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Example 2. If $x, y$ are real numbers, and
$$
y=\frac{\left(1-x^{2}\right)^{\frac{1}{2}}+\left(x^{2}-1\right)^{\frac{1}{6}}}{4 x-5} \text {, }
$$
find the value of $\log _{\frac{1}{7}}(x+y)$.
|
To make the expression for $y$ meaningful, $x$ must simultaneously satisfy the following three inequalities.
$$
\begin{array}{c}
1-x^{2} \geqslant 0, \\
x^{2}-1 \geqslant 0, \\
4 x-5 \neq 0 . \\
\therefore x=1, \quad \text { thus } y=0 . \\
\therefore \log _{\frac{1}{7}}(x+y)=\log _{\frac{1}{7}} 1=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, find the smallest positive integer $n$ (where $n>1$) such that the square mean of the first $n$ natural numbers is an integer.
(Note: The square mean of $n$ numbers $a_{1}, a_{2}, \cdots, a_{i}$ is defined as $\left.\left[\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}\right]^{\frac{1}{2}}\right)$
|
$$
\begin{array}{c}
\text { Three, let } \frac{1^{2}+2^{2}+\cdots+n^{2}}{n} \\
=\frac{n(n+1)(2 n+1)}{6 n}=\frac{(n+1)(2 n+1)}{6} \text { be a }
\end{array}
$$
perfect square $m^{2}$, then $6 m^{2}=(n+1)(2 n+1)$. Since $6 m^{2}$ is even, it must be odd. We can set $n=3 p+q$, where $q=-1, 1$ or 3.
If $q=2$, then $6 m^{2}=(6 p+4)(12 p+7)$ $=72 p^{2}+90 p+28$. Since $6|72, 6| 90$, but $6 \nmid 28$, this case is impossible.
If $q=-1$, then $6 m^{2}=(6 p)(12 p-1)$, i.e., $m^{2}=p(12 p-1)$. Since $(p, 12 p-1) = 1$, both $p$ and $12 p-1$ must be perfect squares. Let $p=s^{2}$, $12 p-1=t^{2}$, we get $t^{2}-12 s^{2}=-1$. Since $t^{2} \equiv 0$ (mod 3), this case is also impossible.
If $q=1$, then $6 m^{2}=(6 p+2)(12 p+3)$, i.e., $m^{2}=(3 p+1)(4 p+1)$. Since $(3 p+1, 4 p+1)=1$, both $3 p+1=u^{2}$ and $4 p+1=v^{2}$, we get $3 v^{2}+1=4 u^{2}$. Clearly, $u=v=1$ is a solution, but it leads to $p=0, m=1, n=1$, which does not meet the requirement. To find a solution for $n>1$, by trying integer values of $u$ from 2 to 12, $v$ has no integer value, but when $u=13$, $v=15$. Therefore, the smallest solution is $p=56, m=195, n=337$. The answer is $n=337$.
|
337
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\therefore 、 m$ distinct positive even numbers and $n$ distinct positive odd numbers have a total sum of 1987, for all such $m$ and $n$, what is the maximum value of $3 m+4 n$? Please prove your conclusion.
The sum of $m$ distinct positive even numbers and $n$ distinct positive odd numbers is 1987, for all such $m$ and $n$, what is the maximum value of $3 m+4 n$? Please prove your conclusion.
|
The maximum value of $3 m+4 n$ is 221. The proof is as follows: Let $a_{1}+\cdots+a_{n}+b_{1}+\cdots+b_{n}=1987$. Here, $a_{i} (1 \leqslant i \leqslant m)$ are distinct positive even numbers, and $b_{j} (1 \leqslant j \leqslant n)$ are distinct positive odd numbers. Clearly, $u$ is always an odd number, and
$$
\begin{array}{l}
a_{1}+\cdots+a_{\mathrm{m}} \geqslant 2+4+\cdots+2 m=m(m+1), \\
b_{1}+\cdots+b_{n} \geqslant 1+3+\cdots+(2 n-1)=n^{2} . \\
\therefore m^{2}+m+n^{2} \leqslant 1987 \text {. (where } n \text { is odd) }
\end{array}
$$
This inequality is equivalent to $\left(m+\frac{1}{2}\right)^{2}+n^{2} \leqslant 1987+\frac{1}{4}$. By the Cauchy-Schwarz inequality,
$$
\begin{array}{c}
3\left(m+\frac{1}{2}\right)+4 n \leqslant \sqrt{3^{2}+4^{2}} \\
\cdot \sqrt{\left(m+\frac{1}{2}\right)^{2}+n^{2}} \\
\leqslant 5 \sqrt{1987+\frac{1}{4}} \\
\therefore \quad 3 m+4 n \leqslant\left[\left(5 \cdot \sqrt{1987+\frac{1}{4}}-\frac{3}{2}\right)\right] .
\end{array}
$$
where $[x]$ denotes the integer part of $x$.
Thus, $3 m+4 n \leqslant 221$.
On the other hand, when $m=27, n=35$, $m^{2}+m+n^{2} \leqslant 1987$ and $3 m+4 n=221$. Therefore, for $m$ and $n$ satisfying the conditions of the proposition, the maximum value of $3 m+4 n$ is 221.
|
221
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3 . Find the integers $x, y$ that satisfy the equation $2 x-3 y=17$, and for which $x^{2}+y^{2}$ is minimized, as well as the minimum value of $x^{2}+y^{2}$.
|
3. Answer: $x=4, y=-3, x^{2}+y^{2}$ has a minimum value of 25.
Hint: The general solution of $2 x-3 y=17$ is $x=3 t+7$, $y=2 t-1$. Substituting into $x^{2}+y^{2}$, we get $x^{2}+y^{2}$ $=13\left(t+\frac{19}{13}\right)^{2}+\frac{289}{13}$. The integer $t$ closest to $t=-\frac{19}{13}$ is -1.
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Indicate one method for each of the following to divide a square with a side length of 1986 mm:
(1) Divide it into 1986 smaller squares with integer side lengths;
(2) Divide it into 1986 smaller squares with not all side lengths being integers.
|
(1) $\because 1986=2 \times 993$, divide the two sides $A B$ and $A D$ of the square into 993 equal parts (each part being $2 \text{ mm}$). As shown in Figure 5, using the discussion from Proof 1, we get: $2 \times 993-1=1985$ small squares with a side length of $2 \text{ mm}$, and one square $F C E E_{1}$ with a side length of $1984 \text{ mm}$. This results in 1986 integer-sided squares, and it is clear that this division method meets the requirements.
( 2) $\because 1986=662 \times 3$ is a $3k$ type number, referring to the discussion of $M_{2}$ in Proof 2: $1^{\circ}$ Perform one $B$-type division in the square, resulting in 5 smaller squares with a side length of $662 \text{ mm}$, and one square with a side length of $1324 \text{ mm}$, totaling 6 smaller squares.
$2^{\circ}$ Perform 662-2=660 $A$-type divisions. Since each $A$-type division reduces the side length of the resulting squares by half, after performing 660 $A$-type divisions in one of the 6 existing squares, the side length of the smallest square is $\frac{1}{2^{660}}$ of the original square's side length. Since $\frac{662}{2^{660}}$ and $\frac{1324}{2^{660}}$ are not integers, the result must include some non-integer-sided squares. At this point, one square is divided into $3 \times 660+1=1981$ smaller squares. Considering that one small square is used up during the $A$-type division, combining $1^{\circ}$ and $2^{\circ}$, we get: a square with a side length of $1986 \text{ mm}$ is divided into $6+1981-1=1986$ smaller squares, not all of which have integer sides.
|
1986
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. Given a three-digit integer that is a multiple of 5, the sum of its digits is 20, and the sum of one digit and the hundreds digit is a multiple of 3, find this integer.
|
Solution: Let the required three-digit integer be
$$
N=100 x+10 y+z \text {. }
$$
Since $N$ is a multiple of 5, it must be that $z=0$ or $z=5$. It is also given that the sum of the digits of $N$ is 20, so
$$
x+y+z=20 \text { . }
$$
If $z=0$, then $x+y=20$.
Given $1 \leqslant x \leqslant 9,0 \leqslant y \leqslant 9$, we have $1 \leqslant x+y \leqslant 18$,
which contradicts (1). Therefore, only $z=5$, in which case $x+y=15, y=15$ $-x \leqslant 9,6 \leqslant x \leqslant 9$. It is also given that $x+z=x+5$ is a multiple of 3, so $x=7$, and thus $y=8$.
Therefore, the required three-digit integer is 785.
|
785
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example. The infinite sequence
$$
a_{1}, a_{2}, a_{3}, \ldots
$$
has the relation
$$
a_{n}=4+\frac{1}{3} u_{n-1}(n=2,3,1, \cdots)
$$
Find $\lim _{n \rightarrow \infty} a_{n}$.
|
$$
\begin{array}{l}
a_{2}=4+\frac{1}{3} a_{1}, \\
a_{3}=4+\frac{1}{3} a_{2}=4+\frac{1}{3}\left(4+\frac{1}{3} a_{1}\right) \\
=4+4 \cdot \frac{1}{3}+\left(\frac{1}{3}\right)^{2} a_{1} \\
a_{4}=4+\frac{1}{3} a_{3} \\
=4+4 \cdot \frac{1}{3}+4\left(\frac{1}{3}\right)^{2}+\left(\frac{1}{3}\right)^{2} a_{1} \\
a_{n}=4+4 \cdot \frac{1}{3}+4\left(\frac{1}{3}\right)^{2}+\cdots \\
+4\left(\frac{1}{3}\right)^{n-2}+\left(\frac{1}{3}\right)^{n-1} a_{1} \\
\therefore \lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty}\left\{4+4 \cdot \frac{1}{3}\right. \\
+4\left(\frac{1}{3}\right)^{2}+\cdots \\
\left.+4\left(\frac{1}{3}\right)^{n-2}\right\} \\
+\lim _{n \rightarrow}\left(\frac{1}{3}\right)^{n-1} a_{1}=6. \\
\end{array}
$$
Above, by simplifying $a_{2}, a_{3}, a_{4}, \cdots$ successively, we find $a_{n}$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Lake 1. Given the arithmetic sequence with the first term $a_{1}=3, a_{n+1}=a_{n}+5$, find up to the 5th term.
|
Solution: From the recursive formula $a_{2}=a_{1}+5=3+5=8$,
$$
\begin{array}{l}
a_{3}=a_{2}+5=8+5=13, \\
a_{4}=a_{3}+5=13+5=18, \\
a_{5}=a_{4}+5=18+5=23 .
\end{array}
$$
From the recursive formula to find the general term, there are commonly four forms:
I $\cdot a_{n+1}=a_{n}+d$,
Arithmetic sequence type.
|
23
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Find the smallest positive integer $n$ that makes $\frac{n-13}{5 n+6}$ a non-zero reducible fraction.
|
Solution: Since $(1,5)=1,|1 \times 6-(-13) \times 5|$ $=71 \neq(1,5),(1,71)=1$, the fraction can be simplified by 71. From $\left.n-13=71 m_{1}, n=71 m_{1}+13\left(m_{1} \in Z\right)\right)$, and $n-13>0$, so we take $m_{1}=1$, obtaining the smallest $n=84$.
|
84
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
{\left[\frac{1^{2}}{1980}\right],\left[\frac{2^{2}}{1980}\right],\left[\frac{3^{2}}{1980}\right], \cdots,} \\
{\left[\frac{1900^{2}}{1980}\right] \text { How many different numbers are there in the sequence? }}
\end{array}
$$
|
Solution: First, note that when $\alpha-\beta>1$, the value of $[a] 1_{j} [\beta]$ is definitely different. Therefore, we solve the inequality
$$
\frac{(k+1)^{2}}{1980}-\frac{k^{2}}{1980}>1,
$$
which simplifies to
$$
\begin{array}{l}
2 k+1>1980, \\
k>989 .
\end{array}
$$
Thus, starting from the 990th term, these 1980 terms, i.e.,
$$
\left[\frac{990^{2}}{1980}\right],\left[\frac{991^{2}}{1980}\right], \cdots,\left[\frac{1980^{2}}{1980}\right]
$$
are all different, totaling 991 distinct numbers.
On the other hand, when $k \leqslant 44$,
$$
\begin{array}{l}
0<\frac{k^{2}}{1980}<1, \\
{\left[\frac{k^{2}}{1980}\right]=0 .}
\end{array}
$$
Thus, the first 44 numbers are all 0.
When $45 \leqslant k \leqslant 62$,
$$
\begin{array}{l}
1<\frac{k^{2}}{1980}<2, \\
{\left[\frac{k^{2}}{1980}\right]=1 .}
\end{array}
$$
Therefore, the first 62 numbers can only be 0 or 1.
Also, since
$$
\left[\frac{989^{2}}{1980}\right]=494, \quad\left[\frac{990^{2}}{1980}\right]=495,
$$
the first 989 terms must include the 495 distinct numbers $0,1,2, \cdots, 494$.
Thus, there are $991+495=1486$ distinct numbers.
|
1486
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Given $\sin \alpha+\cos \alpha=a$.
(1) Find the value of $\sin ^{5} \alpha+\cos ^{5} \alpha$;
(2) If $a=1$, find the value of $\sin ^{n} \alpha+\cos ^{n} \alpha$.
|
Let $f(n)=\sin ^{n} \alpha+\cos ^{n} \alpha$, then (1) $=a$.
Given $f(2)=\sin ^{2} \alpha+\cos ^{2} \alpha=1$,
$f(3)=a f(2)-\frac{a^{2}-1}{2} \cdot a$
$=\frac{-a^{3}+3 a}{2}$,
$f(4)=a f(3)-\frac{a^{2}-1}{2} f(2)$
$=-a^{4}+2 a^{2}+1$
2
$\therefore f(5)=\sin ^{5} \alpha+\cos ^{5} \alpha$
$=a f(4)-\frac{a^{2}-1}{2} f(3)$
$=\frac{-a^{5}+5 a}{4}$
(2) $\because \sin \alpha+\cos \alpha=1$,
$\therefore \sin \alpha \cdot \cos \alpha=0$
$\therefore f(n)=\sin ^{n} \alpha+\cos ^{n} \alpha=(\sin \alpha$
$+\cos \alpha) f(n-1)-(\sin \alpha \cdot \cos \alpha) f(n-2)$
$=f(n-1)$
Similarly, $f(n-1)=f(n-2)=\cdots=f(1)$ $=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Learn, there are 2 teachers who can teach mathematics, and there are 4 teachers who can teach both English and Japanese. Now, 3 mathematics teachers and 3 Japanese teachers are being dispatched to teach outside the school during the holiday. How many ways are there to select them?
保留源文本的换行和格式，直接输出翻译结果。
|
Let the 3 teachers competent in English teaching be set $A$, 2 teachers competent in Japanese teaching be set $B$, and 4 teachers competent in both English and Japanese teaching be set $C$.
Method 1: Classify set $A$
(1) Select 3 English teachers (choose 3 Japanese teachers from $B$ and $C$), total $C_{3}^{3} \cdot C_{8}^{3} = 20$ (ways);
(2) Select 2 English teachers (then choose 1 English teacher from $C$, and 3 Japanese teachers from $B$ and $C$),
total $C_{3}^{2} \cdot C_{3}^{1} \cdot C_{5}^{3} = 120$ (ways);
(3) Select 1 English teacher and 2 English teachers from $C$ (choose 3 Japanese teachers from the remaining 4 in $B$ and $C$),
total $C_{3}^{1} \cdot C_{4}^{2} \cdot C_{4}^{3} = 72$ (ways);
(4) Do not select any teachers from $A$ (only choose 3 English teachers from $C$, and 3 Japanese teachers from the remaining 3 in $B$ and $C$),
total $C_{4}^{3} \cdot C_{3}^{3} = 4$ (ways).
According to the principle of addition: the total number is $20 + 120 + 72 + 4 = 216$ (ways).
Method 2: Classify set $B$
(1) Select 2 teachers to teach Japanese (then choose 1 teacher from $C$ to teach Japanese, and 3 teachers from the remaining 6 in $A$ and $C$ to teach English),
total $C_{2}^{2} \cdot C_{4}^{1} \cdot C_{6}^{3} = 80$ (ways);
(2) Select 1 teacher to teach Japanese (then choose 2 teachers from $C$ to teach Japanese, and 3 teachers from the remaining 5 in $A$ and $C$ to teach English),
total $C_{2}^{1} \cdot C_{4}^{2} \cdot C_{5}^{3} = 120$ (ways);
(3) Do not select any teachers from $B$ (only choose 3 teachers from $C$ to teach Japanese, and 3 teachers from the remaining 4 in $A$ and $C$ to teach English),
total $C_{4}^{3} \cdot C_{4}^{3} = 16$ (ways).
According to the principle of addition: the total number is $80 + 120 + 16 = 216$ (ways).
This problem can also be solved by classifying set $C$.
In permutation and combination problems, errors such as missing solutions or overlaps often occur, or the conditions may seem messy and difficult to handle. One reason for this is the inability to correctly classify the objects involved in the problem.
(Author's unit: Li Jiashen High School, Baodi County)
|
216
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11. Find the number of consecutive zeros at the end of 1987!.
untranslated text:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
translated text:
Example 11. Find the number of consecutive zeros at the end of 1987!.
