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6. Given the sets $A=\{1,2,3,4,5,6\}, B=$ $\{6,7,8,9\}$, by selecting 3 elements from $A$ and 2 elements from $B$, we can form $\qquad$ new sets with 5 elements.
|
6. Selecting 3 elements from set $A$ has $C_{6}^{3}$ ways, and selecting 2 elements from set $B$ without 6 has $C_{3}^{2}$ ways, resulting in a total of $C_{6}^{3} \cdot C_{3}^{2}$ ways; selecting 3 elements from set $A$ without 6 has $C_{5}^{3}$ ways, and selecting 1 element from set $B$ with 6 has $C_{3}^{1}$ ways, resulting in a total of $C_{5}^{3} \cdot C_{3}^{1}$ ways. Therefore, the number of sets that meet the conditions is $C_{6}^{3} \cdot C_{3}^{2} + C_{5}^{3} \cdot C_{3}^{1} = 90$ (sets).
|
90
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $a_{1}, a_{2}, \cdots, a_{n}$ represent any permutation of the integers 1, 2, $\cdots, n$. Let $f(n)$ be the number of such permutations that satisfy: (1) $a_{1}=1$; (2) $\left|a_{i}-a_{i+1}\right| \leqslant 2, i=1$, $2, \cdots, n-1$. Determine whether $f(1996)$ is divisible by 3.
|
Solution: It is easy to find that $f(1)=f(2)=1, f(3)=2$.
When $n \geqslant 4$, we have $a_{1}=1, a_{2}=2$ or 3.
(a) When $a_{2}=2$, we proceed as follows: delete $a_{1}$, $b_{1}=a_{2}-1, b_{2}=a_{3}-1, \cdots, b_{n-1}=a_{n}-1$, to get a permutation $b_{1}, \cdots, b_{n-1}$ that meets the conditions, and the number of such permutations is $f(n-1)$.
(b) When $a_{2}=3$, for $a_{3}$ we have two cases:
( $\left.1^{\circ}\right) a_{3}=2$, then $a_{4}=4$. By the same process as (a), the number of permutations that meet the conditions is $f(n-3)$;
$\left(2^{\circ}\right) a_{3} \neq 2$, then 2 must be after 4. This leads to the conclusion that the odd numbers are arranged in reverse order, i.e., $1357 \cdots 8642$, which is one permutation.
Therefore, $f(n)=f(n-1)+f(n-3)+1$.
For $n \geqslant 1$, let $r(n)$ be the remainder when $f(n)$ is divided by 3, then we have
$$
\begin{array}{l}
r(1)=r(2)=1, r(3)=2, \\
r(n)=r(n-1)+r(n-3)+1 .
\end{array}
$$
Thus, $r(n+8)=r(n+7)+r(n+5)+1$
$$
\begin{aligned}
= & {[r(n+6)+r(n+4)+1] } \\
& +[r(n+4)+r(n+2)+1]+1 \\
= & r(n+6)+2 r(n+4)+r(n+2)+3 \\
= & {[r(n+5)+r(n+3)+1] } \\
& +2 r(n+4)+r(n+2)+3 \\
= & \{[(r(n+4)+r(n+2)+1] \\
& +[r(n+2)+r(n)+1]+1\} \\
& +2 r(n+4)+r(n+2)+3 \\
= & 3 r(n+4)+3 r(n+2)+6+r(n),
\end{aligned}
$$
which means $r(n+8)$ has the same remainder as $r(n)$ when divided by 3, i.e., $r(n+8)=r(n)$.
Therefore, $r(1996)=r(4+249 \times 8)=r(4)$.
And $r(3)+r(1)+1=4$,
so $r(4)=1$,
which means the remainder when $f(1996)$ is divided by 3 is 1. Therefore, $f(1996)$ is not divisible by 3.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Find the largest integer $n$, such that all non-zero solutions of the equation $(z+1)^{n}=z^{n}+1$ lie on the unit circle.
|
$$
\begin{array}{l}
\text { Solution: Using the binomial theorem, the equation can be transformed into } \\
z\left(C_{n}^{1} z^{n-2}+C_{n}^{2} z^{n-3}+C_{n}^{3} z^{n-4}+\cdots+C_{n}^{n-2} z+\right. \\
\left.C_{n}^{n-1}\right)=0(n>3) .
\end{array}
$$
Let the non-zero solutions of the equation be $z_{i}(i=1,2, \cdots, n-2)$. By Vieta's formulas, we have
$$
\begin{aligned}
S_{1} & =\sum_{i=1}^{n-2} z_{i}=-\frac{C_{n}^{2}}{C_{n}^{1}}=-\frac{n-1}{2}, \\
S_{2} & =\sum_{1 \leqslant i4$ satisfies the given conditions, then $z_{i} \cdot \bar{z}_{i}=\left|z_{i}\right|^{2}$ $=1, x_{i}=z_{i}+\bar{z}_{i}$ is a real number $(i=1,2, \cdots, n-2)$. Since the coefficients of the equation are all real, the roots appear in conjugate pairs, so the non-zero solutions can also be represented as $\bar{z} i=1,2, \cdots, n-2)$. Therefore,
$$
\begin{array}{l}
t_{1}=\sum_{i=1}^{n-2} x_{i}=\sum_{i=1}^{n-2}\left(z_{i}+\bar{z}_{i}\right)=2 S_{1}=1-n, \\
t_{2}=\sum_{1: i<j \leqslant n-i} x_{i} x_{i}:=\frac{1}{2}\left(t_{1}^{2}-\sum_{i=1}^{n-2} x_{i}^{2}\right), \\
\sum_{i=2}^{n-2} x_{i}^{2}=\sum_{i=1}^{n-2}\left(z_{i}^{2}+\bar{z}_{i}^{2}+2 z_{i} \cdot \bar{z}_{i}\right) \\
=\sum_{i=1}^{n-2}\left(z_{i}^{2}+\bar{z}_{i}^{2}\right)+2 \sum_{i=1}^{n-2} z_{i} \cdot \bar{z}_{i} \\
=2 \sum_{i=1}^{n-2} z_{i}^{2}+2 \sum_{i=1}^{n-2} 1 \\
=2\left(S_{1}^{2}-2 S_{2}\right)+2(n-2) .
\end{array}
$$
Therefore, $t_{2}=\frac{1}{2}\left[\left(2 S_{1}\right)^{2}-2\left(S_{1}^{2}-2 S_{2}\right)\right.$
$$
\begin{aligned}
& -2(n-2)] \\
= & \frac{1}{12}\left(7 n^{2}-30 n+35\right) .
