C).
Assume the monkey climbs as follows: 0 $\nearrow 16 \searrow 7 \nearrow 23 \searrow 14 \searrow 5$ $\nearrow 21 \searrow 12 \searrow 3 \nearrow 19 \searrow 10 \searrow 1, \nearrow 17 \searrow 8 \nearrow 24 \searrow 15 \searrow 6 \nearrow$ $22 \searrow 13 \searrow 4$ Л $20 \searrow 11 \searrow 2$ オ゙ $18 \searrow 9 \searrow 0$. We get that the minimum value of $n$ is 24.
In fact, when $n<24$, the monkey's climbing method is unique. At the latest, when the monkey climbs to the 8th step, it can neither climb up nor down, which contradicts the problem statement. Therefore, the minimum value of $n$ is 24.
|
24
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { 1. } \lim _{n \rightarrow \infty} \frac{1}{\sqrt[3]{n}}\left(\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}+ \\ \frac{1}{\sqrt[3]{4}+\sqrt[3]{6}+\sqrt[3]{9}}+ \\ \left.\frac{1}{\sqrt[3]{(n-1)^{2}}+\sqrt[3]{n(n-1)}+\sqrt[3]{n^{2}}}\right)=\end{array}$
|
$$
\approx .1 .1 \text {. }
$$
Since
$$
\begin{array}{l}
\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}+\cdots+ \\
\frac{1}{\sqrt[3]{(n-1)^{2}}+\sqrt[3]{n(n-1)}+\sqrt[3]{n^{2}}} \\
=\frac{\sqrt[3]{2}-1}{2-1}+\cdots+\frac{\sqrt[3]{n}-\sqrt[3]{n-1}}{n-(n-1)} . \\
=\sqrt[3]{n}-1,
\end{array}
$$
Therefore, the required limit is $\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n}-1}{\sqrt[3]{n}}=1$.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. On each face of an opaque cube, a natural number is written. If several (one, two, or three) faces of the cube can be seen at the same time, then find the sum of the numbers on these faces. Using this method, the maximum number of different sums that can be obtained is _.
翻译结果如下：
5. On each face of an opaque cube, a natural number is written. If several (one, two, or three) faces of the cube can be seen at the same time, then find the sum of the numbers on these faces. Using this method, the maximum number of different sums that can be obtained is _.
|
5.26 .
There are 6 cases where a single face can be seen, 12 cases where two faces sharing a common edge can be seen simultaneously, and 8 cases where three faces sharing a common vertex can be seen simultaneously, thus yielding 26 sums. The numbers on the faces can be filled in such a way that all 26 sums are distinct (for example, the numbers $1, 10, 10^{2}, 10^{3}, 10^{4}, 10^{5}$ can be written on the 6 faces).
|
26
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let the sequence of positive integers $a_{1} 、 a_{2} 、 a_{3} 、 a_{4}$ be a geometric sequence, with the common ratio $r$ not being an integer and $r>1$. The smallest value that $a_{4}$ can take in such a sequence is $\qquad$ .
|
6. 27 .
According to the problem, $r$ is a rational number, so there exist coprime positive integers $p$ and $q (q > p \geqslant 2)$, such that $r=\frac{q}{p}$. Therefore, $a_{4}=a_{1} r^{3}=\frac{a_{1} q^{3}}{p^{3}}$.
Since $a_{4}$ is an integer, $a_{1}$ must be a multiple of $p^{3}$.
Thus, let $a_{1}=k p^{3}$ ($k$ is a positive integer).
At this point, $a_{4}=k q^{3}$. Note: Given $q > p \geqslant 2$, to make $a_{4}$ the smallest, we have $k=1, q=3$.
|
27
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Simplify $\frac{(x+b)(x+c)}{(a-b)(a-c)}+\frac{(x+c)(x+a)}{(b-c)(b-a)}$ $+\frac{(x+a)(x+b)}{(c-a)(c-b)}$.
|
(Answer: 1).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Let $x=b y+c z, y=c z+a x, z=a x$ $+b y$. Find the value of $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}$.
|
Solution: $\frac{a}{a+1}=\frac{a x}{a x+x}=\frac{a x}{a x+b y+c z}$.
Similarly, $\frac{b}{b+1}=\frac{b y}{a x+b y+c z}$,
$$
\frac{c}{c+1}=\frac{c z}{a x+b y+c z} \text {. }
$$
Adding them up, the value of the desired expression is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 If $\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=$ $\frac{t}{x+y+z}$, let $f=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}$. Prove: $f$ is an integer.
(1990, Hungarian Mathematical Competition)
|
Proof: If $x+y+z+t \neq 0$, by the property of proportion, we have
$$
\begin{array}{l}
\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y} \\
=\frac{t}{x+y+z}=\frac{x+y+z+t}{3(x+y+z+t)}=\frac{1}{3} . \\
\text { Then we have }\left\{\begin{array}{l}
y+z+t=3 x, \\
z+t+x=3 y, \\
t+x+y=3 z, \\
x+y+z=3 t .
\end{array}\right.
\end{array}
$$
Solving, we get $x=y=z=t, f=4$ is an integer.
If $x+y+z+t=0$, it is easy to see that $f=-4$ is also an integer.
|
-4
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given $4 x-3 y-6 z=0, x+2 y-7 z$ $=0(x y z \neq 0)$. Find the value of $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 y^{2}+7 z^{2}}$.
(1992, Sichuan Province Junior High School Mathematics League Preliminary)
|
Solution: Let $y=k_{1} x, z=k_{2} x$. Then
$$
\left\{\begin{array} { l }
{ 4 - 3 k _ { 1 } - 6 k _ { 2 } = 0 , } \\
{ 1 + 2 k _ { 1 } - 7 k _ { 2 } = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
k_{1}=\frac{2}{3}, \\
k_{2}=\frac{1}{3} .
\end{array}\right.\right.
$$
Then $\frac{2 x^{2}+3 y^{2}+6 z^{2}}{x^{2}+5 y^{2}+7 z^{2}}=\frac{2+3 k_{1}^{2}+6 k_{2}^{2}}{1+5 k_{1}^{2}+7 k_{2}^{2}}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given the theorem: "If three prime numbers greater than 3, $a, b$, and $c$, satisfy the equation $2a + 5b = c$, then $a + b + c$ is a deficient number of the integer $n$." What is the maximum possible value of the integer $n$ in the theorem? Prove your conclusion.
|
Three, the maximum possible value of $n$ is 9.
First, prove that $a+b+c$ can be divided by 3.
In fact, $a+b+c=a+b+2a+5b=3(a+2b)$, so $a+b+c$ is a multiple of 3.
Let the remainders when $a$ and $b$ are divided by 3 be $r_{a}$ and $r_{b}$, respectively, then $r_{a} \neq 0, r_{b} \neq 0$.
If $r_{a} \neq r_{b}$, then $r_{a}=1, r_{b}=2$ or $r_{a}=2, r_{b}=1$. In this case, $2a+5b$ must be a multiple of 3, i.e., $c$ is a composite number, which is a contradiction.
Therefore, $r_{a}=r_{b}$, then $r_{a}=r_{b}=1$ or $r_{a}=r_{b}=2$,
In this case, $a+2b$ must be a multiple of 3, thus $a+b+c$ is a multiple of 9.
Next, prove that 9 is the largest.
$$
\because 2 \times 11+5 \times 5=47 \text {, and } 11+5+47=63 \text {, }
$$
While in $2 \times 13+5 \times 7=61$, $13+7+61=81$, and $(63, 81)=9$, hence 9 is the largest possible value.
|
9
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. In an equilateral $\triangle ABC$, $P$ is a point on side $AB$, $Q$ is a point on side $AC$, and $AP = CQ$. It is measured that the distance between point $A$ and the midpoint $M$ of line segment $PQ$ is $19 \mathrm{~cm}$. Then the distance from point $P$ to point $C$ is $\qquad$ $\mathrm{cm}$.
|
3. 38 .
|
38
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (Full marks 15 points) A natural number $a$ is exactly equal to the square of another natural number $b$, then the natural number $a$ is called a perfect square (for example, $64=8^{2}$, so $64$ is a perfect square).
If $a=1995^{2}+1995^{2} \cdot 1996^{2}+1996^{2}$, prove that $a$ is a perfect square, and write down the square root of $a$.
|
Let $x=1995$, then $x+1=1996$.
$$
\begin{aligned}
a= & 1995^{2}+1995^{2} \cdot 1996^{2}+1996^{2} \\
= & x^{2}+x^{2}(x+1)^{2}+(x+1)^{2} \\
= & (x+1)^{2}-2 x(x+1)+x^{2}+2 x^{2}(x+1) \\
& +x^{2}(x+1)^{2} \\
= & (x+1-x)^{2}+2 x(x+1)+[x(x+1)]^{2} \\
= & 1^{2}+2 x(x+1)+[x(x+1)]^{2} \\
= & {[1+x(x+1)]^{2} } \\
= & (1+1995 \times 1996)^{2}=(3982021)^{2} .
\end{aligned}
$$
Therefore, $a$ is a perfect square.
According to the definition of square roots, the square roots of $a$ are $\pm 3982021$.
|
3982021
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks: 15 points) From the 91 natural numbers $1,2,3, \cdots, 90,91$, select $k$ numbers such that there must be two natural numbers $p, q$ satisfying $\frac{2}{3} \leqslant \frac{q}{p} \leqslant \frac{3}{2}$. Determine the minimum value of the natural number $k$, and explain your reasoning.
|
Five, divide the 91 natural numbers from 1 to 91 into nine groups, such that the ratio of any two natural numbers in each group is no less than $\frac{2}{3}$ and no more than $\frac{3}{2}$, and the division is as follows:
$$
\begin{array}{l}
A_{1}=\{1\}, A_{2}=\{2,3\}, A_{3}=\{4,5,6\}, \\
A_{4}=\{7,8,9,10\}, \\
A_{5}=\{11,12,13,14,15,16\}, \\
A_{6}=\{17,18,19,20,21,22,23,24,25\}, \\
A_{7}=\{26,27,28,29, \cdots, 39\}, \\
A_{8}=\{40,41,42,43, \cdots, 58,59,60\}, \\
A_{9}=\{61,62,63, \cdots, 89,90,91\} .