Note: The note at the end is not part of the original text and is provided for context.
|
To solve this problem, it is equivalent to finding the exponent of $1 \mathrm{C}$ in 1987!, which is the same as finding the exponent of 5 in 1987!, since
$$
\begin{array}{l}
{\left[\frac{1987}{5}\right]+\left[\frac{1987}{5^{2}}\right]+\left[\begin{array}{c}
1987 \\
5^{3}
\end{array}\right]} \\
+\left[\frac{1987}{5^{4}}\right]=494,
\end{array}
$$
thus 1987! ends with exactly 494 zeros.
|
494
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Let $x, y$ satisfy $3 x^{2}+2 y^{2}=6 x$, find the maximum value of $x^{2}+y^{2}$:
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
|
If not analyzed, from the known conditions:
$$
\begin{array}{l}
x^{2}+y^{2}=-\frac{1}{2} x^{2}+3 x \\
=-\frac{1}{2}(x-3)^{2}+\frac{9}{2},
\end{array}
$$
it would be incorrect to say that when $x=3$, $x^{2}+y^{2}$ achieves its maximum value $\frac{9}{2}$. This is because from $y^{2}=-\frac{3}{2} x^{2}+3 x \geqslant 0$, we get $0 \leqslant x \leqslant 2$.
The correct answer is: when $x=2$, $x^{2}+y^{2}$ achieves its maximum value 4.
Through geometric analysis, the error and the correct reasoning can be clearly explained: the constraint condition is transformed into
$$
(x-1)^{2}+\frac{y^{2}}{3 / 2}=1 \text{. }
$$
This is an ellipse. The maximum value we are seeking is the square of the radius of the largest circle
$$
x^{2}+y^{2}=R^{2}
$$
that has a common point with this ellipse. The point on the ellipse farthest from the origin $O(0, 0)$ is $(2,0)$, and the point $(3, y)$ is outside the ellipse.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Find the maximum distance between two points, one on the surface of a sphere centered at $(-2$, $-10,5)$ with a radius of 19, and the other on the surface of a sphere centered at $(12,8,-16)$ with a radius of 87.
|
2. Let $O$ and $O_{1}$ be the centers of two spheres, and $P, P_{1}$ be the intersection points of the extended line segment $O_{1}$ with the two spherical surfaces, such that $O$ is inside $P O_{1}$ and $O_{1}$ is inside $O P_{1}$. Clearly, the maximum distance between these two points is $P P_{1}=P O+O O_{1}+O_{1} P_{1}=19+31+87=137$.
(Where $\left.O O_{1}=\sqrt{ }[12-(-2)] 2+[8-(-10)] 21\right)-(16-5)^{2}=\sqrt{961}=31$ ).
|
137
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. A natural number greater than 1, if it is exactly equal to the product of its distinct proper divisors (factors excluding 1 and itself), then it is called "good". Find the sum of the first ten "good" natural numbers.
|
3. Let $k$ be a positive integer, and let $1, d_{1}, d_{2}, \cdots$, $d_{n_{-1}}, d_{n}, k$ be all its divisors, arranged in increasing order $\Rightarrow 1 \cdot k=d_{1} \cdot d_{n}=d_{2} \cdot d_{n_{-1}}=\cdots$. If $k$ is "good," then by definition, these products are also equal to $d_{1} \cdot d_{2} \cdots \cdots d_{n-1} \cdot d_{n}$. Thus, $n$ can only be 2 and $d_{1}$ must be a prime number (otherwise, the prime factors of $d_{1}$ would appear between 1 and $d_{1}$ in the sequence). Similarly, $d_{2}$ must be a prime number or the square of $d_{1}$ (otherwise, there would not be only one prime factor $d_{1}$ between 1 and $d_{2}$). Therefore, $k$ is either the product of two different primes or the product of a prime and its square, i.e., the cube of a prime. The first 10 good numbers are easily listed as: $6, 8, 10, 14, 15$, $21, 22, 26, 27$ and 33, and their sum is 182.
|
182
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Find the area of the region enclosed by the graph of the equation $|x-60|+|y|=\left|\frac{x}{4}\right|$.
untranslated text remains the same as requested.
|
4. First, the graph of this equation is symmetric about the $x$-axis, so we only need to find the area enclosed by
$$
\left\{\begin{array}{l}
\left.y=\left|\frac{x}{4}\right|-1 x-60 \right\rvert\,, \\
y \geqslant 0
\end{array}\right.
$$
The region enclosed by the difference of $0 \cdots \times 1$ and the $x$-axis in the upper half-plane is the triangle $\triangle D A B$. It is easy to find that the coordinates of the three vertices are $D$ (60, 15), $A(48,12), B(80,20)$. Thus, $S_{\triangle D A B}=\frac{A B \times C D}{2}=240$. Therefore, the area of the region enclosed by the graph of the original equation is 480.
|
480
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$5 . x 、 y$ are integers that satisfy the equation $y^{2}+3 x^{2} y^{2}=30 x^{2} +517$, find the value of $3 x^{2} y^{2}$ .
|
5. The original equation transforms to $\left(y^{2}-10\right)\left(3 x^{2}+1\right)$ $=3 \times 13^{2}>y^{2}-10=1 ， 3 ， 13 ， 39 ， 169$ or $507=>y^{2}=11,13,23,42,179$ or 517. Since $y$ is an integer $>y^{2}=49 \rightarrow y^{2}-10=39 \Rightarrow 3 x^{2}+1=13$ $\therefore 3 x^{2}=12>3 x^{2} y^{2}=12 \times 49=588$.
|
588
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in the figure below, rectangle $A B C D$ is divided into 4 equal-area parts by 5 line segments. Given $X Y=Y B+B C$ $+C Z=Z W=W D+D A+A X, P Q / / A B$. If $B C=19 \mathrm{~cm}, P Q=87 \mathrm{~cm}$, find the length of $A B$ (in cm).
|
6.18 The trapezoids $X Y Q P$ and $Z W P Q$ have equal areas, $H A Y-\| Z$. Both are equal to $\frac{B C}{2}$. Also, $X Y$ is $\frac{1}{4}$ of the perimeter of rectangle $A B C D$, so
$$
\begin{array}{l}
X Y=\frac{A B+B C}{2} \text { by } \frac{(P Q+X Y)}{2} \times \frac{B C}{2} \\
=\frac{A B \cdot B C}{4} \Rightarrow P Q+X Y=A B \Rightarrow P Q \\
+\frac{A B+B C}{2}=A B \Rightarrow A B=B C+2 P Q \\
=193(\mathrm{~cm}) .
\end{array}
$$
|
193
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Find the largest positive integer $n$, such that the inequality $\frac{8}{15}<$ $-\frac{n}{n+k}<\frac{7}{13}$ holds for exactly one integer $k$.
|
8 . Transform the original inequality to $\frac{15}{8}>\frac{n+k}{n}>\frac{13}{7}$, which is equivalent to $\frac{7}{8}>\frac{k}{n}>\frac{6}{7} \Leftrightarrow 49 n>56 k>48 n$. Therefore, the problem is converted to finding the largest open interval ( $48 n$, $49 n$ ) that contains only one multiple of 56. Since the interval length is $n$, it contains $n-1$ integers. If $n-1 \geq 2 \times 56$, then the interval must contain at least 2 multiples of 56 $\Rightarrow n-1<2 \times 56 \Rightarrow n \leqslant 112$. When $n=112$, we have $48 \times 112=56 \times 96<56 \times 97<56 \times 98=49 \times 112$. Here, $k=97$. It is clearly the only integer that satisfies this inequality, so $n=112$ is the solution.
|
112
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Pay for an escalator moving upwards. A walks down from its top to its bottom, totaling 150 steps, B walks up from its bottom to its top, totaling 75 steps. Assuming A's speed (number of steps walked per unit time) is 3 times B's speed, how many steps of the escalator are visible at any given moment? (Assume this number is a constant)
|
10. Let $v_{1}, v_{2}, v$ represent the speeds (number of steps walked per unit time) of $A, B$, and the automatic escalator, respectively, and let $t_{1}, t_{2}, t$ represent the time taken by $A, B$, and the automatic escalator, respectively. From the problem, we have: $v_{1}=3 v_{2}, v_{1} t_{1}=150, v_{2} t_{2}=75$ $\rightarrow t_{2}: t_{1}=3: 2$. Let $x$ be the number of visible steps on the automatic escalator $\Rightarrow x=\left(v_{2}+v\right) t_{2}=\left(v_{1}-v\right) t_{1}$
$$
\begin{array}{l}
\Rightarrow x=75+v t_{2}=150-v t_{1} \Rightarrow t_{2}=\frac{x-75}{v}, \\
t_{1}=\frac{150-x}{v}>t_{2}: t_{1}=(x-75):(150-x) \\
=3: 2 \rightarrow 2(x-75)=3(150-x) \Rightarrow x=120 .
\end{array}
$$
|
120
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Find the maximum value of $k$ such that $3^{11}$ can be expressed as the sum of $k$ consecutive positive integers.
|
i. Find the maximum positive integer $k$ such that $3^{11}=(n+1)+(n+2)+\cdots+(n+k)$ holds (where $n$ is a non-negative integer). From the right side, we have $=\frac{k(k+2 n+1)}{2}$ $->K \cdot(k+2 n+1)=2 \cdot 3^{11}$. To make the smaller factor $k$ as large as possible, $n$ must be non-negative $\Rightarrow k=2 \cdot 3^{5}=$ 486. (At this point, $n=121,3^{11}=122+123+\cdots+607$ )
|
486
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. The square root of a number without $m$ is $\cdots$ numbers of the form $n+r$, where $n$ is a positive integer, and $r$ is a real number less than $\frac{1}{1000}$. If $m$ is the smallest positive integer satisfying the above condition, find the value of $n$ when $m$ is the smallest positive integer.
|
12. From the problem: $\sqrt[3]{m}=n+r, 0\frac{1000}{3}
\end{array}
$
$\rightarrow n^{2} \approx \frac{1000}{3}$. Since $18^{2}<\frac{1000}{3}<19^{2}$, we can guess $n=18$ or $n=19$. Upon verification, when $n=18$, the inequality does not hold, but when $n=19$, the inequality is satisfied. Therefore, $n=19$ is the smallest positive integer that meets the problem's requirements (at this time, $m=19^{3}+1=6860$ is the smallest positive integer whose cube root has a positive fractional part less than $\frac{1}{1000}$).
|
19
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. For a known sequence of distinct real numbers $r_{1}, r_{2}, r_{3}, \cdots, r_{\text {n}}$, a single operation involves comparing the second term with the first term, and swapping them if and only if the second term is smaller; then comparing the third term with the new second term, and swapping them if and only if the third term is smaller; and so on, until the last term is compared with its new preceding term, and they are swapped if and only if the last term is smaller. For example, the following diagram shows how the sequence 1, 9, 8, 7 is transformed into the sequence $1, 8, 7, 9$ through one operation. Each pair of numbers being compared is marked with a “-” below them. Clearly, any known sequence can be sorted into a non-decreasing sequence through one or more such operations.
$$
\begin{array}{llll}
1 & 9 & 8 & 7 \\
\hline 1 & 9 & 8 & 7 \\
1 & 8 & 5 & 6 \\
1 & 8 & 7 & 9
\end{array}
$$
Now assume $n=40$, and $r_{1}, r_{2}, \cdots, r_{40}$ are distinct and randomly arranged. Let $\frac{p}{q}$ (in lowest terms) represent the probability that a single operation will move the 20th term ($r_{20}$) to the 30th position (with 29 terms to its left and 10 terms to its right). Find $p+q$.
|
13. Notice that the operation defined in the problem, when applied to any sequence $r_{1}, r_{2}, \cdots, r_{k}$ once, will always result in the last number being the largest in the sequence. Therefore, for the initial sequence $r_{1}, r_{2}, \cdots, r_{20}, \cdots, r_{30}, r_{31}, \cdots, r_{40}$, $r_{20}$ can be moved to the 30th position through one operation if and only if in the first 31 numbers $r_{1}, r_{2}, \cdots, r_{20}, \cdots, r_{30}, r_{31}$, $r_{3}$ is the largest number, and $r_{20}$ is the second largest number. Hence, the required probability $\frac{p}{q}=\frac{29!}{31!}=\frac{1}{930} \Rightarrow p+q=931$.
|
931
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Find the minimum value of $|x-1|+|x-3|+|x-5|$.
|
To solve this type of problem, we can use the method of "fixing points, dividing segments, and discussing."
Points. Let $x-1=0, x-3=0, x-5=0$, to determine $x_{1}=1, x_{2}=3, x_{3}=5$ three points.
Dividing segments. The above three points divide the number line into four segments: $(-\infty, 1]$, $(1,3]$, $(3,5]$, $(5,+\infty)$.
Discussion. When $x \in(-\infty, 1]$, the original expression $=9-3 x$. In this case, the minimum value of the original expression is 6;
When $x \in(1,3]$, the original expression $=7-x$. In this case, the minimum value of the original expression is 4;
When $x \in(3,5)$, the original expression $=x+1$. In this case, the minimum value of the original expression is 4;
When $x \in(5,+\infty)$, the original expression $=3 x-9$. In this case, the minimum value of the original expression is 6.
In summary, the minimum value of the original expression is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Let $y=\sqrt[3]{1+3 x}, x=\frac{1}{3}\left(y^{3}-1\right)$, when $x \rightarrow 0$, $y \rightarrow 1$.
Original expression $=\frac{2}{3} \lim _{y \rightarrow 1} \frac{y^{3}-2 y^{3}+1}{y^{3}-3 y+2}$ is still of the form $\frac{0}{0}$. Let $z=y-1$, then,
Original expression $=\frac{2}{3} \lim _{z \rightarrow 0} \frac{\left(z^{2}+3 z+3\right)}{z+3}=2$.
Second, for indeterminate forms of rational expressions in trigonometric functions of the form $\lim _{x \rightarrow x_{0}} R(\sin x, \cos x)$, the universal substitution can be used.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Some people stand in a row, among them, A does not stand at the head, and B does not stand at the end. How many ways are there to arrange them?
|
(1) 䒴
$P_{4}^{4}$
etc. transmission
(2) one of the middle three positions $P_{3}^{\frac{1}{3}} P \frac{1}{3} P_{3}^{3}$, so the number of ways for not standing at the head and not standing at the tail is $P_{4}^{4}+P{ }_{3}^{1} P_{3}^{1} P_{3}^{3}=78$ (benefit).
Solution two (using the properties 2 and 3 mentioned earlier)
Let $A=$ \{arrangements where A stands at the head\},
$B=\{$ arrangements where B stands at the tail\},
$E=\{$ arrangements where five people stand in a row\}.
It is easy to see that $N(A)=N(B)=P{ }_{4}^{4}$,
$$
N(E)=P_{5}^{5}, N(A B)=P_{3}^{3} \text {. }
$$
The number of arrangements required by the problem should be $N(\bar{A} \cap \bar{B})$, by the principle of duality, we know $A \cap B=\overline{A \cup B}$, so
$$
\begin{array}{l}
N(\bar{A} \cap \bar{B})=N(\overline{A \cup B}) \\
=N(E)-N(A \cup B) \\
=N(E)-N(A)-N(B)+N(A B) \\
=P_{5}^{5}-P_{4}^{4}-P_{4}^{4}+P_{3}^{3}=78 \text { (ways). }
\end{array}
$$
|
78
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Five people stand in a row, requiring that A does not stand at the head, B does not stand at the end, and C and D do not stand together. How many ways are there to arrange them?
|
Let $E=\{$ permutations of five people standing in a row $\}$,
$A_{1}=\{$ permutations where A stands at the head $\}$,
$A_{2}=\{$ permutations where B stands at the tail $\}$,
$A_{3}=\{$ permutations where C and D stand together $\}$.
Then the number of permutations required in the problem is
$$
\begin{aligned}
& N\left(\overline{A_{1}} \cap \overline{A_{2}} \cap \overline{A_{3}}\right)=N\left(\overline{\left.A_{1} \cup A_{2} \cup A_{3}\right)}\right. \\
= & N(E)-N\left(A_{1}\right)-N\left(A_{2}\right)-N\left(A_{3}\right) \\
& +N\left(A_{1} A_{2}\right)+N\left(A_{1} A_{3}\right)+N\left(A_{2} A_{3}\right) \\
& -N\left(A_{1} A_{2} A_{3}\right) \\
= & P_{5}^{5}-P_{4}^{4}-P_{4}^{4}-P_{4} P_{2}^{2}+P_{3}^{3}+P_{3}^{2} P_{2}^{2} \\
& +P_{3}^{3} P_{2}^{2}-P_{2}^{2} P_{2}^{2} \\
= & 120-24-24-48+6+12+12-4 \\
= & 50(\text { ways }) .