\end{aligned}
$$
By Vieta's formulas, the real numbers $x_{i}(i=1,2, \cdots, n-2)$ are the $n-2$ roots of the real-coefficient equation $x^{n-2}-t_{1} x^{n-3}+t_{2} x^{n-4}+\cdots+$ $t_{n-3} x+t_{n-2}=0$. Using Theorem (1), we have
$$
\Delta_{1}=(n-3)\left(-t_{1}\right)^{2}-2(n-2) t_{2} \geqslant 0,
$$
which is $(n-3)(n-1)^{2}-2(n-2) \cdot$
$$
\frac{7 n^{2}-30 n+35}{12} \geqslant 0 \text {. }
$$
Simplifying, we get
$$
(n-4)\left[(n-5)^{2}-12\right] \leqslant 0 .
$$
Solving, we get $n \leqslant 5+\sqrt{12}<9$, i.e., $n \leqslant 8$.
When $n=8$, the equation becomes
$$
\begin{array}{l}
\left(8 z^{6}+28 z^{5}+56 z^{4}+70 z^{3}+56 z^{2}\right. \\
+28 z+8) z=0 .
\end{array}
$$
Its non-zero solutions are the 6 roots of the following equation (*):
$$
\begin{array}{l}
4 z^{6}+14 z^{5}+28 z^{4}+35 z^{3}+28 z^{2}+14 z+4 \\
=0 . \\
\text { The equation }(*) \text { can be transformed into } \\
4\left(z^{3}+z^{-3}\right)+14\left(z^{2}+z^{-2}\right)+28\left(z+z^{-1}\right) \\
+35=0, \\
4\left(z+z^{-1}\right)^{3}+14\left(z+z^{-1}\right)^{2}+16\left(z+z^{-1}\right) \\
+7=0 .
\end{array}
$$
The equation (*) can be transformed into
$4\left(z+z^{-1}\right)^{3}+14\left(z+z^{-1}\right)^{2}+16\left(z+z^{-1}\right)$
$\Delta_{2}=(3-1) \cdot 16^{2}-2 \cdot 3 \cdot 14 \cdot 7=$ $-76<0$, by Theorem (2), the roots of the equation $4 x^{3}+14 x^{2}+$ $16 x+7=0$ are not all real, i.e., there exists a root $z_{i}$ of equation (*) such that $z_{i}+z_{i}^{-1}$ is not a real number. Thus, $z_{i}+z_{i}^{-1} \neq z_{i}+\bar{z}_{i}$, i.e., $\left|z_{i}\right| \neq 1$. Therefore, $n=8$ does not satisfy the given conditions.
When $n=7$, the equation becomes
$$
\begin{aligned}
0 & =(z+1)^{7}-\left(z^{7}+1\right) \\
& =(z+1) \sum_{i=0}^{6} C_{6}^{i} z^{i}-(z+1) \sum_{i=0}^{6}(-1)^{i} z^{6} \\
& =(z+1) \sum_{i=0}^{6}\left[C_{6}^{i}-(-1)^{i}\right] z^{i} \\
& =7(z+1) \approx\left(z^{2}+z+1\right)^{2} .
\end{aligned}
$$
Its non-zero solutions are -1 and $\cos 120^{\circ} \pm i \sin 120^{\circ}$, all on the unit circle.
In summary, the largest integer $n$ that satisfies the given conditions is
$$
\boxed{7}
$$
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. On the right
side, in a column of squares, each square except for 9.7 represents a number. It is known that the sum of any 3 consecutive squares is 19. Then $A+H+M+O$ equals ( ).
(A) 21
(B) 23
(C) 25
(D) 26
|
-1. D.
From the given, we easily know that $A+H=10$.
$$
\begin{array}{l}
\because M+O+X=O+X+7, \\
\therefore M=7 .
\end{array}
$$
Similarly, $O=9$.
$$
\therefore A+H+M+O:=26 \text {. }
$$
|
26
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 A square piece of paper contains 100 points. Together with the vertices of the square, there are 104 points, and no three points among these 104 points are collinear. Now, the square piece of paper is entirely cut into triangles, with each vertex of these triangles being one of the 104 points, and each of the 104 points being a vertex of these triangles. If cutting along a line segment counts as one cut, how many cuts are needed to obtain these triangles, and how many triangles can be cut in total?
|
Analysis: If there is only one point $P_{0}$ inside the square, connecting $P_{0}$ to the four vertices results in four triangles, which clearly require the first 4 cuts. If another point $P_{1}$ is added, this point must fall within one of the triangles, let's assume it falls within $\triangle P_{0} C D$ (as shown in Figure 5), at this point, two more triangles are added, requiring 3 cuts. If another point $P_{2}, \cdots \cdots$ is added, and so on, the total number of triangles is $4 + 2 \times 99 = 202$, and the total number of cuts is $4 + 3 \times 99 = 301$.
Explanation: This problem can also examine the total degree sum of all the internal angles of the triangles, to find the number of triangles $=$ $\frac{360^{\circ} \times \text{number of points inside the square} + 90^{\circ} \times 4}{180^{\circ}}$.
|
301
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $S=1^{2}-2^{2}+3^{2}-1^{2}+\cdots+99^{2}-$ $100^{2}+101^{2}$. Then the remainder when $S$ is divided by 103 is $\qquad$
|
2. 1 .
$$
\begin{aligned}
S= & 1+\left(3^{2}-2^{2}\right)+\left(5^{2}-4^{2}\right)+\cdots+\left(99^{2}-98^{2}\right) \\
& +\left(101^{2}-100^{2}\right) \\
= & 1+2+3+\cdots+100+101 \\
= & \frac{101 \times 102}{2}=5151=103 \times 50+1 .
\end{aligned}
$$
Therefore, the required remainder is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Locations A and B are respectively upstream and downstream on a lake. Each day, there is a boat that departs on time from both locations and travels at a constant speed towards each other, usually meeting at 11:00 AM on the way. ... Due to a delay, the boat from location A left 40 minutes late, and as a result, the two boats met at 11:15 AM on the way. It is known that the boat departing from location A travels at a speed of 44 kilometers per hour in still water, and the boat departing from location B travels at a speed of $v$ kilometers per hour in still water. Then $v=$ $\qquad$ kilometers per hour.
|
3. 6 .
As shown in the figure, the two ships usually meet at point $A$. On the day when ship B is late, they meet at point $B$. According to the problem, ship A takes $15^{\prime}$ to navigate segment $AB$, and ship C takes
$$
\begin{array}{l}
(40-15)^{\prime}=25^{\prime} . \\
\therefore \frac{15}{60}(44+v)=\frac{25}{60}\left(v^{2}-v\right),
\end{array}
$$
which simplifies to $5 v^{2}-8 v-132=0$.
$$
(v-6)(5 v+22)=0 \text {. }
$$
Solving this, we get $v=6$ (the negative root is discarded).