\end{array}
$$
If any nine numbers are taken from these 91 numbers, for example, one number from each of $A_{1}, A_{2}, \cdots, A_{9}$, such as $1,3,6,10,16,25,39,60,91$, the ratio of any two of these nine numbers is less than $\frac{2}{3}$ or greater than $\frac{3}{2}$. Therefore, $k > 9$. So $k \leqslant 10$. When $k=10$, if any 10 numbers are taken from the 91 natural numbers from 1 to 91, according to the pigeonhole principle, there must be two numbers in the same $A_{i}$. For example, these two numbers are $p$ and $q$, and $p<q$, then $\frac{2}{3} \leqslant \frac{q}{p} \leqslant \frac{3}{2}$ holds.
Therefore, the minimum value of $k$ is 10.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 If $x, y, z$ are real numbers, and
$$
\begin{aligned}
(y-z)^{2} & +(z-x)^{2}+(x-y)^{2} \\
= & (y+z-2 x)^{2}+(z+x-2 y)^{2} \\
& +(x+y-2 z)^{2},
\end{aligned}
$$
find the value of $M=\frac{(y z+1)(z x+1)(x y+1)}{\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right)}$.
|
Solution: The condition can be simplified to
$$
x^{2}+y^{2}+z^{2}-x y-y z-z x=0 .
$$
Then $(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=0$,
which implies $x=y=z$.
Therefore, $M=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 20 points) There is a 14-digit number, the digit in the units place is 3 less than the digit in the tens place, and the new four-digit number formed by reversing its digits differs from the original four-digit number by 8987. Find this four-digit number and write out the reasoning process.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Five, let the thousands digit of this four-digit number be $a$, the hundreds digit be $b$, the tens digit be $c$, and the units digit be $c-3$.
This number is $1000a + 100b + 10c + (c-3)$.
The new four-digit number is $1000(c-3) + 100c + 10b + a$.
According to the problem, we have
$$
\begin{array}{l}
1001(a+c-3) + 110(b+c) = 8987 . \\
1001(a+c) + 110(b+c) = 11990 . \\
91(a+c) + 10(b+c) = 1090 .
\end{array}
$$
The right side of the equation has a factor of 10, the second term on the left side has a factor of 10, and the first term $91(a+c)$ on the left side should also have a factor of 10. Since $91 = 7 \times 13$, it follows that $a+c$ is a multiple of 10.
Also, $1 \leqslant a \leqslant 9, 4 \leqslant c \leqslant 9$,
so $5 \leqslant a+c \leqslant 18$, which gives $a+c=10$.
Substituting (2) into (1) yields $b+c=18$.
Solving, we get $a=1, b=c=9, c-3=6$.
Answer: The four-digit number is 1996.
|
1996
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, in a chess tournament, there are an odd number of participants, and each participant plays one game against every other participant. The scoring system is as follows: 1 point for a win, 0.5 points for a draw, and 0 points for a loss. It is known that two of the participants together scored 8 points, and the average score of the others is an integer. How many participants are there in the tournament?
Will the text above be translated into English, please keep the original text's line breaks and format, and output the translation result directly.
|
Five, suppose there are $(n+2)$ players in total, except for 2 people who get 8 points, $n$ people on average get $k$ points each ($k$ is an integer).
$\because$ Each person plays one match with everyone else, and there are $(n+2)$ people,
$\therefore$ A total of $\frac{(n+1)(n+2)}{2}$ matches are played.
Since each match results in 1 point, we can set up the equation
$$
\frac{1}{2}(n+1)(n+2)=8+n k \text {, }
$$
Rearranging gives $n^{2}+(3-2 k) n-14=0$.
$\because$ The number of people is a positive integer, and $3-2 k$ is an integer,
$\therefore$ The number of people can only be $1, 2, 7, 14$.
Also, $\because$ the number of people is odd, $\therefore n=1$ or 7,
But when $n=1$, $k<0$ which is a contradiction, $\therefore n=7$.
At this time, $n+2=9$, i.e., a total of 9 people participate in the competition.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 As shown, in Rt $\triangle ABC$, the hypotenuse $AB=5, CD \perp AB$. It is known that $BC, AC$ are the two roots of the quadratic equation $x^{2}-(2 m-1) x+4(m-1)=0$. Then the value of $m$ is $\qquad$.
|
Solution: Let $A C=b$, $B C=a$. By Vieta's formulas, we get $a+b=2 m-$
$$
\begin{array}{l}
1, a b=4(m-1) . \\
\begin{aligned}
\therefore A B^{2} & =a^{2}+b^{2}=(a+b)^{2}-2 a b \\
& =(2 m-1)^{2}-2 \times 4(m-1)=5^{2},
\end{aligned}
\end{array}
$$
i.e., $m^{2}-3 m-4=0$.
$$
\therefore m=4 \text { or } m=-1 \text {. }
$$
Since $a$ and $b$ are the lengths of the sides of a triangle, we have $a>0, b>0$. Therefore, $a+b=2 m-1>0$, i.e., $m>\frac{1}{2}$.
$$
\therefore m=4 \text {. }
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The train is $400 \mathrm{~m}$ long, and it takes 10 minutes to pass through the tunnel (from the front of the train entering the tunnel to the rear of the train leaving the tunnel). If the speed increases by 0.1 kilometers per minute, then it will take 9 minutes, and the length of the tunnel is $\qquad$ $ـ$.
|
5. 8600 meters.
Let the full length be $x$ meters, and the original speed of the train be $v$ meters per minute. According to
$$
\left\{\begin{array}{l}
\frac{x+400}{v}=10, \\
\frac{x+400}{v+100}=9 .
\end{array} \text { Solving, we get } v=900\right. \text { (meters/minute.) }
$$
|
8600
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (Full marks 14 points) Given that the roots of the equation $x^{2}+p x+q=0$ are 1997 and 1998, when $x$ takes the integer values $0,1,2, \cdots, 1999$, the corresponding values of the quadratic trinomial $y=x^{2}+p x+q$ are $y_{0}$. Find the number of these values that are divisible by 6.
|
Let $y=(x-1997)(x-1998)$ be divisible by 6. Since when $x$ takes integer values, all $y$ values can be divisible by 2, we only need to examine the cases where the factors are divisible by 3.
$$
\begin{array}{l}
\text { (1) When } x-1997=3 k \text {, then } x-1998=3 k-1 \text {. } \\
\because 0 \leqslant x \leqslant 1999, \therefore-1997 \leqslant 3 k \leqslant 2 .
\end{array}
$$
In this case, $k$ can take 666 values from -665 to 0.
（2）When $x-1998=3 k$, then $x-1997=3 k+1$.
$$
\because 0 \leqslant x \leqslant 1999, \therefore-1998 \leqslant 3 k \leqslant 1 \text {. }
$$
In this case, $k$ can take 667 values from -666 to 0.
Thus, the total number of values that can be divisible by 6 is
$$
666+667=1333 \text {. }
$$
|
1333
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 13 Given that $x_{1}, x_{2}$ are the two real roots of the equation $x^{2}-(k-2) x + (k^{2}+3 k+5)=0$ (where $k$ is a real number). Then the maximum value of $x_{1}^{2}+x_{2}^{2}$ is $\qquad$
|
According to Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=k-2, \quad x_{1} x_{2}=k^{2}+3 k+5 . \\
\therefore \quad x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \\
=(k-2)^{2}-2\left(k^{2}+3 k+5\right) \\
=9-\left(k+5\right)^{2} .
\end{array}
$$
From the discriminant theorem, we get
$$
\begin{aligned}
\Delta & =(k-2)^{2}-4\left(k^{2}+3 k+5\right) \\
& =-3 k^{2}-16 k-16 \geqslant 0 .
\end{aligned}
$$
Solving this, we get $-4 \leqslant k \leqslant-\frac{4}{3}$.
Therefore, when $k=-4$, $x_{1}^{2}+x_{2}^{2}$ has a maximum value of
18.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in the figure, in the right trapezoid $A B C D$, the base $A B=13$, $C D=8, A D \perp A B$, and $A D=$ 12. Then the distance from $A$ to the side $B C$ is ( ).
(A) 12
(B) 13
(C) 10
(D) $\frac{12 \times 21}{13}$
|
6. (A).
Draw $C C^{\vee} \perp A B$, then
$$
B C=13-8=5 \text {. }
$$
Let the distance from $A$ to $B C$ be $h$,
Connect AC.
$$
\begin{aligned}
& \because S_{\triangle A C D}+S_{\triangle A B C} \\
& =S_{\text {UEABCD }}, \\
\therefore & \frac{1}{2} \times 8 \times 12+\frac{1}{2} \times h \times 13=\frac{1}{2}(8+13) \times 12
\end{aligned}
$$
Therefore, $h=12$.
|
12
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $|x| \leqslant 2$, the sum of the maximum and minimum values of the function $y=x-|1+x|$ is $\qquad$ .
|
2. -4 .
$$
\begin{array}{l}
\because|x| \leqslant 2, \text { i.e., }-2 \leqslant x \leqslant 2, \\
\therefore y=x-|1+x| \\
\quad=\left\{\begin{array}{ll}
2 x+1, & \text { when }-2 \leqslant x<-1, \\
-1, & \text { when }-1 \leqslant x \leqslant 2.
\end{array}\right.
\end{array}
$$
As shown in the figure.
Indeed, when $x=-2$, $y$
has the minimum value -3; when $-1 \leqslant$ $x \leqslant 2$, $y$ has the maximum value -1.