\end{aligned}
$$
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. Five people stand in a row. When they reline up, none of them stands in their original position. How many different formations are there?
|
Let the original formation of the five people be
$$
a_{1} a_{2} a_{3} a_{4} a_{5}
$$
$E=\{$ all permutations of five people standing in a line $\}$,
$A_{i}=\{a$ standing in his original position in the permutations $\}$
$$
\text { ( } i=1,2,3,4,5) \text {, }
$$
Then the number of desired permutations is
$$
\begin{array}{l}
-N\left(\bigcap_{i=1}^{5} A_{i}\right) \\
=P_{5}^{5}-C_{5}^{1} P_{4}^{4}+C_{5}^{2} P_{5}^{3}+C_{5}^{5} P_{2}^{2}-C_{5}^{4} P_{1}^{1} \\
-C_{5}^{5} \\
=120-120+60-20+5-1 \\
=44 \text { (ways). } \\
\end{array}
$$
This solution can be generalized to any $n$ elements arranged in a line, where each element is not in its original position. The number of different arrangements is
$$
\begin{array}{l}
P_{n}^{n}-C_{n}^{1} P_{n-1}^{n-1}+C_{n}^{2} P_{n-2}^{n-2}+\cdots \\
+(-1)^{k} C_{n}^{k} P_{n-k}^{n-k}+\cdots+(-1)^{n} C_{n}^{n} \\
=P_{n}^{n}\left[1-1+\frac{1}{2!}-\frac{1}{3!}+\cdots+\frac{(-1)^{k}}{k!}\right. \\
\left.+\cdots+\frac{(-1)^{n}}{n!}\right] \\
\end{array}
$$
$\approx P_{n}^{n} e^{-1}$ (take $e=2.71828$, then round to the nearest integer).
|
44
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
甽6. If $x+3y=10$, find the extremum of $x^{2}+3y^{2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6. If $x+3y=10$, find the extremum of $x^{2}+3y^{2}$.
|
Solve $x+3 y$
$=10$ as the line in Figure 6. Let $x^{2}+3 y^{2}=c$, then
$$
\frac{x^{2}}{(\sqrt{c})^{2}}+\frac{y^{2}}{\left(\sqrt{\frac{c}{3}}\right)^{2}}=1 \text {. }
$$
From Figure 6, we know that when the line $x+3 y=10$ is tangent to the ellipse $x^{2}+3 y^{2}=c$, the semi-axes of the ellipse are the shortest. We have
$$
\left\{\begin{array}{l}
x+3 y=10 \\
x^{2}+3 y^{2}=c
\end{array}\right.
$$
We get $c=25$, i.e., the minimum value of $x^{2}+3 y^{2}$ is 25. (Author's affiliation: No. 100 High School, Tianjin City)
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. To complete a certain project, the number of days required for $A$ to work alone is $m$ times that of $B$ and $C$ working together, the number of days required for $B$ to work alone is $n$ times that of $A$ and $C$ working together, and the number of days required for $C$ to work alone is $p$ times that of $A$ and $B$ working together. Try to prove that the ratio of the number of days each person needs to work alone is $(m+1):(n+1):(p+1)$, and that $-\frac{n}{m+1}+\frac{n}{n+1} \div \frac{n}{p+1}=2$.
|
Let $x$, $y$, $z$ be the number of days required for $A$, $B$, $C$ to complete the work individually, respectively, then we have
$$
\frac{m}{x}=\frac{1}{y}+\frac{1}{z}
$$
Therefore, $\frac{m+1}{x}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.
Similarly, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{n+1}{y}=\frac{p+1}{z}$.
Thus, the ratio of the number of days each person works individually is
$$
(m+1):(n+1):(p+1) \text {. }
$$
Additionally, by $\frac{1}{m+1}=\frac{1}{x} \div\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$,
$$
\begin{array}{l}
\frac{1}{n+1}=\frac{1}{y} \div\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right), \\
\frac{1}{p+1}=\frac{1}{z} \div\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right),
\end{array}
$$
Adding the three equations yields
$$
\frac{1}{m+1}+\frac{1}{n+1}+\frac{1}{p+1}=1 \text {. }
$$
liij
$$
\begin{array}{l}
\frac{m}{m+1}=1-\frac{1}{m+1}, \\
\frac{n}{n+1}=1-\frac{1}{n+1}, \\
\frac{p}{p+1}=1-\frac{1}{p+1},
\end{array}
$$
Thus, $\frac{m}{m+1}+\frac{n}{n+1}+\frac{p}{p+1}$
$$
=3-\left(\frac{1}{m+1}+\frac{1}{n+1}+\frac{1}{p+1}\right)=2 .
$$
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
14. A number is the product of three prime factors. The squares of these three prime factors; the sum is 2331, the number 7560 is less than this number and is coprime with it, and the sum of its divisors (excluding 1 itself) is 10560. Find this number.
|
Let $c, b, c$ be the prime factors of a number, then
$$
a^{2}+b^{2}+c^{2}=2331.
$$
In addition, we know that the number of integers less than this number and coprime to it is
$$
\begin{array}{l}
a b c\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right). \\
\text { That is, }(a-1)(b-1)(c-1)=7560.
\end{array}
$$
According to the problem, we also have
$$
(a+1)(b+1)(c+1)=10560.
$$
From (2) and (3), we get:
$$
a b c+a+b+c=9060.
$$
And $b c+c a+a b+1=1500$.
From (1) and (5), we get
$$
\begin{array}{l}
(a+b+c)^{2}=5329. \\
\text { Hence } a+b+c=73.
\end{array}
$$
From (4), we get
$$
a b c=8987=11 \cdot 19 \cdot 43.
$$
The solution to this problem uses two theorems in number theory: the first, the number of positive integers less than a given number and coprime to it, if $N$ represents this number, and $N=a^{p} b^{q} c^{r} \cdots$, where $a, b, c, \cdots$ are different prime numbers, and $p, q, r, \cdots$ are positive integers, then $\phi(N)=N\left(1-\frac{1}{a}\right) \cdot\left(1-\frac{1}{b}\right) \cdot\left(1-\frac{1}{c}\right) \cdots$. The second, "the number of divisors of a number", if $N$ represents this number, and $N=a^{p} b^{q} c^{r} \cdots$, where $a, b, c, \cdots$ are different prime numbers, and $p, q, r, \cdots$ are positive integers, then the number of divisors is $(p+1)(q+1)(r+1) \cdots$.
|
8987
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. Given $f\left(1-x^{2}\right)=\frac{1-x^{2}}{x^{2}}$. Find $f\left(\frac{1}{2}\right)$.
|
$\begin{array}{l}\text { Solve } f\left(1-x^{2}\right)=\frac{1-x^{2}}{1-\left(1-x^{2}\right)}, \\ \text { hence } f(x)=\frac{x}{1-x}, f\left(\frac{1}{2}\right)=1.\end{array}$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Given an integer $n$, if $n$ plus 38 is a perfect square, prove that $n$ plus 38 is a perfect square, find $n$.
The text seems to have some typographical errors or unclear phrasing. Here is a corrected version for clarity:
Example 2. Given an integer $n$, if $n$ plus 38 is a perfect square, prove that $n$ plus 38 is a perfect square, and find $n$.
|
Let $n-51=x^{2},(x, y \in Z)$
$$
n+38=y^{2} .
$$
Subtracting the equations yields
$$
\begin{array}{l}
(y-x)(y+x)=89=1 \cdot 89 \text { ? } \\
=(-1)(-89) .
\end{array}
$$
Thus, $\left\{\begin{array}{l}y+x=1,89,-1,-89, \\ y-x=89,1,-89,-1 .\end{array}\right.$
Therefore, $\left\{\begin{array}{l}x=-44,41,41, \quad 41, \\ y=45,45,-45,-45,\end{array}\right.$ so $n=1987$.
|
1987
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The page numbers of a book range from 1 to $n$. When all these page numbers are added together, one of the page numbers was mistakenly added twice. The incorrect sum obtained is 1987. What is the page number that was added twice?
|
Answer: 34
|
34
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9. Given the dihedral angle $C-A B=D=120^{\circ}$, $\angle C B A=60^{\circ}, \angle D A B=30^{\circ}, A B=\sqrt{37} \mathrm{~cm}$ (Figure 17). Find the distance $d$ between the skew lines $B C, A D$.
|
Let the angle between $BC$ and $AD$ be $\theta$. By the triple product formula,
$$
\begin{aligned}
\cos \theta= & \cos 60^{\circ} \cos 150^{\circ} \\
& +\sin 60^{\circ} \sin 150^{\circ} \cos 120^{\circ} \\
= & -\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{8}=-\frac{3 \sqrt{3}}{8} .
\end{aligned}
$$
Thus, $\sin \theta=\frac{\sqrt{37}}{8}$.
Therefore, $d=\frac{\sqrt{37} \sin 60^{\circ} \sin 150^{\circ} \sin 120^{\circ}}{\sin \theta}$
$=3(\text{cm})$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Find a three-digit number such that the ratio of the number to the sum of its digits is minimized.
untranslated text remains unchanged:
例2. 求一个三位数, 便它与它的各位数字之和的比为最小.
However, for a proper translation, it should be:
Example 2. Find a three-digit number such that the ratio of the number to the sum of its digits is minimized.
|
Solve: Arrange the ratios of all three-digit numbers to the sum of their respective digits in a $9 \times 100$ grid (Table 1).
We examine the numbers in each row and select the smaller ones. By the inequality $\frac{a}{b}>\frac{a+1}{b+1}(a>b)$, from left to right, every 10 numbers form a group, and the last number in each group is the smallest. Extract these numbers and arrange them in a $9 \times 10$ grid.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline \begin{tabular}{l}
$\frac{100}{1}$
\end{tabular} & \begin{tabular}{l}
$\frac{101}{2}$
\end{tabular} & \begin{tabular}{l}
$\frac{102}{3}$
\end{tabular} & $\cdots$ & \begin{tabular}{l}
$\frac{109}{10}$
\end{tabular} & \begin{tabular}{l}
$\frac{110}{2}$
\end{tabular} & \begin{tabular}{l}
$\frac{111}{3}$
\end{tabular} & $\cdots$ & \begin{tabular}{l}
$\frac{199}{19}$
\end{tabular} \\
\hline \begin{tabular}{l}
$\frac{200}{2}$
\end{tabular} & \begin{tabular}{l}
$\frac{201}{3}$
\end{tabular} & \begin{tabular}{l}
$\frac{202}{4}$
\end{tabular} & $\cdots$ & \begin{tabular}{l}
$\frac{209}{11}$
\end{tabular} & \begin{tabular}{l}
$\frac{210}{3}$
\end{tabular} & \begin{tabular}{l}
$\frac{211}{4}$
\end{tabular} & $\cdots$ & \begin{tabular}{l}
$\frac{299}{20}$
\end{tabular} \\
\hline $\mathbf{\vdots}$ & $\vdots$ & $\mathbf{\vdots}$ & & $\mathbf{\vdots}$ & : & $\vdots$ & & $\mathbf{\vdots}$ \\
\hline \begin{tabular}{l}
$\frac{900}{9}$
\end{tabular} & \begin{tabular}{l}
$\frac{901}{10}$
\end{tabular} & \begin{tabular}{l}
$\frac{902}{11}$
\end{tabular} & $\cdots$ & \begin{tabular}{l}
$\frac{909}{18}$
\end{tabular} & \begin{tabular}{l}
$\frac{910}{10}$
\end{tabular} & \begin{tabular}{l}
$\frac{911}{11}$
\end{tabular} & $\cdots$ & $\frac{999}{27}$ \\
\hline
\end{tabular}
Table 1
From Table 2, it is easy to see that the numbers in each row from left to right are
\begin{tabular}{|c|c|c|c|c|}
\hline \begin{tabular}{l}
$\frac{109}{10}$
\end{tabular} & \begin{tabular}{l}
$\frac{119}{11}$
\end{tabular} & \begin{tabular}{l}
$\frac{129}{12}$
\end{tabular} & $\cdots$ & \begin{tabular}{l}
$\frac{199}{19}$
\end{tabular} \\
\hline \begin{tabular}{l}
$\frac{209}{11}$
\end{tabular} & \begin{tabular}{l}
$\frac{219}{12}$
\end{tabular} & \begin{tabular}{l}
$\frac{229}{13}$
\end{tabular} & $\cdots$ & \begin{tabular}{l}
$\frac{299}{20}$
\end{tabular} \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline \begin{tabular}{l}
$\frac{909}{18}$
\end{tabular} & \begin{tabular}{l}
$\frac{919}{19}$
\end{tabular} & \begin{tabular}{l}
$\frac{929}{20}$
\end{tabular} & $\cdots$ & \begin{tabular}{l}
$\frac{999}{27}$
\end{tabular} \\
\hline
\end{tabular}
Table 2
gradually decreasing, while the rightmost column increases from top to bottom. Therefore, the smallest number is $\frac{190}{19}$, and the required three-digit number is 159.
|
159
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. For the equation $(1984 x)^{2}-1983 \times 1985 x-1$ $=0$, the larger root is $r$, and for the equation $x^{2}+1983 x-1984=0$, the smaller root is $s$. What is $r-s$? (1984 Beijing Junior High School Mathematics Competition Question)
|
Solve $\because 1984^{2}+(-1983) \times 1985$ $+(-1)=0$, according to property 1, we can get the roots of the first equation $x_{1}=1, \quad x_{2}=-\frac{1}{1984^{2}}$.
Also $\because 1+1983-1984=0$, we can get the roots of the second equation $x_{1}^{\prime}=1, x_{2}^{\prime}=-1984$.
$$
\text { Therefore } r-s=1985 \text {. }
$$
|
1985
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Given that 1 is a root of the equation $a x^{2}+b x+c=0$, find the value of $\frac{a^{2}+b^{2}+c^{2}}{a^{3}+b^{3}+c^{3}}+\frac{2}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.
|
Given that $1$ is a root of the equation $a x^{2}+b x+c=0$,
$$
\begin{array}{c}
\therefore a+b+c=0 . \\
\text { Also, } a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c) \\
.\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) .
\end{array}
$$
From (1) and (2), we can deduce
$$
\begin{array}{c}
a^{3}+b^{3}+c^{3}=3 a b c . \\
\text { Therefore, the original expression }=\frac{a^{2}+b^{2}+c^{2}}{3 a b c} \\
+\frac{2(a b+b c+c a)}{3 a b c} \\
=\frac{(a+b+c)^{2}}{3 a b c}=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the sets $M=\{x, x y, \lg (x y)\}$ and $N=\{0,|x|, y\}$, and $M=N$. Then, $\left(x+\frac{1}{y}\right)+\left(x^{2}+\frac{1}{y^{2}}\right)+\left(x^{3}+\right.$ $\left.\frac{1}{y^{3}}\right)+\cdots+\left(x^{2001}+\frac{1}{y^{2001}}\right)$ is equal to
|
Given $\because M=N$,
$\therefore 0 \in M$, and $x, y$ cannot both be 0, then
$\lg (x y)=0, x y=1$.
Thus, $1 \in N$. If $y=1$, then $x=1$, which contradicts the distinctness of the elements in the set. Therefore, $|x|=1$, and $x=-1$, so $y=-1$.
Therefore, $x+\frac{1}{y}=-2 \cdot x^{2}+\frac{1}{y^{2}}=2$, $x^{3}+\frac{1}{y^{3}}=-2, \cdots, x^{2000}+\frac{1}{y^{2000}}=2$, $x^{2001}+\frac{1}{y^{3} \frac{1}{301}}=-2$, and the sum is -2.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, given $x, y \in N$, find the largest $y$ value such that there exists a unique $x$ value satisfying the following inequality:
$$
\frac{9}{17}<\frac{x}{x+y}<\frac{8}{15} \text {. }
$$
|
Solve the original system of inequalities $\Longleftrightarrow\left\{\begin{array}{l}8 x-9 y>0, \\ 7 x-8 y>0\end{array}\right.$ $\longleftrightarrow \frac{9 y}{8}<x < \frac{8 y}{7}$.
For the interval $\left(\frac{9 y}{\circ}, \frac{8 y}{7}\right)$ to contain a unique integer, the necessary and sufficient condition is either $1^{\circ}$ or $2^{\circ}$:
$1^{\circ} \quad$ If $\frac{8 y}{7} \in \mathbb{N}$, then $\frac{8 y}{7}=\left[\frac{9 y}{8}\right]+2$;
$2^{\circ}$ If $\frac{8 y}{7} \notin \mathbb{N}$, then $\left[\frac{8 y}{7}\right]=\left[\frac{9 y}{8}\right]+1$.
For convenience, let $y=8 t+r, t \in \mathbb{N}, r=0$, $1,2, \cdots, 7$.
From $1^{\circ}$, we have $9 t+\frac{t+8 r}{7}=9 t+\left[\frac{9 r}{8}\right]+2$, which simplifies to $\frac{t+r}{7}=\left[\frac{r}{8}\right]+2$.
Here, $\left[\frac{r}{8}\right]=0$, and $\frac{t+r}{7} \in \mathbb{N}$. In this case, $r=0$. When $t$ is 14, $y$ reaches its maximum value of 112. The corresponding $x=127$, which is unique.
From $2^{\circ}$, similarly, we have $\left[\frac{t+r}{7}\right]=\left[\frac{r}{8}\right]+1$ $=1$. Since $\left(\frac{t+r}{7}\right) \in \mathbb{N}$, we can take $t+r=13$, i.e., $t=13, r=0$, at which point $y_{\max }=104$. The corresponding $x=118$.
Combining the above, we find that the required $y_{\max }=112$.
|
112
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $\mathrm{f}$ be a function satisfying the following conditions:
(i) If $x>y$, and $f(y)-y \geqslant v \geqslant f(x)-x$, then for some number $z$ between $x$ and $y$, $f(z)=v+z$;
(ii) The equation $\mathrm{f}(\mathrm{x})=0$ has at least one solution, and among the solutions, there is one that is not less than all other solutions;
(iii) $\mathrm{f}(0)=1$;
(iv) $\mathrm{f}(1987) \leq$ 588;
(v) $f(x) \cdot f(y)=f(x \cdot f(y)+y \cdot f(x)$ $-x y)$.