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $a$ is an integer, the two real roots of the equation $x^{2}+(2 a-1) x+$ $a^{2}=0$ are $x_{1}$ and $x_{2}$. Then $\left|\sqrt{x_{1}}-\sqrt{x_{2}}\right|=$
|
4. 1.
From the problem, the discriminant $\Delta=-4 a+1 \geqslant 0$, so $a \leqslant 0$.
$$
\begin{array}{l}
\left(\sqrt{x_{1}}-\sqrt{x_{2}}\right)^{2}=\left(x_{1}+x_{2}\right)-2 \sqrt{x_{1} x_{2}} \\
=1-2 a+2 a=1 \\
\therefore\left|\sqrt{x_{1}}-\sqrt{x_{2}}\right|=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $x_{1}, x_{2}, \cdots, x_{7}$ be natural numbers, and $x_{1}<x_{2}$ $<\cdots<x_{6}<x_{7}$, also $x_{1}+x_{2}+\cdots+x_{7}=159$. Then the maximum value of $x_{1}+x_{2}+x_{3}$ is $\qquad$
|
5. 61.
Given that $x_{7} \geqslant x_{6}+1 \geqslant x_{5}+2 \geqslant x_{4}+3 \geqslant x_{3}+4 \geqslant$ $x_{2}+5 \geqslant x_{1}+6$.
Similarly, $x_{6} \geqslant x_{1}+5, x_{5} \geqslant x_{1}+4, x_{4} \geqslant x_{1}+3, x_{3} \geqslant x_{1}+2, x_{2} \geqslant x_{1}+1$.
$$
\begin{array}{l}
\therefore 159=x_{1}+x_{2}+\cdots+x_{7} \geqslant 7 x_{1}+(1+2+\cdots+6) \\
\quad=7 x_{1}+21 . \\
\therefore x_{1} \leqslant \frac{138}{7}<20 .
\end{array}
$$
Therefore, the maximum value of $x_{1}$ is 19.
Taking $x_{1}=19$, then $140=x_{2}+x_{3}+\cdots+x_{7} \geqslant 6 x_{2}+(1+2+\cdots+5)$. Thus, $x_{2} \leqslant 20$.
Taking $x_{1}=19, x_{2}=20$, then $120=x_{3}+\cdots+x_{7} \geqslant 5 x_{3}+(1+2+3+4)$. Thus, $x_{3} \leqslant 22$.
Following this, we can get the 7 numbers that meet the conditions: $19,20,22,23,24,25,26$. Therefore, the maximum value sought is 61.
|
61
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Try to design a method to divide a square into 8 smaller squares without repetition or omission (the sizes of the smaller squares can be different); also, how to divide a square into 31 smaller squares under the same requirements?
|
Three, 1. It is easy to divide a square into $4^{2}=16$ smaller squares, then combine 9 of them located at one corner into one square, resulting in a total of $16-9+1=8$ squares.
Divide into 16 smaller squares, then further divide any 5 of them into 4 smaller squares each, resulting in a total of $16-5+5 \times 4=31$ squares.
|
31
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 - 8 smaller cubes are combined to form a larger cube with an edge length of 10 units, which is then painted and subsequently divided into the original smaller cubes. The number of smaller cubes that have at least one face painted is ( ).
(A) 600
(B) 512
(C) 488
(D) 480
|
Very complicated. We can consider the opposite situation, that is, the $8 \times 8 \times 8$ cube in the very center of the cube with each side measuring 10 units, then the answer is not hard to find: $1000-512=488$ units.
|
488
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3: In the sequence of natural numbers $1,2,3, \cdots, n, \cdots$, the sum of numbers from the $m$-th $(m$ $<n$ ) to the $n$-th number is 1997. Then $m$ equals $\mathrm{f}$
|
3. 998 .
$(m+n)(n-m+1)=2 \cdot 1997, 1997$ is a prime number.
(1) $\left\{\begin{array}{l}m+n=1997, \\ n-m+1=2 ;\end{array}\right.$
(2) $\left\{\begin{array}{l}m+n=2 \cdot 1997, \\ n-m+1=1\end{array}\right.$
| (1), $n=999, m=998$.
| (2), $m=n=1997$ (discard).
|
998
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 On the square $A B C D$: there are 10 points, 8 of which are inside $\triangle A B C$, and 2 points are on the side $\mathrm{I}$ of the square (not at the vertices). [L These 10 points, together with points $A, B, C, D$, are not collinear. Find how many small triangles these 10 points, along with the 4 vertices of the square, can divide the square into.
|
Solution: Since adding a point on the plane is equivalent to adding a full angle, the total angle of the figure formed by 14 vertices is $360^{\circ} \times 14=5040^{\circ}$. We have $\}^{\circ}, B, C, D$ as the vertices of the square, so $\angle A+\angle B+\angle C+\angle D=$ $90^{\circ} \times 4=360^{\circ}$. Two of the points are on the sides of the square, so the sum of the interior angles of the figure formed by these two points is $180^{\circ} \times 2$. Thus,
$$
\begin{aligned}
& 5040^{\circ}-360^{\circ} \times 4+360^{\circ}-360^{\circ} \times 2 \\
+ & 180^{\circ} \times 2 \\
= & 3600^{\circ} .
\end{aligned}
$$
Therefore, there are $20\left(=\frac{3600^{\circ}}{180^{\circ}}\right)$ small
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the equation $\left(a^{2}-1\right) x^{2}-2(5 a+1) x+24=0$ has two distinct negative integer roots. Then the integer value of $a$ is $\qquad$ .
（The 1st Zu Chongzhi Cup Junior High School Mathematics Competition）
|
(Solution: $a=-2$ )
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $k$ is an integer, and the equation $\left(k^{2}-1\right) x^{2}-$ $3(3 k-1) x+18=0$ has two distinct positive integer roots. Then $k=$ $\qquad$
（4th Hope Forest Junior High School Mathematics Competition）
|
(Solution: $k=2$ )
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given that $a$ is an integer, the equation $x^{2}+(2 a+1) x+a^{2}=0$ has integer roots $x_{1} 、 x_{2}, x_{1}>x_{2}$. Try to find $\sqrt[4]{x_{1}^{2}}-\sqrt[4]{x_{2}^{2}}$.