Therefore, $-3+(-1)$ $=-4$.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In $\triangle A B C$, $\angle B C A=90^{\circ}$, a perpendicular $C D \perp A B$ is drawn from $C$ intersecting $A B$ at $D$. Suppose the side lengths of $\triangle A B C$ are all integers, and $B D=29^{3}, \cos B=\frac{m}{n}$, where $m$ and $n$ are coprime positive integers. Then $m+n=$ $\qquad$
|
4. 450 .
Let the three sides of $\triangle ABC$ be $a, b, c$. It is easy to prove that $BC^2 = BD \cdot BA$. Therefore, $a^2 = 29^3 c$.
Since 29 is a prime number, $29^2 \mid a$.
Let $a = 29^2 k$ ($k$ is a positive integer), then $c = 29 k^2$.
Since $b = \sqrt{c^2 - a^2} = 29 k \sqrt{k^2 - 29^2}$ is an integer, we know that $k^2 - 29^2 = t^2$ ($t$ is a positive integer).
$$
\therefore (k-t)(k+t) = 29^2 \text{. }
$$
Solving this, we get $k = 421$.
$$
\therefore BC = 29^2 \times 421 \text{. }
$$
Thus, $\cos B = \frac{BD}{BC} = \frac{29}{421}$.
$$
\therefore m+n = 450 \text{. }
$$
|
450
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $M=\cos 5^{\circ} \sin 15^{\circ} \sin 25^{\circ} \sin 35^{\circ}, N=$ $\sin 5^{\circ} \cos 15^{\circ} \cos 25^{\circ} \cos 35^{\circ}$. Then $\frac{M}{N}=$ $\qquad$ .
|
$=1.1$.
$$
\begin{aligned}
\frac{M}{N} & =\frac{\frac{1}{2}\left(\sin 20^{\circ}+\sin 10^{\circ}\right) \cdot \frac{1}{2}\left(\cos 10^{\circ}-\cos 60^{\circ}\right)}{\frac{1}{2}\left(\sin 20^{\circ}-\sin 10^{\circ}\right) \cdot \frac{1}{2}\left(\cos 10^{\circ}+\cos 60^{\circ}\right)} \\
& =\frac{\left(\sin 20^{\circ}+\sin 10^{\circ}\right)\left(\cos 10^{\circ}-\frac{1}{2}\right)}{\left(\sin 20^{\circ}-\sin 10^{\circ}\right)\left(\cos 10^{\circ}+\frac{1}{2}\right)} \\
= & \frac{\sin 20^{\circ} \cos 10^{\circ}-\frac{1}{2} \sin 20^{\circ}+\sin 10^{\circ} \cos 10^{\circ}-\frac{1}{2} \sin 10^{\circ}}{\sin 20^{\circ} \cos 10^{\circ}+\frac{1}{2} \sin 20^{\circ}-\sin 10^{\circ} \cos 10^{\circ}-\frac{1}{2} \sin 10^{\circ}} \\
= & \frac{\sin 20^{\circ} \cos 10^{\circ}-\frac{1}{2} \sin 10^{\circ}}{\sin 20^{\circ} \cos 10^{\circ}-\frac{1}{2} \sin 10^{\circ}}=1 .
\end{aligned}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $a+\lg a=10, b+10^{b}=10$. Then $a+b$
|
4. 10.
Thought 1: From the given information,
$$
\begin{array}{l}
a=10^{10-a}, \\
10-b=10^{b} .
\end{array}
$$
Subtracting, we get $10-a-b=10^{b}-10^{10-a}$.
If $10-a-b>0$, then $10-a>b$. By the monotonicity of the exponential function, we get $10^{10-a}>10^{b}$. Substituting into (1), we get
$$
010-a-b=10^{b}-10^{10-a}>0 \text{. }
$$
Contradiction.
Therefore, $10-a-b=0$, which means $a+b=10$.
Thought 2: The geometric meaning of this problem is: $a$ is the x-coordinate of the intersection point $A$ of $y=\lg x$ and $y=10-x$, and $b$ is the x-coordinate of the intersection point $B$ of $y=10^{x}$ and $y=10-x$. By the symmetry of the graphs of two inverse functions, $A$ and $B$ are symmetric about the line $y=x$. Solving the system of equations
$$
\left\{\begin{array}{l}
y=x \\
y=10-x
\end{array}\right.
$$
we get $C(5,5)$. Since $C$ is the midpoint of $A$ and $B$, we have $\frac{a+b}{2}=5$, which means $a+b=10$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
On September 4, 1996, scientists used a supercomputer to find the 33rd Mersenne prime, which is the largest prime number known to humans so far. It is: $2^{125787}-1$ (378632 digits). Try to find the last two digits of this prime number.
|
Solution: First, find the unit digit.
$$
\begin{array}{l}
\because 2^{1257787}-1=2 \cdot 4^{628893}-1=8 \cdot 16^{31446}-1 \\
\quad \equiv 8 \times 6-1(\bmod 10) \equiv 7(\bmod 10),
\end{array}
$$
$\therefore 2^{1257787}-1$ has a unit digit of 7.
Next, find the tens digit.
$$
\begin{aligned}
\because & \frac{1}{10}\left[\left(2^{1257887}-1\right)-7\right] \\
= & \frac{1}{10}\left[2 \cdot 4^{628895}-8\right]=\frac{1}{5}\left(4^{628893}-4\right) \\
= & \frac{4}{5}\left(4^{628892}-1\right)=\frac{4}{5}\left(16^{314446}-1\right) \\
= & \frac{4}{5} \times(16-1) \times\left(16^{314445}+16^{314444}+\cdots+16\right. \\
& \quad+1) \\
\equiv & 12(6 \times 314445+1)(\bmod 10) \\
\equiv & 2(\bmod 10),
\end{aligned}
$$
$\therefore 2^{1257787}-1$ has a tens digit of 2.
Therefore, the last two digits of $2^{1257787}-1$ are 27.
|
27
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
In the plane: There are 212 points all located within or on the boundary of a unit circle. Connect any two points to form a line segment. Prove: The number of line segments with length not greater than 1 is at least 1996.
|
Proof: (1) First, prove that in a circle or on any $\epsilon$ points, there must be two points whose distance is not greater than:
Let these 6 points be $A_{1}, A_{2}, \cdots, A_{6}$, and the center be $O$. If one point is at the center, the proposition holds. Otherwise, connect $O A_{i},(i=1,2, \cdots, 6)$, to get 6 angles with $O$ as the vertex. It is easy to see that at least one angle is not greater than $60^{\circ}$. The two points corresponding to this angle, say $A_{i}$ and $A_{j}$, must have a distance not greater than 1.
(2) Now, prove the original proposition.
Let $A_{1}$ be the point among all points with the fewest lines of length not greater than 1 to other points, and assume $A_{1}$ is connected to $A_{2}, A_{3}, \cdots, A_{m+1}$ with lines of length not greater than 1; let the set $M=\left\{A_{1}, A_{2}, \cdots, A_{m+1}\right\}$, and the set $N$ be the set of the remaining points. Take any 5 points from $N$, and add $A_{i}$ to make 6 points. By (1) and the assumption about $A_{1}$, there must exist two points $C_{i_{1}}, C_{i_{2}} \in N$ whose distance is not greater than 1. Thus, there are $C_{211-m}^{5}$ lines of length not greater than 1. However, some of these lines are repeated. For this, consider any line of length not greater than 1, $C_{1} C_{2}, C_{1} 、 C_{2} \in N$. Take another 3 points from $N$, and add $A_{1}$, $C_{1}$, and $C_{2}$ to form 6 points. By (1), $C_{1} C_{2}$ can appear at most $C_{209-m}^{3}$ times in the $C_{211-m}^{5}$ lines. Therefore, the points in $N$ can produce at least
$$
\frac{C_{211-m}^{5}}{C_{209-m}^{3}}=\frac{1}{20}(211-m)(210-m)
$$
lines of length not greater than 1.
Additionally, in $M$, by the assumption about $A_{1}$, each $A_{i}$ can form at least $m$ lines of length not greater than 1. Thus, the number of lines of length not greater than 1 that contain points from $M$ is at least $\frac{1}{2} m(m+1)$. Let
$$
\begin{aligned}
f(m) & =\frac{1}{2} m(m+1)+\frac{1}{20}(211-m)(210-m) \\
& =\frac{1}{20}\left(11 m^{2}-411 m+211 \times 210\right) .
\end{aligned}
$$
It is easy to see that when $m=18$ or $m=19$, $f(m)$ reaches its minimum integer value $2023>1996$.
|
1996
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $a, b$ be unequal real numbers, and $a^{2}+2 a-5$ $=0, b^{2}+2 b-5=0$. Then $a^{2} b+a b^{2}=$ $\qquad$ .
|
（Answer:10）
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, 610 people go to the bookstore to buy books, it is known that
(1) each person bought three books;
(2) any two people have at least one book in common.
How many people at most bought the book that was purchased by the fewest people?
|
Solution: Let the number of people who bought the most popular book be $x$. Among the 10 people, person A bought three books. Since the other 9 people each have at least one book in common with A, and $9 \div 3=3$, it follows that among A's three books, the most popular one must have been bought by at least 4 people, so $x \geqslant 4$.
If $x=4$, then each of A's three books was bought by 4 people. Similarly, each book bought by the other 9 people was also bought by 4 people. Therefore, the total number of books bought by the 10 people should be a multiple of 4, i.e., $4 \mid 30$, which is a contradiction. Thus, $x \geqslant 5$.