Find $f(1987)$.
|
1. Solution
(1) From (ii), we know that $f(x)=0$ has at least one solution. Let's assume $u$ is a solution of $f(x)$.
(2) Assume $u>0$.
(3) From (i) and (iii), we know that there exists a number $z$ between 0 and $u$ such that $f(z)=z$.
(4) From (v), we get:
$$
\begin{array}{l}
0=f(z) \cdot f(u)=f(z \cdot f(u)+u \cdot f(z)-u z \\
=f(0)=1, \text{ contradiction. }
\end{array}
$$
(5) Therefore, from (iii), we have $u<0$.
(6) From (ii), we know that among the solutions of $\mathrm{f}(\mathrm{x})=0$, there is a solution $\mathbf{x}_{0}$ that is not less than all other solutions.
(7) From (5), we get $x_{0}<0$.
(8) From (v), we get $f(x) \cdot f\left(x_{0}\right)=f\left(x \cdot f\left(x_{0}\right)\right.$
$$
\left.+x_{0} \cdot f(x)-x x_{0}\right),
$$
which means $0=f\left(x_{0} \cdot(f(x)-x)\right)$, and $x$ can be any number.
(9) From (6), we get $x_{0} \geqslant x_{0} \cdot(f(x)-x)$. Since $x < 0$, we can rewrite the inequality as $f(x)-x \geqslant 1$.
(10) From (iv) and (9), we get $1987 \leqslant 1988 \leqslant f(1987)$. Therefore, we have $f(1987)=1988$.
|
1988
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
List 4. (IMO-4-1) Find the smallest natural number $n$ whose decimal representation ends with 6, such that when the last digit 6 is deleted and written at the beginning of the remaining digits, it becomes four times $n$.
|
Solve for $n$ whose decimal representation ends with 6, meaning that it leaves a remainder of 6 when divided by 10. By the division algorithm, we have
$$
n=10 m+6 \text {. }
$$
where $m$ is a positive integer, and we assume it is an $l$-digit number. According to the problem,
$$
4(10 m+6)=6 \times 10^{l}+m .
$$
Thus,
$$
m=\frac{2\left(10^{2}-4\right)}{13} .
$$
For $m$ to be an integer, we need $13 \mid 2\left(10^{\imath}-4\right)$. Since 13 is a prime number, and $13 \times 2$, it follows that $13 \mid 10^{2}-4$ (here we use the property of prime numbers). Furthermore, to make $n$ the smallest, from the above equation, this is equivalent to making $m$ the smallest. Therefore, our problem is reduced to finding the smallest positive integer $l$ such that $13 \mid 10^{2}-1$.
By taking $l=1,2, \cdots$, and verifying one by one, we find that the smallest $l$ that satisfies the condition is $l=5$ (we used the "smallest" constraint). At this point, $m=15381$, and the smallest $n$ we seek is $153815$.
Comment: The numbers involved in this problem are relatively large, so verifying them is not difficult. This kind of "trial method" can be easily solved with the help of a computer for large number calculations. However, for certain theoretical problems, a computer is still powerless.
The following problem is closely related to Example 4 and is provided for readers to think about:
Problem. Prove: There exist infinitely many natural numbers $n$ that satisfy the conditions of the problem.
|
153815
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (IMO-20-1) The last three digits of the numbers $1978^{n}$ and $1978^{m}$ are equal. Find the positive integers $n$ and $m$ such that $n+m$ is minimized, where $n>m \geqslant 1$.
|
From the given, we have
$$
\begin{array}{l}
1978^{m}-1978^{m}=1978^{m}\left(1978^{n-m}-1\right) \\
=2^{m} \times 989^{m} \times\left(1978^{m}-1\right) \\
\equiv 0\left(\bmod 10^{3}\right) .
\end{array}
$$
Since
$$
10^{3}=2^{3} \times 5^{3},
$$
and $989^{m}$ and $1978^{\mathrm{m}}-1$ are both odd, it follows that $m \geqslant 3$. Also, 5 is coprime with $2^{m}$ and $989^{m}$, so
$$
1978^{n-m}-1 \equiv 0\left(\bmod 5^{3}\right) \text {. }
$$
That is, $1978^{n-m} \equiv 1(\bmod 125)$.
Thus, $3^{n-m} \equiv 1978^{\mathrm{n}-\mathrm{m}} \equiv 1(\bmod 25)$.
Using the method from Example 10 (now classifying by $\bmod 4$), we can derive from $3^{n-m} \equiv 1(\bmod 5)$ that
$$
n-m=4 k \text {. }
$$
Thus, $3^{4 k}=81^{k} \equiv 1 \quad(\bmod 25)$.
That is, $(80+1)^{\mathbf{k}}=1+k \times 80+C_{\mathbf{k}}{ }^{2} \times 80^{2} \equiv 1$
$(\bmod 25)$. Therefore, (since $25 \mid 80^{2}$) it is easy to see that $5 \mid k$. From (2) we get
$$
\begin{aligned}
1 \equiv 1978^{4 k}= & (3+1975)^{4 k} \equiv 3^{4 k}+4 k \\
& \times 3^{k-1} \times 1975 \equiv 3^{4 k}(\bmod 125) .
\end{aligned}
$$
That is, $1 \equiv 3^{4 k}=81^{k} \equiv 1+k \times 80+\frac{k(k-1)}{2}$
$$
\times 80^{2}=1+80 k(\bmod 125) \text {. }
$$
(Please note that $5 \mid \boldsymbol{k}$).
Thus, $125|80 k, 25| k$. Therefore, $k$ is at least 25, and $n-m$ is at least $4 \times 25=100$.
$$
\begin{array}{l}
n+m=(n-m)+2 m \\
\geqslant 100+2 \times 3=106 .
\end{array}
$$
Hence, the minimum value of $n-m$ is 106 (the above derivation can show that (1) holds in this case).
|
106
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. In the corner of the room, there are several identical cubes piled up. How many of the cubes are not visible?
|
In the second layer, there is 1 that is invisible,
in the third layer, there are $(1+2)$ that are invisible:
in the fourth layer, there are $(1+2+3)$ that are invisible, in the fifth layer, there are $(1+2+3+4)$ that are invisible, so the total number of invisible ones is $1+3+6+10=20$.
|
20
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For a regular $n$-sided polygon, construct $n$ squares outside the polygon, each sharing one side with the polygon. It is known that the $2n$ outer vertices of these $n$ squares form a regular $2n$-sided polygon. For what value of $n$ is this possible?
|
Solve As shown in Figure 1, $P, Q, R$ are three vertices of a regular $n$-sided polygon; $M, N, K, L$ are four vertices of a $2n$-sided polygon. According to the given conditions, we have $MN = NQ, QK = KL, NK = KL = MN$, which means $\triangle NQK$ is an equilateral triangle. Therefore, $\angle NQK = 60^{\circ}$. From the internal angle of the regular $n$-sided polygon $\angle PQR = 120^{\circ}$, we get the exterior angle as $60^{\circ}$.
$$
n \cdot 60^{\circ} = 360^{\circ} \text{, i.e., } n = 6 \text{. }
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. On the blackboard are the numbers $1,2, \cdots, 1987$. Perform the following transformation: erase some of the numbers on the blackboard and add the remainder when the sum of the erased numbers is divided by 7. After several such transformations, only two numbers remain on the blackboard, one of which is 987. Find the other number.
|
Note the fact that with each transformation: the sum of all numbers on the blackboard, when divided by 7, leaves the same remainder. $1+2+\cdots+1987=1987 \cdot 7 \cdot 142$, which indicates that the original sum on the blackboard is divisible by 7, so the sum of the two final numbers can also be divided by 7. Among the two final numbers, one of them (which cannot be 987) is the remainder when divided by 7, i.e., $0,1,2,3,4,5,6$. Since 987 is divisible by 7, the other number can only be 0.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Find the smallest natural number such that when the last digit of this number is moved to the first position, the resulting number is 5 times the original number.
|
Let the required number be $a_{1} a_{2} \cdots a_{n-1} a_{n}$ - According to the given conditions, we have
$$
\overline{a_{n} a_{1} a_{2} \cdots a_{n-1}}=5 \cdot \overline{a_{1} a_{2} \cdots a_{n-1} a_{n}} \text {. }
$$
Then
$$
\begin{array}{l}
a_{n} \cdot 10^{n-1}+\overline{a_{1} a_{2} \cdots a_{n-1}} \\
=5\left(\overline{a_{1} a_{2} \cdots a_{n-1}} \cdot \vdots 0+a_{n}\right) \text {, } \\
a_{n} \cdot\left(10^{n-1}-5\right)=49 a_{1} a_{2} \cdots a_{n-1} \text {, } \\
a_{n} \cdot 99 \cdots 95=45 \cdot a_{1} a_{2} \cdots a_{n-1} \text {, (1) } \\
n-21 \\
\end{array}
$$
$\because a_{n}$ is a digit, $\therefore 49 \times a_{n}$, then $7 \mid 99 \cdots 95$. $n-2$ times
In numbers of the form $99 \cdots 95$, the smallest number divisible by 7 is 99995, at this time $n=6$. Thus (1) becomes $a _ { 0 } \cdot 9 9 9 9 5 = 4 9 \cdot \longdiv { a _ { 1 } a _ { 2 } a _ { 3 } a _ { 4 } a _ { 5 } }$, i.e., $a_{0} \cdot 14285$ $=7 \cdot \overline{a_{1} a_{2} a_{3} a_{4} a_{5}}$. From $7 \times 14285$, we get $\alpha_{0}=7$, and $a_{1} a_{2} a_{3} a_{4} a_{5}=14285$. Therefore, 142857 is the required number.
|
142857
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. At least how many circles with a radius of 1 are needed to cover a circle with a radius of 2.
|
Let $O$ be the center of a circle with radius 2, and let $A B C D E F$ be a regular hexagon inscribed in the circle (Figure 8). The side length of the hexagon is 2. Construct six circles with the sides of the hexagon as diameters. The intersection points of these six circles, other than $A, B, C, D, E, F$, are denoted as $A_{1}, B_{1}, C_{1}, D_{1}, E_{1}, F_{1}$. These points are the vertices of a regular hexagon with side length 1. Use the above circles with radius 1 to cover the circle with radius 2. The remaining part (the shaded part in Figure 8) is contained within the hexagon $A_{1} B_{1} C_{1} D_{1} E_{1} F_{1}$, and can be completely covered by a circle with center $O$ and radius 1. Therefore, 7 circles can cover the circle with radius 2.
On the other hand, each small circle can only cover one-sixth of the large circle, so at least six circles are needed to cover the large circle. However, these six small circles still cannot cover the point $O$, so at least seven small circles are needed to cover the large circle.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9. Primary School One and Primary School Two have the same number of students participating in the Golden Cup Competition. The schools use cars to transport the students to the examination site. Primary School One uses cars that can seat 15 people each; Primary School Two uses cars that can seat 13 people each. As a result, Primary School Two has to send one more car than Primary School One. Later, each school adds one more student to the competition, and this way, the number of cars needed by both schools becomes the same. Finally, it is decided that each school will add one more student to the competition, and once again, Primary School Two has to send one more car than Primary School One. How many students from both schools will participate in the competition in the end?
|
Let's assume that at the beginning, both schools had $x$ participants. According to the problem, the first school sent $\frac{x}{15}=a \cdots 0$ ($a$ cars),
the second school sent $\frac{x}{13}=a \cdots 12(a+1$ cars $)$; the second time, the first school sent $\frac{x+1}{15}=a \cdots 1(a+1$ cars $)$,
the second school sent $\frac{x+1}{13}=a+1 \cdots 0(a+1$ cars $)$; the third time, the first school sent $\frac{x+2}{15}=a \cdots 2(a+1$ cars $)$,
the second school sent $\frac{x+2}{13}=a+1 \cdots 1(a+2$ cars). Therefore, $x+1$ is not a multiple of 13.
Thus, $x=90$.
Answer: In total, both schools had 184 participants.
|
184
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
In a hand clapping competition, the winner of each match gets 2 points, the loser gets 0 points; if it's a draw, each player gets 1 point. Now the scores are: $1979$, $1980$, $198$, $1985$. After verification, how many players participated in the competition?
|
Since $n(n-1)$ is the product of two consecutive natural numbers, it must be even, so 1979 and 1985 are impossible. And the last digit of $n(n-1)$, being the product of two consecutive natural numbers, can only be 0, 2, or 6, so 1984 is also impossible. The only possibility left is 1980. For $n(n-1)=1980$, solving gives $n=45$.
|
45
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. If $a>1, b$ is a positive rational number, $a^{b}+a^{-0}$ $=2 \sqrt{2}$, find the value of $a^{b}-a^{-b}$.
|
Consider the following solution:
Let $a^{b}=x$, then $x+\frac{1}{x}=2 \sqrt{2}$.
Transform it into $x^{2}-2 \sqrt{2} x+1=0$.
Solving yields $x_{1}=\sqrt{2}+1, x_{2}=\sqrt{2}-1$.
$$
\text { When } \begin{aligned}
x & =\sqrt{2}+1, \\
& a^{b}-a^{-b} \\
& =\sqrt{2}+1-\frac{1}{\sqrt{2}+1} \\
& =\sqrt{2}+1-(\sqrt{2}-1) \\
& =2
\end{aligned}
$$
When $x=\sqrt{2}-1$,
$$
\begin{aligned}
& a^{b}-a^{-b} \\
= & \sqrt{2}-1-\frac{1}{\sqrt{2}-1} \\
= & \sqrt{2}-1-(\sqrt{2}+1) \\
= & -2 .
\end{aligned}
$$
This solution is incorrect. The error lies in neglecting the conditions $a>1$ and $b$ being a positive integer - if these conditions were considered, it would be clear that $a^{b}>1$. Therefore, the correct answer should be 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. If $\lg ^{2} x \lg 10 x<0$, find the value of $\frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+\lg 10 x^{2}}$.
|
From the known condition $\lg ^{2} x \lg 10 x<0$, it is not difficult to conclude that $\lg x \neq 0$ and $\lg 10 x<0$, which also implicitly indicates the existence of $\lg x$.
$$
\text { Hence } \begin{aligned}
& \frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+\lg 10 x^{2}} \\
= & \frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+2 \lg x+\lg 10} \\
= & \frac{1}{\lg 10 x} \sqrt{(\lg x+1)^{2}} \\
= & \frac{1}{\lg 10 x} \sqrt{\lg ^{2} 10 x} \\
= & \frac{-\lg 10 x}{\lg 10 x}=-1 .
\end{aligned}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. (1MO-27 Preliminary Question) Let $A, B, C$ be three points on the edge of a circular pool, with $B$ due west of $C$, and $A B C$ forming an equilateral triangle with side lengths of 86 meters. A swimmer starts from $A$ and swims directly to $B$. After swimming $x$ meters, he reaches point $\boldsymbol{E}$, then turns and swims due west for $y$ meters, arriving at point $D$. If $x, y$ are both integers, find $y$.
---
The translation preserves the original text's formatting and line breaks.
|
Given the figure, since $\triangle AEF$ is an equilateral triangle, we have
$$
AF = AE = x.
$$
By symmetry, $FG = DE = y$. Also, since
$$
AE \cdot EB = DE \cdot EG,
$$
we have
$$
\begin{aligned}
& x(86 - x) \\
= & y(x + y).
\end{aligned}
$$
Notice that if $x$ is odd, then $x(86 - x)$ is also odd, while $y(x + y)$ is always even (if $y$ is even, $y(x + y)$ is clearly even; if $y$ is odd, then $x + y$ is even, and $y(x + y)$ is still even).
Therefore, $x$ must be even. From (1), we know that $y(x + y)$ is even, so $y$ must also be even.
The above argument is essentially taking $\left(1; \mathbf{x}^{2}\right)$ modulo 2 to determine $x$. We have $\left(x + \frac{y}{2} - 43\right)^{2} + y^{2} = \left(43 - \frac{y}{2}\right)^{2}$.
This indicates that $\left|x + \frac{y}{2} - 43\right|, y, 43 - \frac{y}{2}$ form a Pythagorean triplet.
Let $d$ be the greatest common divisor of $\left|x + \frac{y}{2} - 43\right|$ and $y$. Then, by the theorem mentioned later, there exist positive integers $a, b, a > b$, such that
$y = 2abd$ (since $y$ is even),
$$
43 - \frac{y}{2} = (a^2 + b^2)d.
$$
From these two equations, we get
$$
(a^2 + ab + b^2)d = 43 \cdot (a > b > 0).
$$
Since 43 is a prime number, we have $d = 1$.
Finally, we solve the equation using the "bounding method"
$$
a^2 + ab + b^2 = 43.
$$
Since $a > b > 0$, we have $3b^2 < 43$, $b^2 < \frac{43}{3}$, $b^2 \leq 14$, so $b \leq 3$. It is easy to see that the only positive integer solutions are $b = 1$ and $a = 6$.