(1991, Nanchang City Junior High School Mathematics Competition)
|
(Given $a>0$, we know $0>x_{1}>x_{2}$, the result is -1)
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 A glasses workshop in a factory has received a batch of tasks, requiring the processing of 6000 type $A$ parts and 2000 type $B$ parts. This workshop has 214 workers, each of whom can process 3 type $B$ parts in the time it takes to process 5 type $A$ parts. These people are divided into two groups, both working simultaneously, with each group processing one type of part. To complete this batch of tasks in the shortest time possible, how should they be divided?
|
Solution: Let the number of people assigned to produce $\mathrm{C} A$ type parts be $x$, then the number of people assigned to produce $B$ type parts is $214-x$. Suppose the respective rates are 5 and 3. The time to complete the entire task is
$$
t(x)=\max \left\{\frac{6000}{5 x}, \frac{2000}{3(214-x)}\right\},(x \in N \text {, }
$$
$1 \leqslant x \leqslant 213)$.
When $1 \leqslant x \leqslant 213, x \in N$, the function $\frac{6000}{5 x}$ is decreasing, and the function $\frac{2000}{3(214-x)}$ is increasing.
Thus, $\min t(x)$ is obtained near the solution $x_{0}=137 \frac{4}{7}$ of the equation $\frac{6000}{5 x}=\frac{2000}{3(214-x)}$.
Since $f(137)<f(138)$, the number of people assigned to produce $A$ type parts is 137, and the number of people in the other group is 77.
|
137
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 The mathematical model of the population dynamics of four types of chicks, where the young can become mature within the same year, estimates the number of a female population after 5 years.
(1) Each female breeds once a year;
(2) Each female breeds only once in the first year of development, and does not breed again afterward, but still survives;
(3) Each female breeds once in the first two years, and then is eliminated (or dies).
(1996, "Mathematical Knowledge Application" Summer Camp)
|
Solution: Let $a_{n}$ represent the number of female birds in the $n$th generation.
(1) According to the problem, $a_{0}=1, a_{n}=2 a_{n-1}, n \geqslant 1$.
It is easy to get $a_{n}=2^{n} a_{0}=2^{n}$.
Therefore, $a_{5}=2^{5}=32$.
(2) According to the problem, $a_{0}=1, a_{n}=a_{n-1}+1$.
Thus, $a_{5}=6$.
(3) Let $b_{n}$ represent the number of young birds in the first year of development in the $n$th year, and $c_{n}$ represent the number of mature birds in the second year of development in the $n$th year. According to the problem,
$$
a_{n}=b_{n}+c_{n}, b_{n}=b_{n-1}+c_{n-1}, c_{n}=b_{n-1}.
$$
From the above equations, we get $a_{n}=a_{n-1}+a_{n-2}, n=2$. Initial conditions are $a_{0}=1, a_{1}=2$.
It is easy to see that $a_{5}=13$.
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. A stationery store that operates both wholesale and retail has the following rule: If you buy 51 pencils or more (including 51), you pay the wholesale price; if you buy 50 pencils or fewer (including 50), you pay the retail price. The wholesale price for 60 pencils is 1 yuan cheaper than the retail price for 60 pencils. Now, the class leader, Xiao Wang, comes to buy pencils. If he buys one pencil for each student, he must pay the retail price, which costs $m$ yuan ($m$ is a positive integer); but if he buys 10 more pencils, he can pay the wholesale price, which also costs exactly $m$ yuan. How many students does Xiao Wang have?
|
(Tip: Let the whole class have $x$ students, the equation is $\frac{60 m}{x}-\frac{60 m}{x+10}$ $=1(40<x \leqslant 50)$, solving it yields $x=50)$
|
50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 If $k$ is a positive integer, and the two roots of the linear equation $(k-1) x^{2}-p x+k=0$ are both positive integers, then the value of $k^{p k}\left(p^{p}+k^{k}\right)+(p+k)$ is $\qquad$ .
|
Solution: From $x_{1} x_{2}=\frac{k}{k-1}$, we know $\frac{k}{k-1}$ is a positive integer.
$$
\begin{array}{l}
\therefore k-1=1, k=2 . \\
\text { So } x_{2}=2 \text {. } \\
\therefore x_{1}=1, x_{2}=2 \text {. }
\end{array}
$$
From $x_{1}+x_{2}=\frac{p}{k-1}$, we get $p=3$.
Therefore, the original expression $=2^{6}\left(3^{3}+2^{2}\right)+(3+2)=1989$.
|
1989
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. There are two flour mills supplying flour to three residential areas. The first flour mill produces 60 tons per month, and the second flour mill produces 100 tons per month. The first residential area requires 45 tons per month, the second residential area requires 75 tons per month, and the third residential area requires 40 tons per month. The distances from the first flour mill to the three residential areas are 10 km, 5 km, and 6 km, respectively. The distances from the second flour mill to the three residential areas are 4 km, 8 km, and 15 km, respectively. How should the flour be allocated to make the transportation most economical?
|
(1960, Shanghai City Mathematics Recontest)(Hint: Let the flour transported by the first flour mill to the first and second residential areas be $x, y$ tons, respectively. Thus, $s=15 x+6 y+840$, where $0 \leqslant x \leqslant 45,0 \leqslant y \leqslant 60 . x+y$ $\geqslant 20$. When $x=20, y=0$, $S_{\text {min }}=960$. The first flour mill delivers $0,20,40$ tons to the three areas, respectively; the second flour mill delivers 45, 55, 0 tons to the three areas, respectively.)
|
960
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Rearrange $1,2,3,4,5,6,7,8$.
(1) such that the even numbers are in their original positions, and the odd numbers are not in their original positions, how many different arrangements are there?
(2) such that the prime numbers are in the odd positions, and the composite numbers are in the even positions, how many different arrangements are there?
(3) such that the even numbers are in the even positions but not in their original positions; the odd numbers are in the odd positions but also not in their original positions, how many different arrangements are there?
(4) if at least one number is in its original position, how many different arrangements are there? How many different arrangements are there?
(6) if at least 4 numbers are in their original positions, how many different arrangements are there?
(7) such that the even numbers are in the even positions but exactly two are not in their original positions; the odd numbers are in the odd positions but exactly two are not in their original positions, how many different arrangements are there?
(8) such that the even numbers are in the even positions and at least two are not in their original positions; the odd numbers are in the odd positions and at least two are not in their original positions, how many different arrangements are there?
|
Solution: (1) In essence, it is the Bernoulli-Euler letter misplacement problem for $n=4$, the number of different arrangements is
$$
u_{1}=9 \text { (ways). }
$$
(2) The different arrangements of even numbers in odd positions and odd numbers in even positions are 4! ways each. By the multiplication principle, the number of different arrangements for this problem is
$(4!)^{2}=576$ (ways).