When $x=5$, let $A_{i}$ represent different books. The scenario where the most popular book is bought by exactly 5 people is as follows:
$$
\begin{array}{l}
\left\{A_{1} A_{2} A_{3}\right\} 、\left\{A_{1} A_{2} A_{6}\right\} 、\left\{A_{2} A_{3} A_{4}\right\} 、 \\
\left\{A_{1} A_{4} A_{6}\right\} 、\left\{A_{1} A_{4} A_{5}\right\} 、\left\{A_{2} A_{4} A_{5}\right\} 、 \\
\left\{A_{1} A_{3} A_{5}\right\} 、\left\{A_{2} A_{5} A_{6}\right\} 、\left\{A_{3} A_{5} A_{6}\right\} 、 \\
\left\{A_{3} A_{4} A_{6}\right\} .
\end{array}
$$
In conclusion, the most popular book was bought by at least 5 people.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given that $x$, $y$, $z$ are positive integers, and $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(y+z)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Explanation: This is a typical example of constructing a geometric figure to solve a problem. As shown in Figure 1, construct $\triangle ABC$, where the lengths of the three sides are
$$
\left\{\begin{array}{l}
a=x+y, \\
b=y+z, \\
c=z+x .
\end{array}\right.
$$
Then its area is
$$
\begin{aligned}
\triangle & =\sqrt{p(p-a)(p-b)(p-c)} \\
& =\sqrt{(x+y+z) x y z}=1 .
\end{aligned}
$$
On the other hand,
$$
(x+y)(y+z)=a b=\frac{2 \triangle}{\sin C} \geqslant 2 .
$$
Therefore, when and only when $\angle C=90^{\circ}(\sin C=1)$, $a b$ takes the minimum value of 2. At this time, by the Pythagorean theorem, we have
$$
(x+y)^{2}+(y+z)^{2}=(x+z)^{2} \text {, }
$$
which means $y(x+y+z)=x z$ when $(x+y)(y+z)$ takes the minimum value. For example, when $x=z=1, y=\sqrt{2}-1$, $(x+y)(y+z)=2$.
This construction is indeed quite ingenious and should be recognized for its training value, but for solving this problem, a direct algebraic method is much simpler.
The following two problems from the former Soviet Union's math competitions are also classic examples of constructing geometric figures to solve problems, and they can both find simpler algebraic solutions.
1. Let $a, b, c$ be positive numbers, and
$$
\left\{\begin{array}{l}
a^{2}+a b+\frac{b^{2}}{3}=25, \\
\frac{b^{2}}{3}+c^{2}=9, \\
a^{2}+a c+c^{2}=16 .
\end{array}\right.
$$
Find the value of $a b+2 b c+3 a c$.
2. Positive numbers $a, b, c, A, B, C$ satisfy the conditions $a+A=$ $b+B=c+C=k$. Prove that $a B+b C+c A<k^{2}$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Color each vertex of a square pyramid with one color, and make the endpoints of the same edge have different colors. If only 5 colors are available, then the total number of different coloring methods is $\qquad$
|
Solution: Vertex $S$ can be colored with any of the $m$ colors, and the color on $S$ cannot appear on the vertices of the polygon $A_{1} A_{2} \cdots A_{n}$. The problem is then reduced to coloring the vertices of the polygon with $m-1$ colors, ensuring that adjacent vertices have different colors. Let there be $a_{n}$ ways to do this. Then,
$$
a_{3}=(m-1)(m-2)(m-3) .
$$
For $n>3$, consider the recurrence relation for $a_{n}$. If starting from $A_{1}$, then $A_{1}$ has $(m-1)$ ways to be colored, and $A_{2}, \cdots, A_{n-1}$ each have $(m-2)$ ways to be colored, and finally to $A_{n}$, if only requiring $A_{n}$ to be different from $A_{n-1}$, there are still $(m-2)$ ways to color it. Thus, there are
$$
(m-1)(m-2)^{n-1}
$$
ways in total. However, this calculation can be divided into two categories: one where $A_{n}$ is different from $A_{1}$, which meets the requirement and is exactly $a_{n}$ ways; the other where $A_{n}$ is the same as $A_{1}$, which does not meet the requirement, but by merging $A_{n}$ and $A_{1}$ into one point, we get $a_{n-1}$ ways. Therefore,
$$
\left\{\begin{array}{l}
a_{n}+a_{n-1}=(m-1)(m-2)^{n-1},(n>3) \\
a_{3}=(m-1)(m-2)(m-3) .
\end{array}\right.
$$
Transforming and recursively,
$$
\begin{array}{l}
a_{n}-(m-2)^{n}=-\left[a_{n-1}-(m-2)^{n-1}\right] \\
=(-1)^{2}\left[a_{n-2}-(m-2)^{n-2}\right] \\
=\cdots \cdots \\
=(-1)^{n-2}(m-2) \\
=(-1)^{n}(m-2) .
\end{array}
$$
Thus, $a_{n}=(m-2)\left[(m-2)^{n-1}+(-1)^{n}\right]$.
Therefore, the total number of coloring methods for the entire pyramid is
$$
N=m(m-2)\left[(m-2)^{n-1}+(-1)^{n}\right] \text {. }
$$
In particular, for $n=4, m=5$, we get
$$
N=5 \cdot 3\left(3^{3}+1\right)=420 \text {. }
$$
Here, we see the shadow of an old problem: dividing a circular surface into $n(n \geqslant 2)$ sectors, denoted as $S_{1}, S_{2}, \cdots, S_{n}$. Each sector can be colored with one of the three different colors: red, white, and blue, with the requirement that adjacent sectors have different colors. How many coloring methods are there?
|
420
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Prove that there exists a unique function $f: N \rightarrow$ $N$ satisfying for all $m, n \in N$, $f[m+f(n)]=n$ $+f(m+95)$. Find the value of $\sum_{k=1}^{19} f(k)$.
|
Solution: Observe and verify that $f(n)=n+95$ meets the requirement, at this time, $\sum_{k=1}^{19} f(k)=1995$.
Now prove the uniqueness.
When $n_{1} \neq n_{2}$,
$$
\begin{array}{l}
f\left[m+f\left(n_{1}\right)\right]=n_{1}+f(m+95) \\
\neq n_{2}+f(m+95)=f\left[m+f\left(n_{2}\right)\right],
\end{array}
$$
i.e., $f\left(n_{1}\right) \neq f\left(n_{2}\right)$.
Let $m=f(a), n=b$, then
$$
\begin{array}{l}
f[f(a)+f(b)] \\
=b+f[f(a)+95]=a+b+f(190) . \\
\text { Also, let } m=f(a-1), n=b+1, \text { then } \\
f[f(a-1)+f(b+1)] \\
=b+1+f[f(a-1)+95] \\
=a+b+f(190) . \\
\therefore f[f(a)+f(b)] \\
=f[f(a-1)+f(b+1)] .
\end{array}
$$
From (1) we get
$$
f(a)+f(b)=f(a-1)+f(b+1) .
$$
Therefore, $\{f(n)\}$ is an arithmetic sequence.
Assume $f(n)=f(1)+(n-1) d$, and $f(n) \in$ $N, d \geqslant 0$. Then
$$
\begin{array}{l}
f(1)+[m+f(n)-1] d \\
=f[m+f(n)]=n+f(m+95) \\
=n+f(1)+(m+95-1) d . \\
\therefore[f(n)-1] d=n+94 d, \text { so } d \mid n .
\end{array}
$$
Since $n$ is any natural number, then $d=1$.
Therefore, $f(n)=n+95$ is the unique function that satisfies the equation.
|
1995
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 The function $f(x)$ is defined on the set of real numbers, and for all real numbers $x$ it satisfies the equations: $f(2+x)=f(2-x)$ and $f(x+7)=f(7-x)$. Suppose $x=0$ is a root of $f(x)=0$, and let $N$ denote the number of roots of $f(x)=0$ in the interval $[-1000,1000]$. Find the minimum value of $N$.
|
$$
\begin{array}{l}
\text { Solution: } \because f(x+2)=f(2-x), \\
\begin{array}{l}
\therefore f(x+4)=f[(x+2)+2] \\
\quad=f[2-(x+2)]=f(-x) . \\
\because f(x+7)=f(7-x), \\
\therefore f(x+14)=f[(x+7)+7]
\end{array} \\
\quad=f[7-(x+7)]=f(-x) .
\end{array}
$$
Thus, $f(x+14)=f(4+x)$,
which means $f(x+10)=f(x)$.
Therefore, $f(x)$ is a periodic function with a period of 10.
$\because x=0$ is a root of $f(x)=0$, then $x= \pm 10$, $\pm 20, \cdots, \pm 1000$ are all roots of $f(x)=0$, totaling 201.
Also, $\because f(4)=f(0+4)=f(-0)=0$, i.e., $x$ $=4$ is a root of $f(x)=0$, then $x=4 \pm 10, \cdots$, $4 \pm 10 k(-1000 \leqslant 4 \pm 10 k \leqslant 1000)$ are all roots of $f(x)$ $=0$, totaling 200.
Therefore, the minimum value of $N$ is 401.
$$
|
401
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$ .
|
Solution: Since $m \neq n$, by the definition of roots, $m, n$ are two distinct roots of the equation $x^{2}-x-1=0$. By Vieta's formulas, we have
$$
\begin{array}{l}
m+n=1, m n=-1 . \\
\because m^{2}+n^{2}=(m+n)^{2}-2 m n \\
\quad=1^{2}-2 \times(-1)=3, \\
\quad m^{3}+n^{3}=(m+n)^{3}-3 m n(m+n) \\
\quad=1^{3}-3 \times(-1) \times 1=4, \\
\therefore m^{5}+n^{5} \\
=-\left(m^{3}+n^{3}\right)\left(m^{2}+n^{2}\right)-(m n)^{2}(m+n) \\
\quad=4 \times 3-(-1)^{2} \times 1=11 .