Thus, we know $y = 12$. Then, from $\left|x + \frac{y}{2} - 43\right| = a^2 - b^2$, we find $x = 2$ or $x = 72$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 14. Find the number of integer solutions $(x, y, z)$ that satisfy $0<x<y<z$, and $\sqrt{1984}=\sqrt{ } x+\sqrt{ } y+\sqrt{ } z$.
|
$$
\text { Stele } \begin{aligned}
& \sqrt{1984}=8 \sqrt{31} \\
& =\sqrt{31}+2 \sqrt{31}+5 \sqrt{31} \\
& =\sqrt{31}+\sqrt{2} \times 31+\sqrt{5} \times 31 \\
& =\sqrt{31}+3 \sqrt{31}+4 \sqrt{31} \\
& =\sqrt{31}+\sqrt{3} \times 31+\sqrt{2} \times 1 .
\end{aligned}
$$
There are two kinds of correct statements in total.
$$
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
oi1. Find the number of positive integer solutions for the equation
$$
x+y+z=20
$$
|
Solving for $z$: $=1$, substituting into (1), we get $x+y=19$. For this boundary, when $x$ takes the values $1, 2, \cdots, 18$ $=2$, there are 17 sets of positive integer solutions. Continuing this way, when $z=18$, there is only 1 set of positive integer solutions. Therefore, the total number of positive integer solutions for equation (1) is
$$
18+17+\cdots+1=171 .
$$
|
171
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4. Find the indefinite solution of the equation
$$
x+y+z=20
$$
|
Find the number of positive integer solutions for (1). For this, we have 20 a's and 19 spaces (denoted by $s$) arranged in one line as follows:
asasasasasasasasasasasasasasasasas asisa.
Now, choose any 2 out of the 19 s's, and remove the rest of the $s$'s, for example:
caaaasaaaaaasaaauaaaaa. This results in three segments. From left to right, the counts are 5, 6, and 9, respectively, so we have
$$
x=5, y=6, z=9 \text {. }
$$
It is easy to see that each way of choosing 2 out of the 19 s's corresponds to a solution of (4), and different choices correspond to different solutions of (4); conversely, each solution of (4) corresponds to a way of choosing 2 out of the 19 s's, and different solutions correspond to different ways of choosing. In summary, the number of positive integer solutions for (4) is equal to the number of ways to choose 2 out of 19 s's, which is
$$
C_{19}^{2}=\frac{19 \times 18}{2}=171 .
$$
To find the number of non-negative integer solutions for (4), we can use 20 white balls and 2 black balls, arranged in any order from left to right on a straight line. These 2 black balls divide the line into 3 segments. Then, count the number of white balls in each segment from left to right, denoted as $\alpha_{1}, \alpha_{2}, \alpha_{3}$, and let
$$
x=\alpha_{1}, y=\alpha_{2}, \quad z=\alpha_{3},
$$
The number of non-negative integer solutions for (4) is equal to the number of arrangements of these 22 balls. However, note that among these 22 balls, 20 white balls are identical, and 2 black balls are identical, so the total number of arrangements is
$$
\frac{22!}{20!2!}=C_{22}^{2}=231 .
$$
$$
x_{1}+x_{2}+\cdots+x_{m}=n \quad(n \geqslant m)
$$
The number of solutions is $C_{n+m-1}^{m-1}$.
|
231
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
List 7. Let $x+y+z+w=20$. (8) Try to find (i) the number of positive integer solutions to equation (8); the number of groups;
(iii) the number of positive integer solutions to (8) satisfying $x \geq 7, y=8$;
(iv) the number of positive integer solutions to (8) satisfying $x \leqslant 6$;
(v) the number of positive integer solutions to (8) satisfying $x \leqslant 6, y \leqslant 7$.
|
(i) From the last statement of Example 4, the number of positive integer solutions of equation (8) is $C_{2}^{4-1}=C_{1}^{3}=969$.
(ii) Let $x^{\prime}=x-6$, then equation (8) becomes $x^{\prime}+y+z+w=14$.
Now, we find the number of positive integer solutions of (9) under the condition that $x \geqslant 7$ for equation (8). Therefore, the number of positive integer solutions is $C_{1}^{4}-1-1=C_{13}^{3}=286$.
(iii) Let $x^{\prime}=x-6, y^{\prime}=y-7$, then equation (8) becomes
$$
x^{\prime}+y^{\prime}+z+w=7.
$$
By the reasoning in (ii), the number of positive integer solutions is $C_{4-1}^{4}=C_{6}^{3}=20$.
(iv) Let $S_{1}=\{(x, y, z, w) \mid x \geq 7$, and $(x, y, z, w)$ is a positive integer solution of (8)\}, then the number of elements in $S_{1}$ (denoted as $\left|S_{1}\right|$) is $\left|S_{1}\right|=286$. From this, we deduce that the number of positive integer solutions of equation (8) satisfying $x \geq 6$ is $969-286=683$.
$$
\text{(v) Let } S_{2}=\{(x, y, z, w) \mid y \geqslant 8 \text{, }
$$
and $(x, y, z, w)$ is a positive integer solution of (8)\}, $S_{3}=\{(x, y, z, w) \mid x \geqslant 7, y \geqslant 8$, and $(x, y, z, w)$ is a positive integer solution of (8)\}, then $\left|S_{2}\right|=C_{13-1}^{1-1}=C_{12}^{3}=220, \quad\left|S_{3}\right|=20$. Let $I$ be the set of positive integer solutions of (8), then $|I|=969$. Now, from Figure 2, the number of positive integer solutions of (8) satisfying $x \geq 7, y \geq 8$ is
$$
|I|-\left|S_{1} \cup S_{2}\right|.
$$
But $\mid S_{1}\left\lfloor\left|S_{2}\right|=\left|S_{1}\right|+\left|S_{2}\right|-\left|S_{1} \cap S_{2}\right|\right|$. Also, note that $\left|S_{3}\right|=\left|S_{1} \cap S_{2}\right|$, so the number of solutions is
$$
\begin{aligned}
& |I|-\left|S_{1}\right|-\left|S_{2}\right|+\left|S_{3}\right| \\
= & 969-286-220+20=483 .
\end{aligned}
$$
As an exercise, the reader can find the number of positive integer solutions of (8) satisfying $x \leq 6, y \leqslant 7, z \leqslant 8, w \leqslant 9$ (Answer: 217).
|
483
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. The boss hands the letters to be printed to the secretary, one at a time, and places each on top of the pile. The secretary, whenever free, takes the top letter to type. One day, there are 9 letters to be typed, and the boss hands them over in the order of the first, second, $\cdots$, to the ninth. During lunch, the secretary tells a colleague that the eighth letter has been typed, but does not reveal any other details about the morning's work. The colleague is very curious to know which letters are left to be typed and in what order they will be printed.
Based on the above information, how many possible sequences are there for the letters typed in the morning? (Not typing any letter is also a possibility).
(Translated by Zhang Bingyi, proofread by Hou Zixin)
|
15. ( 704$)$
We consider two cases.
The first case: The ninth letter is finally typed by the secretary before noon.
This may be the order of the subset of the set $T=\{1,2$, ..., 6, 7, 9\} of possible subsets. In fact, the secretary can type a letter as soon as it arrives if its number is not in the subset, and not type other letters. $T$ has 8 elements, so there are $2^{8}=256$ subsets (including the empty set).
The second case: The ninth letter arrives at the secretary's office after noon.
Now the question is where this ninth letter will be "inserted" among the others? For each subset of $v=\{1,2, \cdots, 6,7\}$, each position is possible. For example, if the letters left at noon are 6, 3, 2, and the boss delivers the 9th letter as soon as the secretary finishes typing, then the typing sequence 6, 3, 9, 2 will occur.
For a permutation of $k$ letters, there are $k+1$ positions where the letter can be placed, before the secretary starts working, the boss delivers the 9th letter,
$$
=448 \text {. }
$$
Therefore, there are a total of $256+448=704$ possible typing sequences.
|
704
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Go to the room, take a card at random, and ask which number is most likely to be the unit digit of the number drawn by Jack and Jill.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
12
I
A
12. The list below, the numbers in the table represent the units digit of the sum of two numbers.
\begin{tabular}{ccccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline 2 & 3 & 4 & 3 & 6 & 7 & 8 & 0 & 3 \\
3 & 4 & 0 & 3 & 7 & 3 & 9 & 1 & 1 \\
4 & 5 & 6 & 7 & 3 & 9 & 0 & 1 & 2 \\
5 & 6 & 1 & 3 & 3 & 0 & 1 & 2 & 3 \\
6 & 7 & 8 & 9 & 0 & 1 & 2 & 3 & 4 \\
7 & 6 & 9 & 0 & 1 & 2 & 3 & 4 & 5 \\
8 & 9 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
9 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
\end{tabular}
The digit 0 appears 9 times, while the digits 1-9 each appear 8 times, so for now
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. A commercial lock with 10 buttons, which can be opened by pressing the correct five numbers, regardless of the order. The figure below is an example using $\{1,2,3,6,9\}$ as its combination. If these locks are reprogrammed to allow combinations of one to nine digits, how many additional combinations (i.e., not using five digits) are possible?
|
1. ( 770$)$
There are $2^{10}$ ways to choose from ten buttons. From this, we need to subtract the case where all 10 digits are selected, the case where no digits are selected, and the case where exactly five digits are selected. Therefore, the total number is
$$
2^{10}-1-1-C_{10}^{5}=1024-2-252=770 .
$$
|
770
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3 . If $\mathrm{iog}_{2}\left(\log _{8} x\right)=\log _{8}\left(\log _{2} x\right)$, find $\left(\log _{2} x\right)^{2}$.
|
$$
\begin{array}{l}
3 .(27) \\
\because \log _{8} x=\frac{1}{\log _{x} 3}=\frac{1}{3 \log _{x} 2}=\frac{1}{3} \log _{2} x, \\
\log _{8}\left(\log _{2} x\right)=\frac{1}{3} \log _{2}\left(\log _{2} x\right),
\end{array}
$$
Let $y=\log _{2} x$, then the equation in the problem is equivalent to
$$
\log _{2} \frac{y}{3}=\frac{1}{3} \log _{2} y .
$$
From (1), we get
$$
\begin{aligned}
& \log _{2}\left(\frac{y}{3}\right)^{3}=\log _{2} y, \\
\therefore \quad & \left(\frac{y}{3}\right)^{3}=y .
\end{aligned}
$$
Simplifying, we get $y\left(y^{2}-27\right)=0$, ( 2 ) $\therefore y^{2}=\left(\log _{2} x\right)^{2}=27$.
|
27
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in the figure, in a 5 $\times$ 5 square, there are 21 empty cells. It is known that positive integers can be filled in the empty cells so that each row and each column forms an arithmetic sequence. What number must be filled in the cell marked with a star?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
6. As shown in the figure, in a 5 $\times$ 5 square, there are 21 empty cells.
It is known that positive integers can be filled in the empty cells so that each row and each column forms an arithmetic sequence.
What number must be filled in the cell marked with a star?
|
$6 .(142)$
Let $a$ and $b \%$ be the numbers that fill in the blanks. From $b-2a$, and the common difference of the first row being $2b-a-74$, we get
$$
2a+4(b-2a)=186
$$
and $a+2(2b-a-74)=103$.
Solving these equations, we can find the numbers that fill in the blanks in the figure. Therefore, the number that should be entered at the starred position is 142.
|
142
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. In $\triangle \mathrm{ABC}$, $\operatorname{tg} \angle \mathrm{CAB}=\frac{22}{7}$, a line is drawn from $\mathrm{A}$ to $\mathrm{BC}$ dividing $BC$ into segments of lengths 3 and 17. What is the area of $\triangle \mathrm{ABC}$?
|
7. (110)
Let $x$ be the length of the altitude, intersecting $B C$ at $D$. Then, from $\angle B A C=\angle B A D+\angle D A C$, we get $\operatorname{arctg} \frac{22}{7}=\operatorname{arctg} \frac{3}{x}+\operatorname{arctg} \frac{17}{x}$.
Taking the tangent of both sides of (1) and using the $\operatorname{tg}(\alpha+\beta)$ formula, we obtain
$$
\frac{22}{7}=\frac{\frac{3}{x}+\frac{17}{x}}{1-\left(\frac{3}{x}\right)\left(\frac{17}{x}\right)}
$$
Simplifying, we get
$$
11 x^{2}-10 x-51=(x-11)(11 x+51)=0 \text {. }
$$
Since $x>0$, we know $11 x+51>0$, thus $x=11$.
Therefore, $S_{\triangle A B C}=\frac{1}{2}(3+17) \times 11=110$.
|
110
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The function $f$ defined on ordered pairs of positive integers satisfies the following three properties:
$f(x, x)=x, f(x, y)=f(y, x)$ and $(x+$ $y) f(x, y)=y f(x, x+y)$. Try to compute $f(14$, 52 ).
|
8. (364)
By property three, we get $f(x, x+y)=\frac{x+y}{y} f(x, y)$. We will repeatedly use this property, applying property two when the first number of the ordered pair is greater than the second, and finally applying property one, to get
$$
\begin{aligned}
& f(14,52)=f(14,14+38)=\frac{52}{38} f(14,38) \\
= & \frac{52}{38} \cdot \frac{38}{24} f(14,24)=\frac{52}{24} f(14,10) \\
= & \frac{26}{5} f(10,14)=\frac{26}{5} \cdot \frac{14}{4} f(10,4) \\
= & \frac{91}{5} f(4,10)=\frac{91}{5} \cdot \frac{10}{6} f(4,6)=\frac{91}{3} \cdot \frac{6}{2} f(4,2) \\
= & 91 f(2,4)=91 \cdot \frac{4}{2} f(2,2)=364 .
\end{aligned}
$$
|
364
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Find the smallest positive integer $n$, such that the last three digits of its cube are 888.
|
9. (192)
If a positive integer ends with 8, then this number itself must end with 2. That is, it can be written in the form $10k + 2$. Thus,
$$
n^{3} = (10k + 2)^{3} = 1000k^{3} + 600k^{2} + 120k + 8.
$$
The term $120k$ determines the tens digit of $n^{3}$, which is 8. Therefore, $12k$ must end with 8, meaning $k$ is 4 or 9. Thus, $k = 5m + 4$. ($m$ is a non-negative integer). Then,
$$
\begin{aligned}
n^{3} = & {[10(5m + 4) + 2]^{3}} \\
= & 125000m^{3} + 315000m^{2} + 26460m \\
& + 70488.
\end{aligned}
$$
Since the last three digits of the first two terms on the right side are 000, and the last three digits of the last term are 088, the last three digits of $26460m$ must be 800. The smallest integer $m$ that satisfies this condition is 3. Thus, $k = 5 \times 3 + 4 = 19$, and $n = 10 \times 19 + 2 = 192$.
|
192
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. The surface of a convex polyhedron is composed of 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex, one square, one octagon, and one hexagon meet. How many line segments connecting the vertices of the polyhedron lie inside the polyhedron, and not on the faces or edges of the polyhedron?
|
10. $(840)$
Let $a, b, c, d$ be the number of vertices, edges, face diagonals, and body diagonals, respectively.
Since at each vertex there is one of each type of regular polygon, and the four vertices of each square must be four different vertices of the polyhedron, we have
$c=4 \times$ the number of squares $=4 \times 12=48$.
(The same result can be obtained by considering the hexagons or octagons)
Since three edges emanate from each vertex, and each edge is counted twice, we have,
$$
b=\frac{1}{2} \times 3 a=72 \text{. }
$$
Also, since each square has 2 diagonals, each hexagon has 9 diagonals, and each octagon has 20 diagonals, we have,
$$
c=12 \times 2+8 \times 9+6 \times 20=216 .
$$
Since the polyhedron is convex, we have
$$
d=\mathrm{C}_{48}^{2}-b-c=1128-72-216=840 .
$$
|
840
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let $w_{1}, w_{4}, \cdots, m_{n}$ be complex numbers. If a straight line $l$ passes through points (complex numbers) $z_{1}: z_{2}, \cdots, z_{n}$, such that $\sum_{k=1}^{n}\left(z_{k}-w_{k}\right)=0$, then $l$ is called the “average line” of $w_{1}, w_{i}$, $\cdots, ~ w w_{\mathrm{n}}$.
For $w_{1}=32+170 i, w_{2}=-7+64 i$, $w_{3}=-9+200 i, w_{4}=1+27 i, w_{5}=-14$ $+43 i$, there is a unique “average line”, whose y-intercept is $y=3$. Find the slope of this line.
|
11. (163)
Let $y=m x+b$ be the "average line" of the complex numbers $w_{\mathrm{K}}=u_{\mathrm{K}}+i v_{\mathrm{K}}$, where $u_{\mathrm{K}}, v_{\mathrm{K}} \in R, k=1,2, \cdots, n$. Assume that on the line $y=m x+b$, we select complex numbers $z_{\mathrm{K}}=x_{\mathrm{K}}+i y_{\mathrm{K}}\left(x_{\mathrm{K}}, y_{\mathrm{K}} \in R, 1 \leqslant k \leqslant n\right)$ such that
$$
\begin{array}{c}
\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(z_{\mathrm{K}}-w_{\mathrm{K}}\right)=0 . \\
\text { Then } \sum x_{\mathrm{K}}=\sum u_{\mathrm{K}}, \Sigma y_{\mathrm{K}}=\Sigma v_{\mathrm{K}}, \text { and } y_{\mathrm{K}}=m x_{\mathrm{K}}+ \\
b(1 \leqslant k \leqslant n) . \text { Therefore, } \Sigma v_{\mathrm{K}}=\Sigma y_{\mathrm{K}}=\Sigma\left(m x_{\mathrm{K}}\right. \\
+b)=m \Sigma x_{\mathrm{K}}+n b=\left(\Sigma u_{\mathrm{K}}\right) m+n b . \\
\text { The problem gives } n=5, b=3, \Sigma u_{\mathrm{K}}=32 \\
+(-7)+(-9)+1+(-14)=3, \\
\sum v_{\mathrm{K}}=170+64+200+27+43=504 . \\
\therefore \quad 504=3 m+5 \text {, so } m=163 .