(3) Similar to the 7th question of Example 1, the number of different arrangements is
$$
\left(u_{4}\right)^{2}=81 \text { (ways). }
$$
(4) Using the exclusion method. For a full permutation of 8 numbers, there are
8! ways of arrangement, minus the arrangements where no number is in its original position, which should be
$$
\begin{aligned}
& 8!-u_{8} \\
= & 8!-8!\quad\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}\right. \\
& \left.+\frac{1}{8!}\right) \\
= & 40320-14833=25487 \text { (ways). }
\end{aligned}
$$
(5) From the numbers $1,2,3,4,5,6,7,8$, select 4 numbers, and they are in their original positions, there are $C_{8}^{4}$ ways of selection, the remaining 4 numbers are not in their original positions, there are $u_{4}$ ways of arrangement. Therefore, by the multiplication principle, the number of different arrangements is
$$
C_{8}^{4} \cdot u_{4}=70 \times 9=630 \text { (ways). }
$$
(6) Using the exclusion method. First, perform a full permutation of 8 numbers, there are 8! ways of arrangement, then subtract the arrangements where $i$ numbers are in their original positions (i=0,1,2,3), hence the number of different arrangements is:
$$
\begin{array}{l}
8!-u_{8}-C_{8}^{1} \cdot u_{7}-C_{8}^{2} \cdot u_{6}-C_{8}^{3} \cdot u_{5} \\
=8!-8!\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}\right. \\
\left.+\frac{1}{8!}\right)-8!\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}\right. \\
\left.-\frac{1}{7!}\right)-\frac{8!}{2!}\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}\right) \\
-\frac{8!}{3!}\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}\right) \text {. } \\
=40320-14833-14832-7420-2464 \\
=771 \text { (ways). } \\
\end{array}
$$
Using the classification method. At least 4 numbers are in their original positions, which can be divided into exactly $i$ numbers in their original positions: $(i=4,5,6, 8$. Note: It is impossible to have exactly 7 numbers in their original positions), hence the number of different arrangements is
$$
\begin{array}{l}
C_{8}^{4} \cdot u_{4}+C_{8}^{5} \cdot u_{3}+C_{8}^{6} \cdot u_{2}+C_{8}^{8} \\
=630+112+28+1 \\
=771 \text { (ways). }
\end{array}
$$
(7) First arrange the even numbers, similar to question (5), the number of different arrangements is $C_{4}^{2} \cdot u_{2}$ ways; then arrange the odd numbers, the number of different arrangements is also $C_{4}^{2} \cdot u_{2}$ ways. By the multiplication principle, the total number of different arrangements is
$$
\left(C_{4}^{2} \cdot u_{2}\right)^{2}=36 \text { (ways). }
$$
(8) Still first arrange the even numbers, at least two numbers are not in their original positions, which includes two cases: exactly two numbers are not in their original positions and all 4 numbers are not in their original positions. For the first case, there are $C_{4}^{2} \cdot u_{2}$ ways of arrangement; for the second case, there are $u_{4}$ ways of arrangement. Therefore, the number of arrangements that meet the condition for even numbers is $\left(C_{4}^{2} \cdot u_{2}\right.$ $\left.+u_{4}\right)$ ways. Similarly, for the odd numbers, the number of arrangements that meet the condition is also $\left(C_{4}^{2} \cdot u_{2}\right.$ $\left.+u_{4}\right)$ ways. By the multiplication principle, the total number of different arrangements is
$$
\left(C_{4}^{2} \cdot u_{2}+u_{4}\right)^{2}=(6+9)^{2}=225 \text { (ways). }
$$
|
225
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 For any $a>1, b>1$,
we have $\frac{a^{2}}{b-1}+\frac{b^{2}}{a-1} \geqslant 8$.
(1992, 26th Commonwealth Mathematics Olympiad (10th grade))
|
To prove for $a:=1+t_{1}, b:=1+t_{2}$, where $t_{1}, t_{2}>0$. The inequality to be proven becomes
$$
\begin{array}{l}
\frac{\left(1+t_{1}\right)^{2}}{t_{2}}+\frac{\left(1+t_{2}\right)^{2}}{t_{1}} \geqslant 8 . \\
\text { The left side of the above equation } \geqslant \frac{2\left(1+t_{1}\right)\left(1+t_{2}\right)}{\sqrt{t_{1} t_{2}}} \\
=\frac{2+2 t_{1} t_{2}+2\left(t_{1}+t_{2}\right)}{\sqrt{t_{1} t_{2}}} \\
\geqslant \frac{2}{\sqrt{t_{1} t_{2}}}+2 \sqrt{t_{1} t_{2}}+4 \geqslant 4+4=8 .
\end{array}
$$
|
8
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 How many ordered quadruples of integers $(a$, $b, c, d)$ satisfy $0<a<b<c<d<500, a+d=b$ $+c$ and $bc-ad=93$?
(11th American Invitational Mathematics Examination)
|
Solution: Let $b=a+t, t \in N$, then $c=d-t$. Substituting into $b c-a d=93$, we get $(d-a-t) t=3 \times 31$.
When $t=1$, we have $b=a+1, c=a+93, d=$ $a+94$. Since $a+94<500$, we have $0<a<406$. Thus, there are 405 sets of $(a, b, c, d)$.
When $t=3$, we have $b=a+3, c=a+31, d=$ $a+34$. In this case, $0<a<466$, so there are 465 sets of $(a, b, c, d)$.
When $t=31,93$, it is easy to see that there are no sets of $(a, b, c, d)$ that satisfy the conditions.
Therefore, there are a total of $405+465=870$ sets of integers $(a, b, c, d)$ that satisfy the conditions.
|
870
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 The roots of the equation $x^{2}+p x+q=0$ are both positive integers, and $p+q=1992$. Then the ratio of the larger root to the smaller root is $\qquad$.
|
Given: $\because x_{1}+x_{2}=-p, x_{1} x_{2}=q$,
$$
\begin{array}{l}
\therefore x_{1} x_{2}-x_{1}-x_{2}=q+p=1992, \\
\left(x_{1}-1\right)\left(x_{2}-1\right)=1993 .
\end{array}
$$
$\because 1993$ is a prime number,
$$
\therefore\left\{\begin{array}{l}
x_{1}-1=1, \\
x_{2}-1=1993 .
\end{array}\right.
$$
Solving, we get $\left\{\begin{array}{l}x_{1}=2, \\ x_{2}=1994 .\end{array}\right.$
$$
\text { Therefore, } \frac{x_{2}}{x_{1}}=997 \text {. }
$$
|
997
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question: There are 12 points on a circle, one of which is painted red, and another is painted blue, with the remaining 10 points unpainted. Convex polygons formed by these points are classified as follows: polygons whose vertices include both the red and blue points are called bicolored polygons; polygons that include only the red (or blue) point are called red (or blue) polygons; polygons that do not include the red or blue points are called colorless polygons.