\end{array}
$$
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given 10 points of the World Team, where 5 of these points lie on a straight line, and no three points lie on another straight line besides these, the number of distinct rays that can be drawn through any 2 of these 10 points is $\qquad$.
|
13. 78
|
78
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 The roots $x_{1}, x_{2}$ of the equation $x^{2}-a x-a=0$ satisfy the relation $x_{1}^{3}+x_{2}^{3}+x_{1}^{3} x_{2}^{3}=75$. Then $1993+5 a^{2}+$ $9 a^{4}=$ $\qquad$
|
Solution: By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=a, x_{1} x_{2}=-a . \\
\therefore x_{1}^{3}+x_{2}^{3}+x_{1}^{3} x_{2}^{3} \\
=\left(x_{1}+x_{2}\right)^{3}-3 x_{1} x_{2}\left(x_{1}+x_{2}\right)+\left(x_{1} x_{2}\right)^{3} \\
=a^{3}-3(-a) a+(-a)^{3}=3 a^{2} .
\end{array}
$$
According to the problem, $3 a^{2}=75$, so $a^{2}=25, a^{4}=625$.
$$
\begin{array}{l}
\therefore 1993+5 a^{2}+9 a^{4} \\
=1993+5 \times 25+9 \times 625=7743 .
\end{array}
$$
|
7743
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
20. For a certain commodity, if you buy 100 (including 100) pieces or less, it is settled at the retail price; if you buy 101 (including 101) pieces or more, it is settled at the wholesale price. It is known that the wholesale price is 2 yuan lower per piece than the retail price. A person originally intended to buy a certain number of pieces of this commodity, and needed to pay $a$ yuan at the retail price, but if they bought 21 more pieces, they could settle at the wholesale price, which would also be exactly $a$ yuan ($a$ is an integer), then the value of $a$ is $\qquad$
|
$\begin{array}{l}20 \\ 840\end{array}$
|
840
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, let $S=\{1,2,3,4\}, n$ terms of the sequence: $a_{1}$, $a_{2}, \cdots, a_{n}$ have the following property, for any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted as $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
|
Three, the minimum value of $n$ is 8.
First, prove that each number in $S$ appears at least 2 times in the sequence $a_{1}, a_{2}, \cdots, a_{n}$. This is because, if a number in $S$ appears only once in this sequence, since there are 3 two-element subsets containing this number, but in the sequence, the adjacent pairs containing this number can have at most two different combinations, it is impossible for all 3 two-element subsets containing this number to appear as adjacent pairs in the sequence.
Therefore, $n \geqslant 8$.
On the other hand, the 8-term sequence: $3,1,2,3,4,1,2,4$ satisfies the condition, so the required minimum value is 8.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The average score of the participants in a junior high school mathematics competition at a certain school is 75 points. Among them, the number of male participants is $80\%$ more than that of female participants, and the average score of female participants is $20\%$ higher than that of male participants. Therefore, the average score of female participants is points.
|
II. 1.84.
Let the number of female participants be $x$, then the number of male participants is $1.8 x$; Let the average score of male participants be $y$ points, then the average score of female participants is $1.2 y$ points. According to the problem, we have $\frac{1.8 x y + 1.2 x y}{x + 1.8 x} = 75$, which simplifies to $1.2 y = 84$.
|
84
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $p+q=96$, and the quadratic equation $x^{2}+p x$ $+q=0$ has integer roots. Then its largest root is
|
2. 98 .
Let the two roots be $x_{1}$ and $x_{2}$. Then $96=(x_{1}-1)(x_{2}-1)-1$, so $(x_{1}-1)(x_{2}-1)=97$. Since 97 is a prime number, then $x_{1}-1= \pm 1$, $x_{2}-1= \pm 97$, hence the largest root is 98.
|
98
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in the figure, the side lengths of $\triangle A B C$ are $A B=14, B C$ $=16, A C=26, P$ is a point on the angle bisector $A D$ of $\angle A$, and $B P \perp A D, M$ is the midpoint of $B C$. Find the value of $P M$ $\qquad$
|
3. 6 .
From the figure, take $B^{\prime}$ on $A C$ such that $A B^{\prime}=A B=14$, then $B^{\prime} C=12$. Since $\triangle A B B^{\prime}$ is an isosceles triangle, we know that the intersection point of $B B^{\prime}$ and $A D$ is $P$ (concurrency of five lines), so $P$ is the midpoint of $B B^{\prime}$.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Find the smallest natural number with the following properties:
(1) Its decimal representation ends with the digit 6;
(2) If the last digit 6 is deleted and this digit 6 is written in front of the remaining digits, then the resulting number is 4 times the original number.
(4th IM)
|
Solution: From (1), we know that the last digit of the required smallest natural number is 6. Combining (2) with the multiplication table, it is not difficult to know that its tens digit is 4, and it is also the last digit of the obtained number.
$46 \times 4=184$.
Thus, the tens digit of the obtained number should be 8, and it is also the hundreds digit of the required number.
$846 \times 4=3384$.
Thus, the hundreds digit of the obtained number should be 3, and it is also the thousands digit of the required number.
$3846 \times 4=15384$.
Thus, the thousands digit of the obtained number should be 5, and it is also the ten-thousands digit of the required number.
$53846 \times 4=215384$.
Thus, the ten-thousands digit of the obtained number should be 1, and it is also the hundred-thousands digit of the required number. Therefore,
$153846 \times 4=615384$.
In this way, the last digit 6 of the required number has appeared for the first time at the beginning of the obtained number, which meets the requirements. Such natural numbers also include $153846153846, \cdots$. Therefore, the smallest required natural number is 153846.
|
153846
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, $1 \overline{a b c d}$ is a four-digit natural number. It is known that $\overline{a b c d}-\overline{a b c}-\overline{a b}-a=1995$. Try to determine this four-digit number $\overline{a b c d}$
(1995, Beijing Middle School Advanced Mathematics Competition for Junior High School Students)
|
Solution: For easier analysis, convert the known horizontal expression into a vertical format, we get
$$
\begin{array}{r}
1000 a+100 b+10 c+d \\
-100 a-10 b-c \\
-10 a-b \\
\text { +) } \begin{array}{r}
-a
\end{array} \\
\hline 1000 \times 1+100 \times 9+10 \times 9+5
\end{array}
$$
By the subtraction rule and borrowing, it is not difficult to see that $a=2$.
From $900+200=1100$, and noting that the tens digit of the difference is 9, we can deduce that $b=2$.
Similarly, $90+20+20=130$, so we can deduce that $c=4$.
Also, $5+2+2+4=13$,
thus $d=3$.
Therefore, $\overline{a b c d}=2243$.
|
2243
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 If the digits of a four-digit number are reversed to form a new four-digit number, the new number is exactly four times the original number. Find the original number.
(1988, Nanjing Mathematical Olympiad Selection Contest)
|
Solution: Let the required four-digit number be $\overline{a b c d}$, then the new four-digit number is $\overline{d c b a}$. According to the problem, we have
$\overline{a b c d}+\overline{d c b a}=5 \overline{a b c d}$.
Obviously, the last digit of $a+d$ can only be 0 or 5.
$\because 4 a \leqslant d \leqslant 9$, and $a, d$ cannot be 0,
$\therefore a=1$ or $a=2$.
When $a=1$, $d$ can only be 4 or 9, and must satisfy
$\overline{4 c b 1}=4 \times \overline{1 b c 4}$,
$\overline{9 c b 1}=4 \times \overline{1 b c 9}$.
Considering the last digit, it is clearly not possible.
When $a=2$, $d$ can only be 3 or 8.
Obviously, $d=3$ does not hold.
When $d=8$, it must satisfy $\overline{8 c b 2}=4 \times \overline{2 b c 8}$.
Considering the last digit that can make the above equation true.
Further find $b, c$.
$\because 4 b \leqslant c \leqslant 9$ (obviously $b, c$ cannot be 0),
$\therefore b=1$ or $b=2$, and the last digit of $4 c+3$ should be the same as $b$.
When $b=2$, $c=8$. But the last digit of $4 \times 8+3$ is not 2.
When $b=1$, $c$ can be $4,5,6,7,8$. It is easy to know that the last digit of $4 \times 7 +3$ is 1. Thus,
$$
b=1, c=7 \text{. }
$$
At this point, $8712=4 \times 2178$.
Therefore, the required four-digit number is 2178.
|
2178
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 A four-digit number has the sum of the squares of its first and last digits equal to 13, and the sum of the squares of its middle two digits equal to 85. If 1089 is subtracted from this number, the result is the same four digits written in reverse order. What is the original four-digit number? $\qquad$
(1994, Dongfang Airlines Cup - Shanghai Junior High School Mathematics Competition)
|
Solution: Let the original four-digit number be $\overline{a b c d}$. According to the problem, we have
$$
\begin{array}{l}
a^{2}+d^{2}=13, \\
b^{2}+c^{2}=85, \\
\overline{a b c d}-1089=\overline{d c b a} .
\end{array}
$$
Analyzing from the first digit, it is easy to know
$$
a-d=1 \text {. }
$$
From (1) and (4), we get
$$
a=3, d=2 \text {. }
$$
Therefore, $\overline{3 b c 2}-1089=\overline{2 c b 3}$.
Performing subtraction in column form, we get
$$
b+8+1=c+10 \text {. }
$$
Thus, $b-c=1$.
From (2) and (5), we get
$$
b=7, c=6 \text {. }
$$
Therefore, the original four-digit number is 3762.
|
3762
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $x, y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{3}+1997(x-1)=-1 \\
(y-1)^{3}+1997(y-1)=1 .
\end{array}\right.
$$
Then $x+y=$ $\qquad$ (Proposed by the Problem Group)
|
$=、 1.2$.
The original system of equations is transformed into
$$
\left\{\begin{array}{l}
(x-1)^{3}+1997(x-1)=-1, \\
(1-y)^{3}+1997(1-y)=-1 .