\end{array}
$$
|
163
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Let $P$ be an interior point of $\triangle A B C$, and extend $A P, B P, C P$ to intersect the opposite sides. In the figure, $a, b, c$, $d$ are the lengths of the respective line segments. Given that $a+b+c=43$, $d=3$, find $a b c=$ ?
|
12. $(441)$
$$
\begin{array}{l}
\because \frac{S_{\triangle B P C}}{S_{\triangle B A C}}=\frac{d}{d+a}, \frac{S_{\triangle C P A}}{S_{\triangle C B A}}=\frac{d}{d+b}, \\
\frac{S_{\triangle A P B}}{S_{\triangle A C B}}=\frac{d}{d+c},
\end{array}
$$
and $S_{\triangle B P C}+S_{\triangle C P A}+S_{\triangle A P B}=S_{\triangle A B C}$,
$$
\therefore \quad \frac{d}{d+a}+\frac{d}{d+b}+\frac{d}{d+c}=
$$
eliminating the denominators and simplifying, we get
$$
\text { collect } \quad \begin{array}{c}
2 d^{3}+\left(a+b+c=2 d^{2}+a b c=0 .\right. \\
+43 \times 3^{2}=441 .
\end{array}
$$
|
441
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. If $a, b$ are integers, and $x^{2}-x-1$ is a factor of $a x^{17}+b x^{10}+1$, try to find the value of $a$.
|
13. ( 987 )
The roots of $x^{2}-x-1=0$ are $p=\frac{1}{2}(1+\sqrt{5})$, $q=\frac{1}{2}(1-\sqrt{5})$. They must be the roots of $\boldsymbol{a} x^{17}$
$+b x^{18}+1=0$.
$$
\therefore a p^{17}+b p^{16}=-1, a q^{17}+b q^{18}=-1 \text {. }
$$
Multiplying the first equation by $q^{16}$, and the second equation by $p^{16}$, and using $p q=-1$, we get
$$
a p+b=-q^{18}, a q+b=-p^{10} .
$$
Subtracting the two equations, we get
$$
\begin{array}{l}
a(p-q)=p^{18}-q^{18} . \\
\therefore a=\frac{p^{16}-q^{16}}{p-q}=\left(p^{8}+q^{8}\right)\left(p^{4}+q^{4}\right) \\
\text { - }\left(p^{2}+q^{2}\right)(p+q) \text {. } \\
\because p+q=1 \text {, } \\
p^{2}+q^{2}=(p+q)^{2}-2 p q=1+2=3 \text {. } \\
p^{4}+q^{4}=\left(p^{2}+q^{2}\right)^{2}-2(p q)^{2}=9-2 \\
=7 \text {, } \\
p^{8}+q^{8}=\left(p^{4}+q^{4}\right)^{2}-2(p q)^{4}=49-2 \\
\end{array}
$$
$=47$.
|
987
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. $C$ is the graph of $x y=1$, and the symmetric image of $C$ about the line $y=2 x$ is $C^{\prime}$. It is known that $C^{\prime}$ can be expressed as
$$
12 x^{2}+b x y+c y^{2}+d=0
$$
Use this form to find the value of $b c$
|
14. $(84)$ points), connect $P^{\prime} P^{\prime}$ (as shown in the figure).
$\because F P^{\prime \prime} \perp$ line $y=2 x$,
$\therefore$ the slope of $P P^{\prime}$ is $-\frac{1}{2}$,
i.e., $-\frac{y-v}{x-u}=-\frac{1}{2}$.
Also, $\because$ the midpoint of $P P^{\prime}$ is on $y=2 x$,
$\therefore \quad \frac{y+v}{2}=2 \cdot \frac{x+u}{2}$.
From (1) and (2), we get
$$
u=\frac{4 y-3 x}{5}, \quad v=\frac{4 x+3 y}{5} .
$$
Substituting them into $u_{v}=1$, we get
$$
12 x^{2}-7 x y-12 y^{2}+25=0 \text {. }
$$
Thus, $b_{c}=(-7)(-12)=84$.
|
84
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. $N$ is the set of positive integers. Define the function $f$ on $N$ as follows:
$f(1)=1, f(3)=3$, and for $n \in N$ we have $f(2 n)=f(n)$,
$f(4 n+1)=2 f(2 n+1)-f(n)$,
$f(4 n+3)=3 f(2 n+1)-2 f(n)$.
Question: How many $n \in N$, and $n \leqslant 1988$ such that $f(n) = n$?
|
3. According to these formulas, we can find
The pattern shown in this table seems to be
$$
\begin{array}{l}
f\left(2^{k}\right)=1, f\left(2^{k}-1\right)=2^{k}-1, \\
f\left(2^{k}+1\right)=2^{k}+1 .
\end{array}
$$
This suggests that we should consider the "binary" expansion of natural numbers. Our conjecture is:
$f(n)$ is the binary number formed by the reverse arrangement of the binary expansion of $n$.
We will prove this conjecture using induction. Since $f(2n) = f(n)$, we only need to consider the case where $n$ is odd. If $n$ is of the form $4m+1$, let
$$
\begin{array}{c}
4 m+1=\varepsilon_{k} 2^{k}+\cdots+\varepsilon_{1} 2+\varepsilon_{0} \cdot 1 \\
\left(\varepsilon_{\mathrm{i}}=0,1\right) \text {. Clearly, we should have } \varepsilon_{0}=1, \varepsilon_{1}=0 \text {. Then }
\end{array}
$$
we have
$$
4 n=\varepsilon_{k} 2^{2}+\cdots+\varepsilon_{2} 2^{2} .
$$
This gives
$$
m=\varepsilon_{k} 2^{k-2}+\cdots+\varepsilon_{3} 2+e_{2} .
$$
Thus,
$$
2 m+1=\varepsilon_{k}^{2 k-1}+\cdots+\varepsilon_{3} 2^{2}+\varepsilon_{2} \cdot 2+1 .
$$
By the induction hypothesis,
$$
\begin{array}{l}
f(2 m+1)=2^{k-1}+\sum_{\mathrm{j}=2}^{k} e_{1} 2^{k-1}, \\
f(m)=\sum_{\mathrm{j}=2}^{k} e_{1} 2^{k-1} .
\end{array}
$$
This leads to
$$
\begin{array}{l}
f(4 m+1)=2 f(2 m+1)-f(m) \\
=2^{k}+\sum_{\mathrm{j}=2}^{k} \varepsilon_{1} 2^{k+1-1}-\sum_{\mathrm{j}=2}^{k} e_{\mathrm{j}} 2^{k-j} \\
=2+\sum_{\mathrm{j}=2}^{k} e_{1} 2^{k-1} \\
=\sum_{j=0}^{k} \varepsilon_{1} 2^{x-1},
\end{array}
$$
which is consistent with our conjecture.
Now, let $n$ be of the form $4m+3$, and assume
$$
4 m+3=\sum_{\mathrm{j}=0}^{\mathrm{k}} \varepsilon_{\mathrm{j}} 21 .
$$
Clearly, in this case, $\varepsilon_{0}=\varepsilon_{1}=1$. Therefore,
$$
\begin{array}{l}
4 m=\sum_{\mathrm{j}=2}^{k} \varepsilon_{1} 2!, \\
m=\sum_{\mathrm{j}=2}^{k} e_{1} 2^{1-2}, \\
2 m+1=1+\sum_{\mathrm{j}=2}^{k} \varepsilon_{\mathrm{j}} 2^{1-1} .
\end{array}
$$
Thus, by definition and the induction hypothesis,
$$
f(4 m+3)=3 f(2 m+1)-2 f(m)
=3 \sum_{j=1}^{k} \varepsilon_{1} 2^{k-1}-2 \sum_{j=2}^{k} \varepsilon_{1} 2^{k-1}
=\varepsilon_{i} 2^{j}+\sum_{j=1}^{k} \varepsilon_{1} 2^{k-1} \text{ (since } \varepsilon_{1}=\varepsilon_{0} \text{ )}
=\sum_{j=0}^{k} \varepsilon_{1} 2^{k-1}.
$$
This completely confirms our conjecture.
Next, we discuss binary numbers with $2m$ digits
$\left(x_{1}, x_{2}, \cdots, x_{n}, y_{1}, y_{2}, \cdots, y_{n}\right)$.
Of course, $x_{1}=1$. If this number has the property:
$\left(x_{1}, x_{2}, \cdots, x_{3}, y_{1}, y_{2}, \cdots, y_{n}\right)
=\left(y_{n}, \cdots, y_{2}, y_{1}, x_{n}, \cdots, x_{2}, x_{1}\right)$.
We call it symmetric. This indicates that a symmetric binary number is completely determined by the first $m$ digits $x_{1}, x_{2}, \cdots, x_{\mathrm{n}}$. Since $x_{1}=1$, and each of $x_{2}, \cdots, x_{\mathrm{m}}$ can be any of the two digits 0,1. Therefore, there are exactly $2^{m-1}$ symmetric binary numbers with $2m$ digits.
Symmetric binary numbers with $2m-1$ digits can also be defined in an obvious way, and the number of such numbers can also be proven to be $2^{m-1}$.
Finally, we look at how many numbers between 1 and 1988 can be expressed as symmetric binary numbers. Since $1024<1988<2048$, i.e., $2^{10}<1988<2^{11}$. Therefore, the number of symmetric binary numbers less than 2048 is
$1+1+2+2+4+4+8+8+16+16+32 = 94$.
In fact, $1988=(11111000100)_{2}$, and we can see that there are only two 11-bit symmetric binary numbers exceeding this number, namely $(1111011111)_{2}$ and $(1111111111)_{2}$.
So there are $94-2=92$ natural numbers $n$ not exceeding 1988 such that $f(n)=n$.
|
92
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
54. (Mongolia 4) Find the smallest natural number $\boldsymbol{n}$ such that if the set $\{1,2, \cdots, n\}$ is arbitrarily divided into two non-intersecting subsets, then one of the subsets contains 3 different numbers, where the sum of two of them equals the third.
|
20. (Mongolia 4)
Solution: The required minimum value is 96.
Let $A_{\mathrm{s}}=\{1,2, \cdots, n\}$ be divided into two subsets $B_{\mathrm{n}}$ and $C_{\mathrm{s}}$, and neither $B_{\mathrm{s}}$ nor $C_{\mathrm{s}}$ contains three distinct numbers where the product of two equals the third. If $n \geqslant 96$, then the divisors of 96, $\{1,2,3,4,6,8,12,16,24,32,48,96\}$, cannot be separated without contradiction. Assume $1,2 \subseteq B_{0}$. This leads to four cases:
$$
\begin{array}{l}
A: 3 \in B_{0}, 4 \in B_{0} ; \\
B: 3 \in B_{0}, 4 \in C_{0} ; \\
C: 3 \in C_{0}, 4 \in B_{0} ; \\
D: 3 \in C_{0}, 4 \in C_{1} .
\end{array}
$$
In case $A$, $2,3,4 \in B_{0}$, so 6, 8, and $k 2 \in C_{0}$. Also, $48,96 \in B_{\mathrm{n}}$, and from $2,48,96 \in B$, a contradiction arises. Similarly, in case $B$, $2,3 \in B_{a}, 4 \in C_{\mathrm{a}}$, leading to $6 \in C_{\mathrm{s}}$, and thus $24 \in B_{\mathrm{s}}$. Consequently, $8,12,48 \in C_{0}$, and since $6 \cdot 8=48$, a contradiction arises. In case $C$, $2,4 \in B_{\mathrm{a}}, 3 \in C_{\mathrm{a}}$, so $8 \in C_{\mathrm{s}}, 24 \in B_{\mathbb{\pi}}$, and thus $6,48 \in C_{\mathrm{a}}$. Again, $6 \cdot 8=48$ leads to a contradiction. Finally, in case $D$, $2 \in B_{\text {: }}$, $3,4 \in C_{\mathrm{s}}$, so $12 \in B_{\mathrm{a}}$, and $6,24 \in C_{\mathrm{a}}$, and since $4 \cdot 6=24$, a contradiction arises.
For example, $B_{\mathrm{a}}=\left\{1, p, p^{2}, p^{2} q r, p^{4} q, p^{3} q^{2}\right\}$
and $C_{0}=\left\{p^{3}, p^{4}, p^{5}, p^{8}, p q, p^{2} q, p q r, p^{3} q \cdot p^{2} q^{2}\right\}$, where $p, q, r$ are different primes. That is, $B_{8 \overline{0}}=\{1,2,3,4,5,7,9,11,13,17,19,23,25,29,31,37,41,43,47,48,49,53,59,60,61,67,71,72,73,79,80,83,84,89,90\}$.
|
96
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In a "Battleship" game on a grid with $7 \times 7$ cells, what is the minimum number of shots needed to guarantee hitting a four-deck ship? It is known that the ship (1) has shape $A$ (Figure 2); (2) consists of four adjacent cells, but is not a square.
|
Solution: (1) Since 12 ships of shape $A$ can be placed in a $7 \times 7$ grid such that no two ships share a common cell, the number of shots required is at least 12. From figures $3, a, b$, it is shown that 12 shots are sufficient.
(2) 20 steps.
As shown in Figure 4, 20 shots can complete the task. Suppose we shoot 19 times, dividing the $7 \times 7$ grid into four non-overlapping $3 \times 4$ rectangles (as in Figure 4), one of which has no more than 4 shots. Assume this rectangle is as shown in Figure 5. Clearly, there should be a shot in one of its columns, and no row contains 3 shots (otherwise, a ship of shape $A$, $B$, or $D$ could be placed in this rectangle without being hit). Considering the central cell $a$ in the rectangle in Figure 5, there can be only one shot configuration in the $3 \times 3$ square (or a symmetric configuration) (Figure 6). However, in that shot configuration, a ship of shape $C$ (Figure 2) can be placed in the $3 \times 3$ square. Therefore, it can be assumed that shots occur in cells $a$ and $b$. In this case, cells 1, 3, 4, 5, 6, and 7 (Figure 5) are not shot. One of the four corner cells must also be unshot. For example, cell 2. However, then a ship can be placed in cells 1, 2, 3, and 4. Therefore, in any configuration of 19 shots, a ship can be placed in the remaining cells of the $7 \times 7$ grid without being hit.
|
20
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Place one more of these shapes inside the square.
Keep the original text's line breaks and format, and output the translation result directly.
Note: The second sentence is a directive for the translation task and should not be included in the translation output. Here is the translation:
Place one more of these shapes inside the square.
|
Solution: For each $2 \times 2$ square, at least two cells should be covered. An $8 \times 8$ square can be divided into 16 non-overlapping $2 \times 2$ squares. Therefore, at least 11 figures should be covered. With 11 figures, it is feasible.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Uncle Qiel Yuemuer every night selects 9 or 10 warriors from 33 warriors to be on duty according to his own wishes. At least how many days are needed to ensure that each warrior has the same number of duty shifts?
|
The minimum number of days is 7 days, during which each warrior is on duty 2 times.
Let the required minimum number of days be $N$, and each warrior is on duty $n$ times within these $N$ days. We number the warriors sequentially as $1,2, \cdots, 33$, and introduce the number $a_{1 \times}(i=1,2, \cdots, N ; k=1,2, \cdots, 33)$ to represent the following: if the $k$-th warrior is not on duty on the $i$-th day, then $a_{1 x}=0$; if the $k$-th warrior is on duty on the $i$-th day, then $a_{i k}=1$. Let $P_{k}$ be the number of people on duty on the $i$-th day, and $Q_{k}$ be the number of days the $k$-th warrior is on duty. Thus, according to the problem conditions, the following relationships hold:
$$
\begin{array}{l}
9 \leqslant a_{i 1}+a_{\mathrm{i} 2}+\cdots+a_{\mathrm{i} 33}=P_{\mathrm{i}} \leqslant \\
10, i=1,2, \cdots, N \text {; } \\
\text { (1) } \\
a_{1_{k}}+a_{2 k}+\cdots+a_{N k}=Q_{k}=n \text {, } \\
i=1,2, \cdots, \dot{N} \text {; } \\
\text { (2) } \\
\end{array}
$$
\begin{tabular}{|c|c|c|}
\hline Day & On-duty Warrior Numbers & \begin{tabular}{l}
On-duty
\end{tabular} \\
\hline 1 & $2,3,4,5,6,7,8,9,28,29$ & 10 \\
\hline 2 & $1,2,3,4,5,6,7,8,9$ & 9 \\
\hline 3 & $11,12,13,14,15,16,17,18,30,31$ & 20 \\
\hline 4 & $10,11,12,13,14,15,16,17,18$ & $\frac{9}{10}$ \\
\hline 5 & $20,21,22,23,24,25,26,27,32,33$ & . 0 \\
\hline 6 & $19,20,21,22,23,24,25,26,27$ & $\theta$ \\
\hline 7 & $1,10,19,28,29,30,31,32,33$ & 9 \\
\hline
\end{tabular}
Figure 12
Summing the inequality (1) for $i=1,2, \cdots, N$, we get the inequality $9 N \leqslant S \leqslant 10 N$, where:
$$
\begin{array}{l}
S=\left(a_{11}+a_{12}+\cdots+a_{1,33}\right)+\left(a_{21}\right. \\
\left.+a_{22}+\cdots+a_{2,33}\right)+\cdots+\left(a_{\mathrm{N} 1}+a_{\mathrm{N} 2}+\cdots\right. \\
\left.+a_{\mathrm{N}, 33}\right) .