How many more bicolored polygons are there compared to colorless polygons, among all convex polygons (with sides ranging from triangles to dodecagons) formed by these 12 points?
(Ninth Zu Chongzhi Cup Mathematics Invitational Competition)
|
Simplified: From the problem, we know that a bicolored polygon must include two points of different colors, while a colorless polygon must not include any of these points. Since all 12 points are on the circumference of the circle, the number of bicolored polygons is $\sum_{i=1}^{10} C_{10}^{i}$, and the number of colorless polygons is $\sum_{i=3}^{10} C_{10}^{i}$. Also, $\sum_{i=1}^{10} C_{10}^{i}-\sum_{i=3}^{10} C_{10}^{i}=55$, so the number of bicolored polygons is 55 more than the number of colorless polygons.
Generalization 1 For $n$ points on the circumference of a circle $(n \geqslant 3)$, then ?.
Generalization 2 If there are $n$ points on the circumference of a circle $(n \geqslant 3)$, and 3 of these points are painted red, blue, and yellow, respectively, the number of colorless, monochromatic, bichromatic, and trichromatic polygons are $2^{n-3}-\frac{n^{2}-5 n+8}{2}$, $3\left(2^{n-3}-n+2\right)$, $3\left(2^{n-3}-1\right)$, and $2^{n-3}$, respectively.
Generalization 3 If there are $n$ points on the circumference of a circle $(n \geqslant 3)$, and $k$ of these points $(0 \leqslant k \leqslant n)$ are painted with $k$ different colors, the number of $m$-colored polygons $(0 \leqslant m \leqslant k)$ is
$$
\left\{\begin{array}{l}
C_{k}^{m} \sum_{i=3-m}^{n-k} C_{n-k}^{i}, 0 \leqslant m \leqslant 2 ; \\
C_{k}^{m} 2^{n-k}, 3 \leqslant m \leqslant n .
\end{array}\right.
$$
Proof omitted.
|
55
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The sum of the first 1997 terms of the sequence $1,1,2,1,2,3,1,2,3,4,1,2 \cdots$ is $\qquad$ .
|
3.42654 .
Sequence grouping: Let $a_{1997}$ be in the $(k+1)$-th group. (1), (1,2), (1,2,3), ..., (1,2,...,k), (1,2,..., $a_{1997}$, ..., k+1). From the sum of terms, we have
$$
\left\{\begin{array}{l}
1+2+\cdots+k=\frac{k(k+1)}{2}<1997, \\
1+2+\cdots+(k+1)=\frac{(k+1)(k+2)}{2} \geqslant 1997 .
\end{array}\right.
$$
Solving this, we get $k=62$. Therefore, $a_{1997}$ is in the 63rd group.
Since $1997-\frac{62 \times 63}{2}=44$, then $a_{1997}=44$.
$$
\begin{aligned}
S_{1997}= & \sum_{k=1}^{62} \frac{k(k+1)}{2}+(1+2+\cdots+44) \\
= & \frac{1}{2}\left[1^{2}+2^{2}+\cdots+62^{2}+1+2+\cdots+62\right] \\
& +\frac{44 \times 45}{2}=42654 .
\end{aligned}
$$
|
42654
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Product $\prod_{k=1}^{7}\left(1+2 \cos \frac{2 k \pi}{7}\right)=$ $\qquad$
|
5.3 .
$$
\begin{array}{l}
\text { Let } \omega=\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7} \text {. Then } \omega^{7}=1 . \\
\omega^{k}=\cos \frac{2 k \pi}{7}+i \sin \frac{2 k \pi}{7}, \\
\omega^{-k}=\cos \frac{2 k \pi}{7}-i \sin \frac{2 k \pi}{7}, \\
\therefore \omega^{k}+\omega^{-k}=2 \cos \frac{2 k \pi}{7} . \\
\text { The original expression }=\prod_{k=1}^{7}\left(1+\omega^{k}+\omega^{-k}\right) \\
=\prod_{k=1}^{7} \omega^{-k}\left(1+\omega^{k}+\omega^{2 k}\right) \\
=\omega^{-1 \cdots 2 \cdots \cdots-7} \cdot \prod_{k=1}^{6}\left(1+\omega^{k}+\omega^{2 k}\right) \cdot\left(1+\omega^{7}\right. \\
\left.\quad+\omega^{14}\right) \\
=\left(\omega^{7}\right)^{-4} \cdot 3 \prod_{k=1}^{6} \frac{1-\omega^{3 k}}{1-\omega^{k}} \\
=3 \cdot \frac{1-\omega^{3}}{1-\omega} \cdot \frac{1-\omega^{6}}{1-\omega^{2}} \cdot \frac{1-\omega^{9}}{1-\omega^{3}} \cdot \frac{1-\omega^{12}}{1-\omega^{4}} \cdot \frac{1-\omega^{15}}{1-\omega^{5}} \\
\cdot \frac{1-\omega^{18}}{1-\omega^{6}}=3 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { 6. Given } P(x)=x^{5}+a_{1} x^{4}+a_{2} x^{3}+ \\ a_{3} x^{2}+a_{4} x+a_{5} \text {, and when } k=1,2,3,4 \text {, } P(k) \\ =k \cdot 1997 \text {. Then } P(10)-P(-5)=\end{array}$
|
6.75315 .
Let $Q(x)=P(x)-1997 x$. Then for $k=1,2,3,4$, $Q(k)=P(k)-1997 k=0$. Hence, $1,2,3,4$ are roots of $Q(x)=0$. Since $Q(x)$ is a fifth-degree polynomial, we can assume
$$
Q(x)=(x-1)(x-2)(x-3)(x-4)(x-r) \text {. }
$$
Therefore,
$$
\begin{array}{l}
P(10)= Q(10)+1997 \cdot 10 \\
= 9 \cdot 8 \cdot 7 \cdot 6(10-r)+1997 \cdot 10 . \\
P(-5)= Q(-5)+1997(-5) \\
=(-6)(-7)(-8)(-9)(-5-r) \\
+1997 \cdot(-5) . \\
\therefore P(10)-P(-5) \\
= 9 \cdot 8 \cdot 7 \cdot 6 \cdot 15+1997 \cdot 15=75315 .