\end{array}\right.
$$
Since $f(t)=t^{3}+1997 t$ is monotonically increasing on $(-\infty,+\infty)$, and $f(x-1)=f(1-y)$, it follows that $x-1=1-y$, i.e., $x+y=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$, a line $l$ intersects the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$ such that there are exactly 3 lines $l$, then $\lambda=$
(Proposed by the Problem Committee)
|
2. 4 .
First, note the following conclusion: For a chord passing through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ and intersecting the right branch at two points, the chord attains its minimum length $\frac{2 b^{2}}{a}=4$ if and only if the chord is perpendicular to the $x$-axis. (In fact, the polar equation of this hyperbola is $\rho=\frac{2}{1-\sqrt{3} \cos \theta}$. Suppose $A B$ is a chord passing through the right focus $F$ and intersecting only the right branch, with $A\left(\rho_{1}, \theta\right), B\left(\rho_{1}, \pi+\theta\right)$ $\left(\rho_{1}>0, \rho_{2}>0\right)$, then $|A B|=\rho_{1}+\rho_{2}=\frac{2}{1-\sqrt{3} \cos \theta}+$ $\frac{2}{1+\sqrt{3} \cos \theta}=\frac{4}{1-\sqrt{3} \cos ^{2} \theta} \geqslant 4$, with equality when $\theta=\frac{\pi}{2}$.)
Second, when there are exactly three lines satisfying the given conditions, there are only two possibilities:
(1) Only one line intersects both the left and right branches of the hyperbola, and two lines intersect only the right branch. In this case, the line intersecting both branches must be the $x$-axis, and the distance between its two intersection points is $2 a=2$. However, the length of the two chords intersecting only the right branch, $\lambda>4$, does not satisfy the given conditions;
(2) Two lines intersect both the left and right branches of the hyperbola, and only one line intersects only the right branch, and this chord must be perpendicular to the $x$-axis (otherwise, by symmetry, there would be two chords intersecting only the right branch). In this case, $|A B|=\lambda=4$, and the length of the chords intersecting both branches can also satisfy this condition, so $\lambda=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $A B C D E F$ be a regular hexagon. A frog starts at vertex $A$ and can randomly jump to one of the two adjacent vertices each time. If it reaches point $D$ within 5 jumps, it stops jumping; if it does not reach point $D$ within 5 jumps, it stops after 5 jumps. How many different jumping sequences can the frog have from the start to the stop? $\qquad$
(Provided by the Problem Group)
|
5. 26 .
As shown in the figure, it is clear that the frog cannot reach point $D$ by jumping 1 time, 2 times, or 4 times. Therefore, the frog's jumping methods are only of the following two scenarios:
(1) The frog reaches point $D$ after 3 jumps, with 2 ways to do so;
(2) The frog stops after a total of 5 jumps. In this case, the number of ways for the first 3 jumps (which definitely do not reach point $D$) is $2^{3}-2$ ways, and the number of ways for the last two jumps is $4^{2}$ ways. Therefore, the number of ways the frog can jump 5 times is $\left(2^{3}-2\right) \cdot 2^{2}=24$ ways.
From (1) and (2), we know that the frog has $2+24=26$ different ways to jump.
|
26
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 What is the smallest positive integer that can be expressed as the sum of 9 consecutive integers, the sum of 10 consecutive integers, and the sum of 11 consecutive integers?
(11th American Invitational Mathematics Examination (AIME))
|
Solution: Let the required positive integer be $A$. According to the problem, $A$ can be expressed as (where $p$, $n$, $k$ are all integers)
$$
\begin{aligned}
A & =(p+1)+(p+2)+\cdots+(p+9) \\
& =9 p+45, \\
A & =(n+1)+(n+2)+\cdots+(n+10) \\
& =10 n+55, \\
A & =(k+1)+(k+2)+\cdots+(k+11) \\
& =11 k+66 .
\end{aligned}
$$
From (1), (2), and (3) we get
$$
\begin{array}{l}
9 p=10(n+1), \\
10 n=11(k+1) .
\end{array}
$$
From (4) and (5), we know that $n$ is a multiple of 11, and when divided by 9, the remainder is 8. Therefore, the smallest positive integer for $n$ is 44.
Thus, the smallest positive integer for $A$ $=10 \times 44+55=$
495.
|
495
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (This question is worth 50 points) In a $100 \times 25$ rectangular table, each cell is filled with a non-negative real number. The number in the $i$-th row and $j$-th column is denoted as $x_{i, j} (i=1,2, \cdots, 100 ; j=1,2, \cdots, 25)$ (as shown in Table 1). Then, the numbers in each column of Table 1 are rearranged in descending order from top to bottom as
$$
\begin{array}{c}
x_{1, j}^{\prime} \geqslant x_{2, j}^{\prime} \geqslant \cdots \geqslant x_{100, j}^{\prime} \\
(j=1,2, \cdots, 25) . \text { (as shown in Table 2) }
\end{array}
$$
Find the smallest natural number $k$, such that if the numbers in Table 1 satisfy
$$
\sum_{j=1}^{25} x_{i, j} \leqslant 1(i=1,2, \cdots, 100),
$$
then when $i \geqslant k$, in Table 2 it can be guaranteed that
$$
\sum_{j=1}^{25} x_{i, j}^{\prime} \leqslant 1
$$
holds.
Table 1
\begin{tabular}{|c|c|c|c|}
\hline$x_{1,1}$ & $x_{1,2}$ & $\cdots$ & $x_{1,25}$ \\
\hline$x_{2,1}$ & $x_{2,2}$ & $\cdots$ & $x_{2,25}$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline$x_{100,1}$ & $x_{100,2}$ & $\cdots$ & $x_{100,25}$ \\
\hline
\end{tabular}
Table 2
\begin{tabular}{|c|c|c|c|}
\hline$x_{1,1}^{\prime}$ & $x_{1,2}^{\prime}$ & $\cdots$ & $x_{1,25}^{\prime}$ \\
\hline$x_{2,1}^{\prime}$ & $x_{2,2}^{\prime}$ & $\cdots$ & $x_{2,25}^{\prime}$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline$x_{100,1}^{\prime}$ & $x_{100,2}^{\prime}$ & $\cdots$ & $x_{100,25}^{\prime}$ \\
\hline
\end{tabular}
(Proposed by the Problem Committee)
|
Three, the minimum value of $k$ is 97.
(1) Take $x_{i, j}=\left\{\begin{array}{ll}0, & 4(j-1)+1 \leqslant i \leqslant 4 j \\ \frac{1}{24}, & \text { otherwise } i .\end{array}\right.$ $(j=1,2, \cdots, 25)$
At this time, $\sum_{j=1}^{25} x_{i, j}=0+24 \times \frac{1}{24} .(i=1,2, \cdots, 100)$
It satisfies the conditions of the problem. After rearrangement, we have
$$
\quad x_{i, j}^{\prime}=\left\{\begin{array}{ll}
\frac{1}{24}, & 1 \leqslant i \leqslant 96, \\
0, & 97 \leqslant i \leqslant 100 . \\
(j=1,2, \cdots, 25)
\end{array}\right.
$$
At this time, $\sum_{j=1}^{25} x_{i, j}^{\prime}=25 \times \frac{1}{24}>1 .(1 \leqslant i \leqslant 96)$
Therefore, the minimum value of $k$ $\geqslant 97$.
(2) First, we prove: In Table 1, there must be a row (let's say the $r$-th row) with all the numbers
$$
x_{r, 1}, x_{r, 2}, \cdots, x_{r, 25}
$$
which must appear in the first 97 rows of the rearranged Table 2.
In fact, if the above conclusion does not hold, then in Table 1, each row must have at least one number that does not appear in the first 97 rows of Table 2, meaning that the first 97 rows of Table 2 contain at most $100 \times 24=2400$ numbers from Table 1. This contradicts the fact that the first 97 rows of Table 2 contain $25 \times 97=2425$ numbers.
Secondly, by the rearrangement requirement, the numbers in each column of Table 2 are arranged from top to bottom in descending order, so when $i \geqslant 97$,
$$
x_{i, j}^{\prime} \leqslant x_{97, j}^{\prime} \leqslant x_{r, j} .(j=1,2, \cdots, 25)
$$
Therefore, when $i \geqslant 97$,
$$
\sum_{j=1}^{25} x_{i, j}^{\prime} \leqslant \sum_{j=1}^{25} x_{r, j} \leqslant 1 .
$$
Combining (1) and (2), the minimum value of $k$ is 97.
(Liu Yujie provided)
|
97
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 A four-digit number has the following property: dividing this four-digit number by its last two digits yields a perfect square (if the tens digit is zero, then divide by the units digit), and this perfect square is exactly the square of the first two digits plus 1. For example, $4802 \div 2=2401=49^{2}=$ $(48+1)^{2}$. The smallest four-digit number with the above property is $\qquad$
(1994, Sichuan Province Junior High School Mathematics League Final)
|
Solution: Let the first two digits of the four-digit number be $c_{1}$, and the last two digits be $c_{2}$. Then $10 \leqslant c_{1} \leqslant 99,1 \leqslant c_{2} \leqslant 99$. The four-digit number can be represented as $100 c_{1}+c_{2}$. According to the problem, we have
$$
\begin{array}{l}
100 c_{1}+c_{2}=\left(c_{1}+1\right)^{2} c_{2}=c_{1}^{2} c_{2}+2 c_{1} c_{2}+c_{2} . \\
\therefore 100 c_{1}=c_{1} c_{2}\left(c_{1}+2\right) . \\
\therefore c_{2}=\frac{100}{c_{1}+2} .
\end{array}
$$
Therefore, $\left(c_{1}+2\right) \mid 100$.
Noting that $10 \leqslant c_{1} \leqslant 99$,
$\therefore c_{1}$ can only be $18,23,48,98$.
Then $c_{2}$ is correspondingly $5,4,2,1$.