\end{array}
$$
Summing the equation (2) for $k=1,2, \cdots, 33$, we get the equation $S=33 n$. This implies $N \leqslant 33 n \leqslant 10 N$, or $\frac{3}{11} N \leqslant n \leqslant \frac{10}{33} N$. An integer $n$ between $\frac{3}{11} N$ and $\frac{10}{33} N$ does not exist for $N=1,2,3,4,5,6$, but when $N=7$, there is an integer $n=2$ that satisfies the inequality $\frac{3}{11} N \leqslant n \leqslant \frac{10}{33} N$. Therefore, the number of duty shifts $n$ is at least 2, and the minimum number of days $N$ is at least 7. The duty schedule (similar to the table in Figure 12) meets the problem's conditions.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A number plus 79 becomes a perfect square, and when this number plus 204 is added, it becomes another perfect square. Then this number is $\qquad$ .
|
$$
\begin{array}{c}
204=y^{2}, \text { where } x \text { and } y \text { are non-negative integers. Thus, } \\
y^{2}-x^{2}=125=5^{3}, \text { i.e., } (y+x)(y-x)=5^{3}. \\
\therefore \quad\left\{\begin{array} { l }
{ y + x = 125, } \\
{ y - x = 1; }
\end{array} \text { or } \left\{\begin{array} { l }
{ y + x = 25, } \\
{ y - x = 5. }
\end{array} \text { Solving, we get } \left\{\begin{array} { l }
{ x = 62, } \\
{ y = 63; }
\end{array} \text { or } \left\{\begin{array}{l}
x=10, \\
y=15.
\end{array}\right.\right.\right. \\
\therefore \quad a=62^{2}-79=3765,
\end{array}
$$
or $a=10^{2}-79=21$.
|
21
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let $E(n)$ denote the greatest integer $k$ that is a divisor of the product $1^{1}, 2^{2}, 3^{3} \cdots n^{n}$. Then $E(150)=$
|
$9.1,2, \cdots, 150$ contains 30 multiples of 5, each having one factor of 5 removed, along with their respective exponents, the exponent of 5 is
$$
\begin{array}{l}
5+5 \cdot 2+5 \cdot 3+\cdots+5 \cdot 30=5(1+2+3 \\
+\cdots+30)=2325 .
\end{array}
$$
1. $2, \cdots, 150$ contains 6 multiples of $5^{2}$, each having one factor of 5 removed, and then another factor of 5 removed, along with their respective exponents, the exponent of 5 is $5^{2}+5^{2} \cdot 2+\cdots+5^{2} \cdot 6=5^{2}(1+2+\cdots+6)$ $=525$.
$1, 2, \cdots, 150$ contains only one multiple of $5^{3}$, which has had two factors of 5 removed, and can have one more factor of 5 removed, along with its respective exponent, the exponent of 5 is $5^{3}$ $=125$.
$1,2, \cdots, 150$ contains no multiples of $5^{4}$.
$\therefore E(150)=2325+525+125=2975$
|
2975
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8 . All three-digit numbers $a b c$ that satisfy $a b c=(a+b+c)^{3}$ are $\qquad$
|
8. $\because 100 \leqslant a b c=(a+b+c)^{3} \leqslant 999$,
$\therefore 5<a+b+c$ S 9 . Directly calculating $5^{3}=125$, $6^{3}=216,7^{3}=343,8^{3}=512,9^{3}=729$, we can see that only the three-digit number 512 meets the conditions of the problem.
|
512
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. In a regular $\triangle A B C$, $D$ and $E$ are the midpoints of sides $A B$ and $A C$ respectively. Fold $\triangle A B C$ along $D E$ to form a dihedral angle $A-D E-C B=60^{\circ}$. If $B C=10 \sqrt{13}$, then the distance between the skew lines $A E$ and $B D$ is $\qquad$
|
10. Let $M$ be the midpoint of $BC$, and the median $AM$ of $\triangle ABC$ intersects $DE$ at $N$. After folding into a dihedral angle, $\angle ANM$ is the plane angle of the dihedral angle $A-DE-CB$, i.e., $\angle ANM = 60^{\circ}$.
Connect $EM, AM, DM. \because EM \parallel DB$, $\therefore$ the distance $d$ between $AE$ and $DB$ is the distance from $DB$ to the plane $AEM$, which is also the distance from $D$ to the plane $AEM$.
Prove $AF = a$, then $EM = a, AN = NM = \frac{\sqrt{3}}{2} a, \triangle ANM$ is an equilateral triangle,
$$
\begin{array}{c}
AM = \frac{\sqrt{3}}{2} a, S_{\triangle AM} = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} a \\
\cdot \sqrt{a^{2} - \left(\frac{\sqrt{3}}{4} a\right)^{2}} = \frac{\sqrt{39}}{16} a^{2} .
\end{array}
$$
Let $O$ be the midpoint of $MN$, connect $AO$, then it is easy to see that $AO \perp$ base $DECB$, and $AO = AN \sin 60^{\circ} = \frac{3}{4} a$, thus
$$
\begin{array}{l}
d \cdot 3_{\triangle AEM} = d \cdot \frac{\sqrt{39}}{10} a = V_{A-DEM} \\
= 4O \cdot S_{\triangle DEM} = \frac{3}{4} a \cdot \frac{\sqrt{3}}{4} a^{2} . \\
d = \frac{3}{\sqrt{13}} a = \sqrt{13} \cdot 5 \sqrt{13} = 15 .
\end{array}
$$
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Divisible by 3, and the digits of each number are limited to $1, 2, 3$ (1, 2, 3 do not have to be all used) all natural numbers less than 200000 are $\qquad$
$\qquad$
|
12. The only one-digit number that can be divided by 3 and is composed of only the digits 1, 2, 3 is 3.
The two-digit numbers that can be divided by 3 and are composed of only the digits $1, 2, 3$ can be obtained as follows: arbitrarily write one of $1, 2, 3$ in the tens place, then configure the units digit based on the remainder of the tens digit divided by 3. If the remainder is 0, write 3 in the units place; if the remainder is 1, write 2; if the remainder is 2, write $1. \therefore$ There are 3 such two-digit numbers.
Similarly, the number of three-digit, four-digit, and five-digit numbers that meet the conditions are $3^{2}, 3^{3}, 3^{4}$ respectively. For six-digit numbers that meet the conditions, the first digit must be 1, thus there are $4^{4}$ such numbers.
Therefore, the total number of natural numbers that meet the conditions is $1+3+3^{2}+3^{3}+2 \times 3^{4}=202$.
|
202
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11. Let $\mathrm{a}$ be a real number, find the minimum value of the quadratic function
$$
y=x^{2}-4 a x+5 a^{2}-3 a
$$
denoted as $\mathrm{m}$.
When $a$ varies in $0 \leqslant a^{2}-4 a-2 \leqslant 10$, find the maximum value of $m$.
|
$$
\text{Solve} \begin{aligned}
y & = x^{2} - 4ax + 5a^{2} - 3a \\
& = (x - 2a)^{2} + a^{2} - 3a
\end{aligned}
$$
The above expression achieves its minimum value when $x = 2a$, so
$$
m = a^{2} - 3a = \left(a - \frac{3}{2}\right)^{2} - \frac{9}{4}.
$$
Furthermore, $0 \leqslant a^{2} - 4a - 2 \leqslant 10$ implies
$$
0 \leqslant a^{2} - 4a - 2,
$$
and $10 \geqslant x^{2} - 4a - 12$.
From (1), we get $a \leqslant 2 - \sqrt{6}$, or $a \geqslant 2 + \sqrt{6}$.
From (2), we get $-2 \leqslant a \leqslant 6$.
From (3) and (4), the range of $a$ that satisfies both (1) and (2) is $-2 \leqslant a \leqslant 2 - \sqrt{6}$ or $2 + \sqrt{6} \leqslant a \leqslant 6$.
In the above range, when $a = 6$, $\left|a - \frac{3}{2}\right|$ is maximized.
Therefore, when $a = 6$, $m$ reaches its maximum value of 18.
$$
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5. (1MO-23-1)
The function $f(n)$ is defined for all positive integers $n$, taking non-negative integer values. For all positive integers $m, n$, $f(m+n)-f(m)-f(n)=0$ or 1;
and $f(2)=0, f(3)>0, f(9999)=3333$. Find $f(1982)$.
|
Solve $0=f(2) \geqslant 2 f(1)$, we get $f(1)=0$.
From $f(3)-f(2)-f(1)=0$ or 1, we get $0 \leqslant f(3) \leqslant 1$.
Given $f(3)>0$, we get $f(3)=1$.
First, we prove that for $k3333,
\end{aligned}
$$
Contradicting the given condition, so it must be that $f(3k)=k$.
It is clear that $660 \leqslant \hat{j}(1332) \leqslant 861$,
If $f(1982)=661$ then
$$
\begin{array}{c}
f(9 \text { 9.9) }=\hat{j}(5 \times 1982+89) \\
\quad \geqslant 5 f(1982)+f(89) \\
\quad \geqslant 5 \times 661+f(87) \\
\quad \geqslant 3305+29>3333,
\end{array}
$$
Contradicting the given condition. So it must be that
$$
f(1982)=660 \text {. }
$$
|
660
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7. (IMO-20-3)
Let $f, g: Z^{+} \rightarrow Z^{+}$ be strictly increasing functions, and $f\left(Z^{+}\right) \cup g\left(Z^{+}\right)=Z^{+}, f\left(Z^{+}\right) \cap g\left(Z^{+}\right)=\phi$, $g(n)=f[f(n)]+1$. Find $f(240)$. Here $Z^{+}$ is the set of positive integers.
|
Let
$$
F=\{f(n)\}, G=\{g(n)\} .
$$
If $F \cup G=Z^{+}, F \cap G=\phi$, then $F$ and $G$ are complementary.
We prove $f(n)=(\sqrt{5}+1) \cdots$ by induction. For: $g(n)=f(f(n))+1>1$. So we can set $f(1)=1=\left(\frac{\sqrt{5}+1}{2}\right)$, and $g(1)$ $=2$.
If $v_{n}<m$, then $f(n)=\left\{\frac{\sqrt{5}+1}{2}\right\}$. For $n$, if $f(n)<m \leqslant f(n+1)$, by the monotonicity of $f$,
$$
\sqrt{5}+1-1<m<\frac{\sqrt{5}+1}{2}(n+1) .
$$
From this, we can see that among the natural numbers up to $f(m)$, there are $m$ numbers belonging to $F$, i.e., $f(1), f(2), \cdots, f(m)$, and $n$ numbers belong to $G$, i.e., $G(1), g(2), \cdots, g(n)=f[f(n)]+1$. Therefore, $f(n)=n+m$. But from (1),
$$
n<\frac{\sqrt{5}-1}{2} m<n+1,
$$
so $\left\{\frac{\sqrt{5}-1}{2} m\right\}=n$. Therefore,
$$
f(m)=\left\{\frac{\sqrt{5}-1}{2} m+m=(\sqrt{5}+1) m\right\} .
$$
This completes the proof by induction. It is not difficult to calculate from this that
$$
f(240)=388 .
$$
|
388
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, solving the equation $\frac{x^{2}-7}{6}=\sqrt{6 x+7}$. This is the result of transitioning from concrete to abstract thinking.
From a geometric perspective, this is equivalent to finding the x-coordinates of the intersection points of the line $y=\frac{x^{2}-7}{6}$ and the curve $y=\sqrt{6 x+7}$. These two happen to be inverse functions of each other. According to the geometric property that the graphs of a function $y=f(x)$ and its inverse $y=f^{-1}(x)$ are symmetric about the line $y=x$, their intersection points must lie on the line $y=x$.
|
Solve $\sqrt{6 x+7}=x$ to get
$$
\begin{array}{l}
x^{2}-6 x-7=0, \\
\therefore x_{1}=-1 \text { (discard), } x_{2}=7 .
\end{array}
$$
Upon verification, the root of the original equation is $x=7$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. There is a four-digit number. It is known that its tens digit minus 1 equals the units digit, and its units digit plus 2 equals the hundreds digit. The sum of this four-digit number and the number formed by reversing the order of its four digits equals 9878. Try to find this four-digit number.
|
Solution: Let the required four-digit number be $a b c d$. According to the problem, we have $\overline{a b c \bar{d}}+\overline{d c b a}=9878$,
which means $\left(10^{3} \cdot a+10^{2} \cdot b+10 \cdot c+d\right)+\left(10^{3} \cdot d\right.$ $\left.+10^{2} \cdot c+10 \cdot b+a\right)=9878$, or $10^{3} \cdot(a+d)+10^{2} \cdot(b+c)+10 \cdot(b+c)$ $+(a+d)=9878$.
By comparing the first and last two digits on both sides of (1), we get
$$
a+d=8 \text {. }
$$
Thus, $b+c=17$.
Also, $c-1=d, d+2=b$,
which means $b-c=1$,
Therefore, the solution is $a=1, b=9, c=8, d=7$.
Hence, the required four-digit number is 1987.
In the solution process, the equations (4) and (5) were not substituted into (1) to solve the indeterminate equation, but instead, the equations (3) and (4) were directly obtained by comparing the first and last two digits of (1). This avoids complex discussions. However, this technique has certain limitations and should not be misused.
|
1987
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Rearrange the digits of a three-digit number to form the largest possible three-digit number, and subtract the smallest digit from it, which is exactly equal to the original number. Find these three digits. (Hua Luo Geng Math Contest 1988 Junior High School Level)
|
Let the three-digit number be $a, b, c$. If $abc$ is the largest, and $cb_{2}$ is the smallest, and
$$
\begin{array}{l}
\overline{a b c}-cba \\
=\left(10^{2} a+10 b+c\right)-\left(10^{2} \cdot c+10 b+a\right) \\
=99(a-c),
\end{array}
$$
i.e., the required three-digit number is a multiple of 99. Among such three-digit numbers $198, 297, 396, 495, 594, 693, 792, 891$, only 495 meets the condition. Therefore, the required three-digit number is 495.
This problem uses the fact that the difference between a three-digit integer and its reverse is a multiple of 99. If this result is not used, various cases need to be discussed separately, making the solution process quite cumbersome.
|
495
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5, find a four-digit number that can be divided by 11, and the quotient obtained is 10 times the sum of the digits of this four-digit number.
|
Let $\overline{a b c d}=110(a+b+c+d)$, then we have
$$
890 a=10 b+100 c+109 d \text {. }
$$
Also, since 11 divides $\overline{a b c d}$, we have $a-b+c-d= \pm 11$ or $a-b+c-d=0$.
If $a-b+c-d=0$,
then from $\left\{\begin{array}{l}a+c=b+d, \\ 890 a=10 b+100 c+109 d\end{array}\right.$
(2) $-10 \times(1)$ we get
$880 a=110 c+99 d \leqslant(110+99) \times 9$
$$
=1881 \text {. }
$$
If $a=1$, substituting into (3) gives $80=10 c+9 d$,
$$
\therefore d=0, c=8, b=9 \text {; }
$$
If $a=2$, substituting into (3) gives $1760=110 c+99 d$
$$
\therefore d=0, c=16 \text { (not valid). }
$$
Therefore, the required four-digit number is 1980.
If $a-b+c-d=11$,
then from $\left\{\begin{array}{l}a+c=11+b+d, \\ 890 a=10 b+100 c+109 d \text { (5) }\end{array}\right.$
(5) $-10 \times$ (4) we get
$110 c=880 a-99 d+110 \geqslant 990$. (6)
$\therefore c \geqslant 9$, i.e., $c=9$. Substituting into (6) gives $a=1$, $d=0, b=-1$ (not valid).
If $a-b+c-d=-11$, similarly, we get no solution.
Therefore, the only solution is 1980.
This problem mainly uses the method of bounding to determine the range of unknowns, which is applicable in a wide range of scenarios.
|
1980
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6. In the six-digit number $31 x y 13$ !, $x 、 y$ are both greater than 5, and this number is divisible by 13, find the four-digit number $\overline{1 x y 9}$.
|
$$
\begin{array}{l}
\text { Solution: Let } n=31 x y 13 \\
=3 \cdot 10^{5}+10^{4}+x \cdot 10^{3}+y \cdot 10^{2} \\
+13 \\
=310013+10^{3} x+10^{2} y, \\
\end{array}
$$
where $14,6 \leqslant x, y \leqslant 9$.
Since the remainders of $310013,10^{3}, 10^{2}$ when divided by 13 are $2,-1,-4$ respectively, we can set $310013=13 k_{1} + 2$, $10^{2} x=13 k_{2}-x, 10^{2} y=13 k_{3}-4 y$. Let $k=k_{1}+k_{2}+k_{3}\left(k_{1}, k_{2}, k_{3} \in \mathbb{N}\right)$, then
$$
n=13 k+2-x-4 y.
$$
$\because 13|n, \quad \therefore 13| 2-x-4 y$.
Let $2-x-4 y=13 t(t \in \mathbb{Z})$.