\end{array}
$$
|
75315
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, on the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$ there are 16 points, sequentially $P_{1}, P_{2}, \cdots, P_{16}, F$ is the left focus, and the angles between each adjacent pair of points and $F$ are equal $\left(\angle P_{1} F P_{2}=\angle P_{2} F P_{3}=\cdots=\angle P_{16} F P_{1}\right)$. Let the distance from $P_{i}$ to the left directrix be $d_{i}(i=1,2, \cdots, 16)$. Find $\sum_{i=1}^{16} \frac{1}{d_{i}}$.
|
$$
\begin{array}{l}
a=5, b \\
=4, c=3 . \text { Let } \angle X F P_{1} \\
=\alpha \cdot, \angle P_{1} F P_{2}= \\
\angle P_{2} F P_{3}=\cdots= \\
\angle P_{16} F P_{1}=\frac{\pi}{8} . \\
F M=d:-P_{i} F \cos \left[(i-1) \frac{\pi}{8}+\alpha\right] \\
\quad=\frac{a^{2}}{c}-c=\frac{16}{3} .
\end{array}
$$
By the definition of an ellipse, $P_{i} F=e d_{i}=\frac{c}{a} d_{i}=\frac{3}{5} d_{i}$, substituting into (1), we have
$$
\begin{array}{l}
d_{i}-\frac{3}{5} d_{i} \cos \left[\frac{(i-1) \pi}{8}+\alpha\right]=\frac{16}{3} . \\
\text { Thus } \frac{1}{d_{i}}=\frac{3}{16} \cdot \frac{1}{5}\left[5-3 \cos \left(\frac{(i-1) \pi}{8}+\alpha\right)\right] \\
\text { ( } i=1,2, \cdots, 16 \text { ). } \\
\therefore \sum_{i=1}^{16} \frac{1}{d}=\frac{3}{80}\left[\sum_{i=1}^{16} 5-3 \sum_{i=1}^{16} \cos \left(\frac{(i-1) \pi}{8}+\alpha\right)\right] \\
=3-\frac{9}{80} \sum_{i=1}^{16} \cos \left[\frac{(i-1) \pi}{8}+\alpha\right] \text {. } \\
\because \sum_{i=1}^{16} \cos \left[\frac{(i-1) \pi}{8}+\alpha\right] \\
=\sum_{i=1}^{i 6} \frac{2 \sin \frac{\pi}{16} \cos \left[(i-1)-\frac{\pi}{3}+\alpha\right]}{2 \sin \frac{\pi}{16}} \\
=\frac{1}{2 \sin \frac{\pi}{16}} \sum_{i=1}^{16}\left(\sin \left[\frac{\pi}{16}+(i-1) \frac{\pi}{8}+\alpha\right]\right. \\
\left.-\sin \left[(i-1) \frac{\pi}{8}+\alpha-\frac{\pi}{16}\right]\right) \\
=\frac{1}{2 \sin \frac{\pi}{16}}\left[\sin \left(\alpha+\frac{16 \pi}{8}-\frac{\pi}{16}\right)-\sin \left(x-\frac{\pi}{16}\right)\right] \\
=0 \text {, } \\
\therefore \sum_{i=1}^{16} \frac{1}{d_{i}}=3 \text {. } \\
\end{array}
$$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
63. Let $a_{1}=1997^{1997^{1997}}{ }^{.197}$ (1997 sevens in total), the sum of the digits in the decimal representation of $a_{1}$ is $a_{2}$, the sum of the digits of $a_{2}$ is $a_{3}$, and so on. Find $a_{2000}$.
|
Solution: Let $x_{n}=10000^{10000^{10000}}$ (with $n$ 10000s), $n=1,2, \cdots$, then for $n \geqslant 2$, we have
$$
\begin{array}{l}
x_{n}=10000^{x_{n-1}}=10^{4 x_{n-1}} . \\
\therefore a_{1}<x_{1997}=10^{4 x_{1996}}, \\
a_{2} \leqslant 9 \times 4 x_{1996}<100 x_{1996} \\
=100 \cdot 10^{4 x 1995}=10^{4 x_{1995}+2}, \\
a_{3} \leqslant 9\left(4 x_{1995}+2\right)<9 \times 6 x_{1995}<100 x_{1995} \\
=10^{4 x_{1994}+2},
\end{array}
$$
Thus, for any $k$, when $1 \leqslant k \leqslant 1996$, we have
$$
a_{k}<10^{4 x} 1997-k{ }^{+2} \text {. }
$$
When $k=1996$, i.e.,
$$
\begin{array}{l}
a_{1996}<10^{4 x_{1}+2}, \\
\therefore a_{1997}<9 \times\left(4 x_{1}+2\right)=9 \times 40002<10^{6}, \\
a_{1998}<9 \times 6=54, \\
a_{1999} \leqslant 13, \\
a_{2000} \leqslant 9 .
\end{array}
$$
Since $a_{2000} \geqslant 1$,
$$
\therefore 1 \leqslant a_{20000} \leqslant 9 \text {. }
$$
From the generation of $a_{2}, a_{3}, \cdots, a_{2000}$, we know that
$$
a_{1} \equiv a_{2} \equiv a_{3} \equiv \cdots \equiv a_{2000}(\bmod 9) \text {. }
$$
It is easy to see that $1997=9 m-1(m=222)$, and the exponent of $a_{1}$ is odd, denoted as $r$.
$$
\begin{aligned}
\therefore a_{1} & =1997^{r}=(9 m-1)^{r} \\
& =9 s+(-1)^{r}=9 s-1, \\
a_{2000} & =a_{1}=-1(\bmod 9) .
\end{aligned}
$$
Therefore, $a_{23000}=8$.
(Qin Zongci, Jiangsu Zhenjiang Teachers College, 212003)
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $m=\sqrt{5}+1$. Then the integer part of $m+\frac{1}{m}$ is $\qquad$ .
|
\begin{array}{l}\text { 1. } m=\sqrt{5}+1, \frac{1}{m}=\frac{1}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{4}, \\ \therefore m+\frac{1}{m}=\frac{5}{4} \sqrt{5}+\frac{3}{4},\left[m+\frac{1}{m}\right]=3 .\end{array}
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $x^{2}-x-1=0$. Then, the value of the algebraic expression $x^{3}$ $-2 x+1$ is $\qquad$ .
|
$\begin{array}{l} \text { 3. } x^{3}-2 x+1 \\ \quad=\left(x^{3}-x^{2}-x\right)+\left(x^{2}-x-1\right)+2 \\ =x\left(x^{2}-x-1\right)+\left(x^{2}-x-1\right)+2=2\end{array}$
The translation is as follows:
$\begin{array}{l} \text { 3. } x^{3}-2 x+1 \\ \quad=\left(x^{3}-x^{2}-x\right)+\left(x^{2}-x-1\right)+2 \\ =x\left(x^{2}-x-1\right)+\left(x^{2}-x-1\right)+2=2\end{array}$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $m$ and $n$ are rational numbers, and the equation $x^{2}+$ $m x+n=0$ has a root $\sqrt{5}-2$. Then the value of $m+n$ is $\qquad$ .