Thus, the four-digit numbers sought are $1805,2304,4802$, 9801. The smallest four-digit number among them is 1805.
|
1805
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $x=\frac{-\sqrt[1]{17}+\sqrt{\sqrt{17}+4 \sqrt{15}}}{2 \sqrt{3}}$. Then, $3 x^{4}-(2 \sqrt{15}+\sqrt{17}) x^{2}+5=$
|
2. 0 .
From the given, we have
$$
2 \sqrt{3} x+\sqrt[4]{17}=\sqrt{\sqrt{17}+4 \sqrt{15}} \text {. }
$$
Squaring and rearranging, we get
$$
\begin{array}{l}
\sqrt{3} x^{2}+\sqrt[4]{17} x-\sqrt{5}=0 . \\
\text { Also, } 3 x^{4}-(2 \sqrt{15}+\sqrt{17}) x^{2}+5 \\
=\left(3 x^{4}-2 \sqrt{3} \cdot \sqrt{5} x^{2}-5\right)-1-\sqrt{17} x^{2} \\
=\left(\sqrt{3} x^{2}-\sqrt{5}\right)^{2}-(\sqrt[4]{17} x)^{2} \\
=\left(\sqrt{3} x^{2}+\sqrt[4]{17} x-\sqrt{5}\right)\left(\sqrt{3} x^{2}-\right. \\
\sqrt[4]{17} x-\sqrt{5}) \\
\stackrel{1}{=} 0 .
\end{array}
$$
Explanation: Generally, if $x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$, then
$$
\begin{array}{l}
a^{2} x^{4}+\left(2 a c-b^{2}\right) x^{2}+c^{2} \\
=\left(a x^{2}+b x+c\right)\left(a x^{2}-b x+c\right) \\
=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that the integer $n$ is not a multiple of 5. Then the remainder when $n^{4}+4$ is divided by 5 is $\qquad$ .
|
4. 0 .
$$
\text { Given } \begin{aligned}
& n^{4}+4=\left(n^{4}-1\right)+5 \\
= & \left(n^{2}+1\right)\left(n^{2}-1\right)+5 \\
= & \left(n^{2}-4\right)\left(n^{2}-1\right)+5\left(n^{2}-1\right)+5 \\
= & (n+2)(n-2)(n+1)(n-1) \\
& +5\left(n^{2}-1\right)+5,
\end{aligned}
$$
and $n$ is not a multiple of 5, so among $n+2, n-2, n+1, n-1$, there must be one that is a multiple of 5, thus $n^{4}+4$ is also a multiple of 5. Therefore, $n^{4}+4 \equiv 0(\bmod 5)$
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. $a, b$ are two positive integers, their least common multiple is 465696. Then the number of such ordered pairs of positive integers $(a, b)$ is ( ) .
(A) 144
(B) 724
(C) 1008
(D) 1155
|
4. (D).
From $[a, b]=465696=2^{5} \cdot 3^{3} \cdot 7^{2} \cdot 11$, we know that
$$
a=2^{a_{1}} \cdot 3^{a_{2}} \cdot 7^{a_{3}} \cdot 11^{\alpha_{4}}, b=2^{\beta_{1}} \cdot 3^{\beta_{2}} \cdot 7^{\beta_{3}} \cdot 11^{\beta_{4}},
$$
where $\alpha_{i}$ and $\beta_{i}$ are non-negative integers, and
$$
\begin{array}{l}
\max \left(\alpha_{1}, \beta_{1}\right)=5, \max \left(\alpha_{2}, \beta_{2}\right)=3, \\
\max \left(\alpha_{3}, \beta_{3}\right)=2, \max \left(\alpha_{4}, \beta_{4}\right)=1 .
\end{array}
$$
It is easy to see that the pairs $\left(\alpha_{1}, \beta_{1}\right)$ have $2 \times 5+1$ possibilities, $\left(\alpha_{2}, \beta_{2}\right)$ have $2 \times 3+1$ possibilities, $\left(\alpha_{3}, \beta_{3}\right)$ have $2 \times 2+1$ possibilities, and $\left(\alpha_{4}, \beta_{4}\right)$ have $2 \times 1+1$ possibilities.
Therefore, the number of ordered pairs of positive integers $(a, b)$ is
$$
\begin{array}{l}
(2 \times 5+1)(2 \times 3+1)(2 \times 2+1)(2 \times 1+1) \\
=1155 .
\end{array}
$$
|
1155
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. For a cube wooden block $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ with an edge length of 1, take points $P, Q, R$ on the three edges passing through vertex $A_{1}$, such that $A_{1} P=A_{1} Q=A_{1} R$. After cutting off the tetrahedron $A_{1}-P Q R$, use the section $\triangle P Q R$ as the base to drill a triangular prism-shaped hole in the cube, making the sides of the prism parallel to the body diagonal $A_{1} C$. When the hole is drilled through, vertex $C$ is cut off, and the exit is a spatial polygon. How many sides does this spatial polygon have?
(A) 3
(B) 6
(C) 8
(D) 9
|
6. (B).
Through point $R$, a line parallel to $A_{1} C$ intersects $A C$ at a point $R^{\prime}$. Through point $Q$, a line parallel to $A_{1} C$ intersects $B_{1} C$ at a point $Q^{\prime}$. A plane through $R Q$ and parallel to $A_{1} C$ intersects the side face of the triangular prism at a point $C_{0}$ on the edge $C_{1} C$. Thus, the plane through $R Q$ and parallel to $A_{1} C$ cuts out two segments $R^{\prime} C_{0}$ and $Q^{\prime} C_{0}$ near point $C$. Similarly, planes through $P Q$ and $R P$ and parallel to $A^{\prime} C$ each cut out two segments near point $C$.
Therefore, the spatial polygon cut out at the exit has a total of 6 sides.
|
6
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $n=\underbrace{111 \cdots 11}_{1999 \uparrow 1}, f(n)=90 n^{2000}+20 n+$ 1997. Then the remainder when $f(n)$ is divided by 3 is
|
Ni.1.1.
A natural number $a$ has the same remainder when divided by 3 as its digit sum $S(a)$ when divided by 3. Therefore, $n=\underbrace{11 \cdots 111}_{1999 \uparrow 1}$ has the same remainder when divided by 3 as 1999, which is 1.
$90 n^{2000}$ has a remainder of 0 when divided by 3, and $20 n$ has a remainder of 2 when divided by 3.
Thus, the remainder of $f(n)=90 n^{2000}+20 n+1997$ when divided by 3 is the same as the remainder of $2+1997=1999$ when divided by 3, which is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given that $a$, $b$, and $c$ are the lengths of the three sides of a right triangle, and for a natural number $n$ greater than 2, the following holds:
$$
\left(a^{n}+b^{n}+c^{n}\right)^{2}=2\left(a^{2 n}+b^{2 n}+c^{2 n}\right) .
$$
Then $n=$
|
6. 4 .
Let $x=a^{\frac{n}{2}}, y=b^{\frac{n}{2}}, z=c^{\frac{\pi}{2}}$, then
$$
\begin{aligned}
0= & 2\left(a^{2 n}+b^{2 n}+c^{2 n}\right)-\left(a^{n}+b^{n}+c^{n}\right)^{2} \\
= & 2\left(x^{4}+y^{4}+z^{4}\right)-\left(x^{2}+y^{2}+z^{2}\right)^{2} \\
= & x^{4}+y^{4}+z^{4}-2 x^{2} y^{2}-2 x^{2} z^{2}-2 y^{2} z^{2} \\
= & -(x+y+z)(x+y-z)(y+z-x)(z+x \\
& -y) .
\end{aligned}
$$
Assume $c$ is the hypotenuse, then $z>x, z>y$. It follows that
$$
x+y+z>0, y+z-x>0, z+x-y>0 \text {. }
$$
$\therefore(*)$ is equivalent to $z=x+y$,
which means $\left(\frac{a}{c}\right)^{\frac{n}{2}}+\left(\frac{b}{c}\right)^{\frac{n}{2}}=1$.
On the other hand, $a^{2}+b^{2}=c^{2}$ holds, or $\left(\frac{a}{c}\right)^{2}+\left(\frac{b}{c}\right)^{2}=$ 1.
Since $0<\frac{a}{c}<1,0<\frac{b}{c}<1, y=\left(\frac{a}{c}\right)^{x}+\left(\frac{b}{c}\right)^{x}$ is a monotonically decreasing function, and it takes $y=1$ at only one point $x$, therefore, $\frac{n}{2}=2, n=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six. (12 points) On the hypotenuse $AB$ of the right triangle $\triangle ABC$, color the points $P$ that satisfy $PC^{2} = PA \cdot PB$ in red. How many red points are there at least, and at most, on the hypotenuse?
|
Six, as shown, let $B P=x$, then $A P=c-x$. Draw the altitude $C H$ on the hypotenuse, then $C H=\frac{a b}{c}, B H$ $=\frac{a^{2}}{b}$. Therefore,
$$
\left.A P=\frac{a^{2}}{c}-x \right| \, \text { ( } P \text { can be between } A \text { and } Y \text {, also please consider } H \text { and }
$$
$B$ points.
In the right triangle $\triangle C$, we have
$$
\begin{aligned}
P C^{2} & =C H^{2}+H P^{2} \\
& =\frac{a^{2} b^{2}}{c^{2}}=\left(\frac{a^{2}}{c}-x\right)^{2} \\
& =x^{2}-\frac{2 a^{2}}{c} x+a^{2} .
\end{aligned}
$$
But from the given, we also have
$$
P C^{2}=P A \cdot P B=x(c-x) .
$$
From (1) and (2), we get
$$
x^{2}-\frac{2 a^{2}}{c} x+a^{2}=x(c-x) \text {, }
$$
which simplifies to $2 x^{2}-\frac{2 a^{2}+c^{2}}{c} x+a^{2}=0$.
Factoring gives $(2 x-c)\left(x-\frac{a^{2}}{c}\right)=0$.