From $6 \leqslant x, y \leqslant 9$ we get
$$
-43 \leqslant 13 t \leqslant-28.
$$
Therefore: $t=-3$. Substituting $-x-4 y=13 t$, we get
$$
x + 4 y=41, \quad \text { and } y=10+\frac{1-x}{4}.
$$
Thus, $x=9, y=8$. The required four-digit number is 1989.
This problem mainly uses the divisibility of numbers, and the method is also quite special.
For such problems, it is not necessary to express integers as polynomials, but since this is not within the scope of this article, it will not be discussed further here.
$$
|
1989
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1. The nine-digit number in the form of $1234 \times \times \times \times \times$ composed of $1,2,3,4,5,6,7,8,9$, where the ten-thousands place is not 5, the thousands place is not 6, the hundreds place is not 7, the tens place is not 8, and the units place is not 9. How many such nine-digit numbers are there?
|
Solution: Obviously, this is a problem of five letters being placed in the wrong envelopes. And
$$
\begin{aligned}
\overline{5} & =5!-C_{5}^{1} 4!+C_{5}^{2} 3!-C_{5}^{3} 2! \\
& +C_{5}^{4} 1!-C_{5}^{5} 0! \\
& =120-120+60-20+5-1 \\
& =44 .
\end{aligned}
$$
The number of derangements it is asking for is 44.
|
44
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Prove: $\quad(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}$ has a minimum value of 8 on $0<u<\sqrt{2}, v>0$.
|
$(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}$ can be seen as the square of the distance between point $P_{1}(u, \sqrt{2-u^{2}})$ and $P_{2}\left(v, \frac{9}{v}\right)$.
$P_{1}$ satisfies the equation of the circle:
$$
x^{2}+y^{2}=2 \text{. }
$$
$P_{2}$ satisfies the equation of the hyperbola:
$$
x y=9, x>0 \text{. }
$$
Obviously, the distance between $M$ and $N$ reaches the minimum distance between $P_{1}$ and $P_{2}$. Let $y=x$, it is easy to get $M(1,1), N(3,3)$, so the minimum value $=(3-1)^{2}+(3-1)^{2}=8$.
|
8
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $\alpha, \beta$ be acute angles. When
$$
-\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta}
$$
takes the minimum value, the value of $\operatorname{tg}^{2} \alpha+\operatorname{tg}^{2} \beta$ is
|
To find the minimum value of the original expression, we need
$$
\sin ^{2} \beta \cos ^{2} \beta=\frac{1}{4} \sin ^{2} 2 \beta
$$
to take the maximum value. Since $\beta$ is an acute angle, we know $\beta=\frac{\pi}{4}$.
When $\beta=\frac{\pi}{4}$,
$$
\begin{aligned}
\text { the original expression } & \geqslant \frac{1}{\cos ^{2} \alpha}+\frac{4}{\sin ^{2} \alpha} \\
& =\left(\frac{1}{\cos ^{2} \alpha}+\frac{4}{\sin ^{2} \alpha}\right)\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right) \\
& =1+\tan^{2} \alpha+\frac{4}{\tan^{2} \alpha}+4 \\
& \geqslant 1+4+4=9 .
\end{aligned}
$$
When the product of the terms is a constant, the minimum value of their sum is
$$
\tan^{2} \alpha+\frac{4}{\tan^{2} \alpha} \geqslant 4
$$
The equality holds if and only if
$$
\tan^{2} \alpha=\frac{4}{\tan^{2} \alpha}
$$
which gives $\tan^{2} \alpha=2$. Therefore,
$$
\tan^{2} \alpha+\tan^{2} \beta=2+1=3 .
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let [x] denote the greatest integer not exceeding $x$, then the sum of the squares of all real roots of the equation $x^{2}-5[x]+4=0$ is
保留源文本的换行和格式，直接输出翻译结果。
|
Solve: The intersection points of the graphs of the functions $y=x^{2}$ and $y=5[x]-4$ (see the figure below) are the solutions to the original equation.
From $5[x]=x^{2}+4>0$, we know that $[x]>0$.
Therefore, we only
need to consider the case where $x \geqslant 1$.
Since $x \geqslant[x]$,
substituting into the original equation gives $[x]^{2}-5[x]+4 \leqslant 0$.
Thus, $[x] \in[1,4]$.
When $[x]=1$, $x^{2}=5-4=1$, and $x=1$ satisfies the requirement $[x]=1$, so $x=1$ is a solution to the original equation;
Similarly, when $[x]=2,3,4$, it is easy to see that $x=\sqrt{6}$, $\sqrt{11}$, $4$ are also solutions to the original equation. Therefore, the sum of the squares of all roots is
$$
1+6+11+16=34 \text {. }
$$
Note: This problem can also be solved using congruence relations.
|
34
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. A bag of peanuts has a total of 1988 peanuts. On the first day, a monkey takes one peanut. Starting from the second day, the number of peanuts taken each day is the sum of all the peanuts taken in previous days. If on a certain day the number of peanuts left in the bag is less than the total number of peanuts already taken, on that day it takes one peanut again and starts a new cycle according to the original rule. If this continues, on which day will the bag of peanuts be completely taken by the monkey?
|
Let's assume the monkey used $k$ days in the first round, meaning that on the $k+1$-th day, it started again by taking one peanut. Therefore, the total number of peanuts taken in the first $k$ days is
$$
1+1+2+2^{2}+2^{3}+\cdots+2^{k-1}=?^{k-1}
$$
peanuts, and $1988-2^{k-1}<2-1$. Also, $1988-2^{k-1}-2^{k-1}<2^{t-1}, t<k$.
$$
1988=2^{10}+2^{0}+2^{3}+2^{7}+2^{0}+2^{2} \text { . }
$$
Therefore, it took
$$
11+10+9+8+7+3=48 \text { days. }
$$
|
48
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. The general term of the sequence is $a_{\mathrm{n}}=b[\sqrt{n+c}]+d$, and the terms are calculated successively as
$$
1,3,3,3,5,5,5,5,5, \cdots \text {. }
$$
where each positive odd number $m$ appears exactly $m$ times consecutively. The above $b, c, d$ are undetermined integers. Then, the value of $b+c+d$ is $\qquad$ where $[x]$ denotes the greatest integer not exceeding $x$.
|
Solve: First determine $b$.
Since $a_{n}$ is odd, and $a_{n}+1 \geqslant a_{n}$, we know that $a_{n}+1-a_{n}$ $\in\{0,2\}$,
i.e., $b[\sqrt{n+1+c}]-b[\sqrt{n+c}]$ is 0 or 2.
For any natural number $n$, it always holds that
$[\sqrt{n+1+c}]-[\sqrt{n+c}] \in\{0,1\}$.
Clearly, $b \neq 0$.
When $-[\sqrt{n+c}])=2$, by (1), it can only be that $[\sqrt{n+1+c}]-[\sqrt{n+c}]=1, b=2$.
Since $b$ is a constant, we have $b=2$.
Next, find $c$.
From $=2[\sqrt{u+c}]+d$ we know $c \geqslant-1$.
Based on the characteristics of the sequence $a_{\mathrm{n}}$, we have $a_{k}^{2}=2 k-1$.
Taking a sufficiently large $k$, ensuring $2 k+3>c$, at this time $(k+1)^{2}+c<(k+1)^{2}+2 k+3=(k+2)^{2}$; we get $\left[\sqrt{(k+1)^{2}+c}\right]<k+2$. $-1=2 k$ or
$$
\left[\sqrt{(k+1)^{2}+c}\right]-[\sqrt{1+c}] \equiv k \text {. }
$$
Substituting (2) gives: $[\sqrt{1+c}]<2$.
Therefore, $c<3$.
If $c=0,1,2$, then in each case $[\sqrt{5+c}]-[\sqrt{4+c}]$
$=2-2=0$ which contradicts $a_{5}-a_{4}=2$.
In summary, $c=-1$.
Finally, $d=a_{1}=1$.
Thus, $b+c+a'=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. $f(n)$ is a function defined on the set of natural numbers, when $p$ is a prime number, $f(p)=1$, and for any natural numbers $r$, $s$, we have
$$
f(r s)=r f(s)+s f(r) .
$$
Then, the sum of all $n$ that satisfy the condition
$$
f(n)=n, 1 \leqslant n \leqslant 10^{4}
$$
is.
|
Given the problem, for any natural numbers $r, s$, we have
$$
\frac{f(r s)}{r s}=\frac{f(r)}{r}+\frac{f(s)}{s} \text {. }
$$
Thus,
$$
\begin{array}{l}
\frac{f\left(p^{m}\right)}{p^{m}}=\frac{f(p)}{p}+\frac{f(p)}{p}+\cdots+\frac{f(p)}{p} \\
\quad=m \cdot \frac{f(p)}{p} .
\end{array}
$$
When $p$ is a prime number, we get
$$
f\left(p^{m}\right)=m p^{m-1},
$$
Suppose the prime factorization of $n$ is $n=p_{1}^{m} p_{2}^{n} \cdots p_{k}^{m_{k}}$. Then,
$$
\begin{array}{l}
\frac{f(u)}{n}=\frac{f\left(p_{1}^{m_{1}^{1}} p_{2}^{\mathrm{m}}{ }_{2}^{2} \cdots p_{k}^{m}\right.}{p_{1}^{m_{1}} p_{2}^{\mathrm{m}_{2}}{ }_{2}^{2} \cdots p_{k}^{\mathrm{m}_{\mathrm{k}}}} \\
=\frac{f\left(p_{1}^{m}{ }^{1}\right)}{p_{1}^{m}{ }_{1}^{2}}+\frac{f\left(p_{2}^{m}{ }_{2}^{m}\right)}{p_{2}^{m}}+\cdots+\frac{f\left(p_{k}^{m_{k}}\right.}{p_{k}^{m}} \\
=\frac{m_{1}}{p_{1}}+\frac{m_{2}}{p_{2}}+\cdots+\frac{m_{\mathrm{k}}}{p_{\mathrm{k}}} \text {. } \\
\end{array}
$$
Therefore, $f(n)=n$ is equivalent to
$$
\frac{m_{1}}{p_{1}}+\frac{m_{2}}{p_{2}}+\cdots+\frac{m_{\mathrm{k}}}{p_{\mathrm{k}}}=1 \text {. }
$$
By finding a common denominator, we get
$$
\begin{array}{r}
m_{1} p_{2} \cdots p_{k}+m_{2} p_{1} p_{3} \cdots p_{k}+\cdots+m_{k} p_{1} p_{2} \\
\cdots p_{k-1}=p_{1} p_{2} \cdots p_{k}
\end{array}
$$
Since any two of $p_{i}, p_{1}, \cdots, p_{k}$ are coprime, $p_{1}$ must be a factor of $m_{1} p_{2} \cdots p_{k}$, thus $p_{1} \mid m_{1}$. Similarly, for $1 \leqslant i \leqslant k$, we have $p_{i} \mid m_{i}$. Therefore, $\frac{m_{i}}{p_{i}} \geqslant 1$.
Thus, (5) holds only if $k=1$. If $f(n)=n$ holds, the prime factorization of $n$ can only be of the form $n=p^{\mathrm{p}}$.
Since $1 \leqslant n \leqslant 10^{4}$, $n$ can take the values $2^{2}, 3^{3}, 5^{5}$. Therefore, $2^{2}+3^{3}+5^{5}=3156$.
|
3156
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Tank A contains 4 liters of liquid A, Tank B contains 2 kilograms of liquid B, and Tank C contains 2 kilograms of liquid C. These liquids can all be mixed. First, 1 liter of liquid from Tank A is poured into Tank B, and then 1 liter of the mixed liquid from Tank B is poured into Tank C. Finally, 1 liter of the mixed liquid from Tank C is poured back into Tank A. This operation is called one mixing. After $n$ mixings, the amount of liquid A in Tank B is denoted as $f(n)$. If $f(n)>0.9999$, what is the minimum value of $n$? $\qquad$ .
|
Let $a_{n}, b_{n}, c_{n}$ be the amounts of liquid A in containers 甲, 乙, and 丙, respectively, after $n$ mixings. Then,
$$
a_{n}+b_{n}+c_{n}=4 \text{. }
$$
By symmetry, we have
$$
b_{n}=c_{n} \text{. }
$$
Now, let's examine the amount of liquid A in container 甲 after the $(n+1)$-th mixing. According to the problem,
$$
\begin{aligned}
a_{n+1} & =\frac{1}{2} a_{n}+\frac{1}{3}\left(\frac{1}{4} a_{n}+b_{n}\right) \\
+ & \left.\frac{1}{3} \div \frac{1}{4} a_{n}+c_{5}\right) \\
& =\frac{1}{2} a_{n}+\frac{1}{6} a_{n}+\frac{2}{3} b_{n} \\
& =\frac{2}{3} a_{n}+\frac{1}{3}\left(4-a_{n}\right) \\
& =\frac{1}{3} a_{n}+\frac{4}{3}
\end{aligned}
$$
Thus, $a_{n+1}-2=\frac{1}{3}\left(a_{n}-2\right)$.
By induction, we get
$$
\begin{array}{l}
a_{n+1}-2=\frac{1}{3}\left(a_{n}-2\right)=\frac{1}{3^{2}}\left(a_{n-1}-2\right) \\
=\cdots=\left(\frac{1}{3}\right)^{n+1}\left(a_{0}-2\right) .
\end{array}
$$
Substituting $a_{0}=4$ yields
$$
\begin{array}{l}
a_{n}=2\left[1+\left(\frac{1}{3}\right)^{n}\right], \\
b_{n}=\frac{1}{2}\left(4-a_{n}\right)=1-\left(\frac{1}{3}\right)^{n} .
\end{array}
$$
Given $f(n)>0.9999$, i.e.,
$1-\left(\frac{1}{3}\right)^{n}>0.9999$ or $3^{n}>10000$. The smallest value of $n$ is 9.
(Li Yongle, Yu Rongpei)
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3. Write the numbers $1, 2, 3, \cdots$, 1986, 1987 on the blackboard. At each step, determine some numbers from those written and erase them, replacing them with the remainder of their sum divided by 7. After several steps, two numbers remain on the blackboard, one of which is 987. What is the second remaining number?
(13th All-Russian Mathematics Competition, 1987)
|
Solution: Clearly, at each step, the sum of all the numbers written down modulo 7 is preserved.
Let the remaining number be $x$, then $x+987$ is congruent to $1+2+\cdots+1987$ modulo 7.
Since $1+2+\cdots+1987=1987 \times 7 \times 142$ is divisible by 7, the remainder is 0, so $x+987$ is also divisible by 7. Since 987 is not the remainder when divided by 7, $x$ is the remainder of the operation when divided by 7, $0 \leqslant x \leqslant 6$. Since 987 is divisible by 7,
thus $x=0$.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 ・At
7:00 AM, depart from port $A$,
take a motorboat at a constant speed of $v_{1}$ kilometers/hour (4
$\leqslant v_{1} \leqslant 20$) to
head to port $B$ 50 kilometers
away from port $A$, then immediately switch to a steamship at a constant speed of $v_{2}$ kilometers/hour $\left(30 \leqslant v_{2} \leqslant 60\right)$ to travel 300 kilometers to city $C$, and it is stipulated to arrive at city $C$ between 4:00 PM and 9:00 PM. Let the time spent on the motorboat and the steamship be $x, y$ hours, respectively, and try to answer:
(1) Find the range of $x, y$ that satisfies the above conditions,
(2) The travel cost $p$ is given by $p=1000+30(5-x)$ $+20(8-y)$, if the travel cost is to be minimized, what values should $v_{1}, v_{2}$ take?
|
(1) From the given conditions, we have
\[
\left\{
\begin{array}{l}
x=\frac{300}{v_{2}}, \quad 30 \leqslant v_{2} \leqslant 100, \\
y=\frac{50}{v_{1}}, \quad 4 \leqslant v_{1} \leqslant 20,
\end{array}
\right.
\]
and
\[
9 \leqslant x+y \leqslant 14,
\]
which implies
\[
\left\{
\begin{array}{l}
3 \leqslant x \leqslant 10, \\
2.5 \leqslant y \leqslant 12.5, \\
9 \leqslant x+y \leqslant 14.
\end{array}
\right.
\]
The range of the points in question is shown in the shaded area (including the boundaries) in Figure 6.
(2) From the given conditions,
\[
y=-\frac{3}{2} x + \frac{1}{20}(1310 - p)
\]
To minimize \( p \), the intercept on the \( y \)-axis must be maximized. From Figure 6, it is clear that the line must pass through point \( A(10, 4) \). Substituting \( x=10 \) and \( y=4 \) into (1) gives \( p=930 \). At this point, \( v_{1}=\frac{50}{4}=12.5 \) and \( v_{2}=\frac{300}{10}=30 \).
|
930
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Calculate $\sqrt{(31)(30)(29)(28)+1}$.
|
\begin{aligned} \text { 1. } & \because(k+1) \cdot k \cdot(k-1) \cdot(k-2)+1 \\ & =[(k+1)(k-2)][k(k-1)]+1 \\ & =\left(k^{2}-k-2\right)\left(k^{2}-k\right)+1 \\ & =\left(k^{2}-k\right)^{2}-2\left(k^{2} \cdots k\right)+1 \\ & =\left[\left(k^{2}-k\right)-1\right]^{2} \\ \therefore \quad & v^{\prime}(31) \cdot(30) \cdot(29) \cdot(28)+1 \\ & =30^{2}-30-1=869 .\end{aligned}
|
869
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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