Thus, $x_{1}=\frac{c}{2}$ (the midpoint of the hypotenuse),
$x_{2}=\frac{a^{2}}{c}$ (the foot of the altitude on the hypotenuse).
Since a quadratic equation has at most two real roots, the hypotenuse has at most two red points; when the right triangle $\triangle A B C$ is isosceles, the midpoint and the foot of the altitude coincide, so the hypotenuse has at least one red point.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
21. A sequence of integers $a_{0}, a_{1}, \cdots, a_{n}$ is called a 二次 sequence if for each $i \in\{1,2, \cdots, n\}$, the equation
$$
\left|a_{i}-a_{i-1}\right|=i^{2} \text {. }
$$
holds.
(a) Prove that for any integers $b$ and $c$, there exists a natural number $n$ and a 二次 sequence such that $a_{0}=b, a_{n}=c$. A 二次 sequence, and in this sequence $a_{0}=0, a_{n}=1996$.
|
Solution: (a) Since $a_{i}-a_{i-1}= \pm i^{2}, 1 \leqslant i \leqslant n$,
and $c-b=a_{n}-a_{0}=\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)=\sum_{i=1}^{n}\left( \pm i^{2}\right)$.
Let $d=c-b$. We only need to prove that for any integer $d$, there exists a natural number $n$, such that we can make:
“ $\square 1^{2} \square 2^{2} \square \cdots \square n^{2} "$
fill “ $\square$ ” appropriately with “+” or “-”, so that the calculation result equals $d$.
For any natural number $k$, we have
$$
\begin{array}{l}
n^{2}-(n+1)^{2}-(n+2)^{2}+(n+3)^{2}=4, \\
-n^{2}+(n+1)^{2}+(n+2)^{2}-(n+3)^{2}=-4 .
\end{array}
$$
Therefore, for any “ $\square 1^{2} \square 2^{2} \square \cdots \square n^{2} $” with any consecutive 4 terms, we can discuss in 4 cases.
(i) $d \equiv 0(\bmod 4)$, in this case, take $n=4 \cdot \frac{|d|}{4}$, which clearly satisfies the requirement.
(If $n \in N$, if $d=0$, we can take $n=8$ so that the sum of the first 4 terms is 4, and the sum of the next 4 terms is -4)
(ii) $d \equiv 1(\bmod 4)$, take $n=1+|d-1|$, and fill the sign before $1^{2}$ to make it equal to $d$.
(iii) $d \equiv 2(\bmod 4)$, take $n=3+|d-14|$, making the first 3 terms “$+1^{2}+2^{2}+3^{2}$”, and the next $4 \cdot \frac{|d-14|}{4}$ terms sum to $d-14$. This proves that the sum of $n$ terms is $d$.
(iv) $d \equiv 3(\bmod 4)$, take $n=1+|d+11|$, and fill the sign before $1^{2}$ to make it equal to $d$.
(b) If $n \leqslant 17$, then
$$
\begin{aligned}
a_{m} & =\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)+a_{0} \\
& \leqslant \sum_{i=1}^{n}\left|a_{i}-a_{i-1}\right|+a_{0} \\
& =\frac{n(n+1)(2 n+1)}{6} \\
& \leqslant \frac{17 \times 18 \times 35}{6}=1785,
\end{aligned}
$$
which is impossible.
If $n=18$, then
$$
\begin{aligned}
a_{n} & =\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)+a_{6} \\
& =\sum_{i=1}^{n}\left|a_{i}-u_{i-1}\right|(\bmod 2) \\
& \equiv \sum_{i=1}^{18} i^{2} \equiv 1(\bmod 2) .
\end{aligned}
$$
Thus, $a_{n}$ cannot be 1996. Therefore, $n \geqslant 19$.
If $1996=1^{2}+2^{2}+\cdots+19^{2}$
$$
-2\left(1^{2}+\dot{2}^{2}+6^{2}+14^{2}\right) .
$$
If $i=1,2,6,14$, then $a_{i}-a_{i-1}=-i^{2}$, and for the remaining $1 \leqslant i \leqslant 19, a_{i}-a_{i-1}=i^{2}$.
When $a_{0}=0$, $a_{19}=1996$.
Therefore, the smallest natural number that satisfies the condition is $n=19$.
|
19
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
2. In the Cartesian coordinate system, points of the form $\left(m, n^{2}\right)$ are painted red (where $m, n$ are integers), referred to as red points, and their surrounding points are not colored. Then, the parabola $y=x^{2}-$ $196 x+9612$ has $\qquad$ red points.
|
2. 2 .
Let $\left(m, n^{2}\right)$ be on the parabola $y=x^{2}-196 x+9612$, then $n^{2}=m^{2}-196 m+9612$.
Completing the square and factoring, we get
$$
\begin{array}{l}
(n+m-98)(n-m+98) \\
=8=2 \times 4=(-2) \times(-4) .
\end{array}
$$
$\because m, n$ are integers,
$\therefore n+m-98$ and $n-m+98$ are of the same parity.
When $n+m-98$ takes $-4,-2,2,4$ in turn, $n-m+98$ takes $-2,-4,4,2$ in turn. Therefore, $\left(m, n^{2}\right)=(97,9)$ or $(99,9)$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given, as shown in the figure, a semicircle $O$ with a diameter of $20 \mathrm{~cm}$ has two points $P$ and $Q$, $P C \perp A B$ at $C, Q D$ $\perp A B$ at $D, Q E \perp$ $O P$ at $E, A C=4 \mathrm{~cm}$. Then $D E=$
|
$3.8 \mathrm{~cm}$.
Take the midpoint $M$ of $O P$ and the midpoint $N$ of $O Q$, and connect $C M, D N$, and $E N$. Then
$$
\begin{array}{c}
M C=P E=\frac{1}{2} O P \\
=\frac{1}{2} O Q=E N=D N, \\
\angle P M C=\angle M C O+ \\
\angle M O C=2 \angle M O C .
\end{array}
$$
Since $O, D, Q, E$ are concyclic, with $N$ as the center,
$$
\begin{array}{l}
\therefore \angle D N E=2 \angle D Q E=2 \angle M O C=\angle P M C . \\
\therefore \triangle P C M \cong \triangle D E N .
\end{array}
$$
Thus $D E=P C=\sqrt{O P^{2}-O C^{2}}$
$$
=\sqrt{10^{2}-(10-4)^{2}}=8 .
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
一、(满分 20 分) Determine the smallest positive integer $n$, whose last digit is 7, such that if the last digit 7 is moved to the front, the resulting number is 5 times the original number.
|
Let $n=\overline{a_{k} a_{k-1} \cdots a_{2} a_{1}}, x=\overline{a_{k} a_{k-1} \cdots a_{3} a_{2}}$, where $a_{1}$ $=7, k$ is a natural number greater than 1. Then
$$
\begin{aligned}
n & =\overline{a_{k} a_{k-1} \cdots a_{2}} \times 10+7=10 x+7, \\
5 n & =\overline{7 a_{k} a_{k-1} \cdots a_{3} a_{2}} \\
& =7 \times 10^{k-1}+\overline{a_{k} a_{k-1} \cdots a_{3} a_{2}} \\
& =7 \times 10^{k-1}+x .
\end{aligned}
$$
Eliminating $n$ from the above two equations, we get
$$
5(10 x+7)=7 \times 10^{k-1}+x .
$$
Thus, $x=\frac{10^{k-1}-5}{7}$.
When $k$ is the smallest, $x$ is the smallest, and $n$ is also the smallest.
When $k$ takes $2,3,4,5$, $x$ is not a natural number.
When $x=6$, $x=14285$.
Therefore, the smallest positive integer $n$ is 142857:
|
142857
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (Full marks 25 points) Does there exist 100 different lines in the plane, such that their intersection points total 1998?
|
Three, there are 1998 intersection points.
First, divide 98 lines into two sets of parallel lines, one set has $k$ lines, and the other set has $(98-k)$ lines. The number of intersection points is $k(98-k)$. When the value of the positive integer $1998-k(98-k)$ is minimized, $k=26$ or 72.
When $k=26$ or 72, $k(98-k)=26 \times 72=1872$ intersection points. This leaves $1998-1872=126$ more intersection points needed.
Now add two parallel lines, and construct them appropriately so that they intersect the first 98 lines at 126 points different from the initial 1872 points. The specific construction is as follows:
In the coordinate system xoy, take the lines $x=0, x=1$, $x=2, \cdots, x=71, y=0, y=1, y=2, \cdots, y=25$. These 98 lines have 1872 intersection points.
The 99th line $y=x+8$ intersects the first 98 lines at 80 points: $A_{0}(-8,0), A_{1}(-7,1), A_{2}(-6,2), \cdots$, $A_{7}(-1,7), A_{8}(0,8), A_{9}(1,9), A_{10}(2,10), \cdots, A_{25}(17,25), A_{25}(18,26), \cdots, A_{79}(71,79)$, where points $A_{8}$, $A_{9}, A_{10}, \cdots, A_{25}$, these 18 points are among the initial 1872 intersection points, i.e., they are repeated intersection points.
The 100th line, parallel to the 99th line, $y=x+9$ intersects the first 98 lines at 81 points: $B_{0}(-9,0), B_{1}(-8, 1), B_{2}(-7,2), \cdots, B_{8}(-1,8), B_{9}(0,9), B_{10}(1,10)$, $\cdots, B_{25}(16,25), B_{25}(17,26), B_{27}(18,27), \cdots, B_{80}(71, 80)$. Points $B_{9}, B_{10}, B_{11}, \cdots, B_{25}$, these 17 points are among the initial 1872 intersection points, i.e., they are repeated intersection points (as shown in the figure).
Thus, the number of intersection points of these 100 lines is:
$$
1872+(80-18)+(81-17)=1998(\text{ points }).
$$
|
1998
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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