|
4. Since $m, n$ are rational, the other root is $-\sqrt{5}-2$, thus by Vieta's formulas,
$$
\begin{array}{l}
w:=4, n=-1 . \\
\therefore m+n=3 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The number of integer pairs $(m, n)$ that satisfy $1998^{2}+m^{2}=1997^{2}+n^{2}(0<m$ $<n<1998)$ is $\qquad$.
|
6. $n^{2}-m^{2}=3995=5 \times 17 \times 47,(n-m)(n+m)=5 \times 17 \times 47$, obviously any integer factorization of 3995 can yield $(m, n)$, given the condition $(0<m<n<1998)$, thus there are 3 integer pairs $(m, n)$ that satisfy the condition.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Distribute 100 apples to several people, with each person getting at least one apple, and each person receiving a different number of apples. Then, the maximum number of people is $\qquad$.
|
10. Suppose there are $n$ people, and the number of apples distributed to each person is $1,2, \cdots, n$. It follows that
$$
1+2+3+\cdots+n=\frac{n(n+1)}{2} \leqslant 100 .
$$
$\therefore n \leqslant 13$, i.e., there are at most 13 people.
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let $a, b$ be real numbers. Then the minimum value of $a^{2}+a b+b^{2}-$ $a-2 b$ is $\qquad$.
|
11 .
$$
\begin{array}{l}
a^{2}+a b+b^{2}-a-2 b \\
=a^{2}+(b-1) a+b^{2}-2 b \\
=\left(a+\frac{b-1}{2}\right)^{2}+\frac{3}{4} b^{2}-\frac{3}{2} b-\frac{1}{4} \\
=\left(a+\frac{b-1}{2}\right)^{2}+\frac{3}{4}(b-1)^{2}-1 \geqslant-1 .
\end{array}
$$
When $a+\frac{b-1}{2}=0, b-1=0$,
i.e., $a=0, b=1$, the equality in the above inequality holds, so the minimum value sought is -1.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Every book has an international book number:
ABCDEFGHIJ
where $A B C D E F G H I$ are composed of nine digits, and $J$ is the check digit.
$$
\text { Let } \begin{aligned}
S= & 10 A+9 B+8 C+7 D+6 E+5 F \\
& +4 G+3 H+2 I,
\end{aligned}
$$
$r$ is the remainder when $S$ is divided by 11. If $r$ is not 0 or 1, then $J=11-r$ (if $r=0$, then $J=0$; if $r=1$, then $J$ is represented by $x$). If a book's number is $962 y 707015$, then $y=$ $\qquad$
|
$$
\text { 15. } \begin{aligned}
S=9 \times 10 & +6 \times 9+2 \times 8+y \times 7+7 \times 6 \\
& +0 \times 5+7 \times 4+0 \times 3+1 \times 2
\end{aligned}
$$
$\therefore S$ the remainder when divided by 11 is equal to the remainder when $7 y+1$ is divided by 11.
From the check digit, we know that the remainder when $S$ is divided by 11 is 11 $-5=6$, so the remainder when $7 y$ is divided by 11 is $6-1=5$. Therefore, $y$ $=7$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given that the line $y=-2 x+3$ intersects the parabola $y=$ $x^{2}$ at points $A$ and $B$, and $O$ is the origin. Then, the area of $\triangle O A B$ is $\qquad$ .
|
7.6 .
As shown in the figure, the line $y=$
$-2 x+3$ intersects the parabola $y$ $=x^{2}$ at points $A(1,1), B(-3,9)$.
Construct $A A_{1}, B B_{1}$ perpendicular to the $x$-axis, with feet of the perpendiculars at
$$
\begin{array}{l}
A_{1} 、 B_{1} . \\
\therefore S_{\triangle O A B}=S_{\text {trapezoid } A A_{1} B_{1} B}-S_{\triangle A A_{1} O}-S_{\triangle B B_{1} O} O \\
=\frac{1}{2} \times(1+9) \times(1+3)-\frac{1}{2} \times 1 \times 1 \\
\quad-\frac{1}{2} \times 9 \times 3=6 .
\end{array}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Let the parabola $y=x^{2}+(2 a+1) x+2 a+$ $\frac{5}{4}$ intersect the $x$-axis at only one point.
(1) Find the value of $a$;
(2) Find the value of $a^{18}+32 a^{-6}$.
|
12. (1) Since the parabola intersects the $x$-axis at only one point, the quadratic equation $x^{2}+(2 a+1) x+2 a+\frac{5}{4}=0$ has two equal real roots. Therefore,
$$
\Delta=(2 a+1)^{2}-4\left(2 a+\frac{5}{4}\right)=0,
$$
which simplifies to $a^{2} \cdots a-1=0$.
$$
\ldots a=\frac{1 \pm \sqrt{5}}{2} \text {. }
$$
(2) From (1), we know that $a^{2}=a+1$. Repeatedly using this identity, we get
$$
\begin{aligned}
a^{4} & =(a+1)^{2}=a^{2}+2 a+1=3 a+2, \\
a^{8} & =(3 a+2)^{2}=9 a^{2}+12 a+4=21 a+13, \\
a^{16} & =(21 a+13)^{2}=441 a^{2}+546 a+169 \\
& =987 a+610, \\
a^{18} & =(987 a+610)(a+1) \\
& =987 a^{2}+1597 a+610=2584 a+1597 .
\end{aligned}
$$
$$
\begin{array}{l}
\text { Also, } a^{-6}=\frac{1}{a^{6}}=\frac{1}{a^{4} \cdot a^{2}}=\frac{1}{(3 a+2)(a+1)} \\
\quad=\frac{1}{8 a+5} . \\
\because a^{2}-a-1=0, \\
\therefore 64 a^{2}-64 a-65=-1,
\end{array}
$$
which means
$$
\begin{array}{l}
(8 a+5)(8 a-13)=-1 . \\
\therefore a^{-6}=\frac{1}{8 a+5}=-8 a+13 . \\
\therefore a^{18}+323 a^{-6} \\
\quad=2584 a+1597+323(-8 a+13) \\
\quad=5796 .
\end{array}
$$
|
5796
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Recently, the Asian football tournament has sparked Wang Xin, a junior high school student, to research football. He found that the football is made by gluing black and white leather pieces together, with black pieces being regular pentagons and white pieces being regular hexagons (as shown in the figure), and he counted 12 black pieces. Then, the number of white pieces is $\qquad$ pieces.
|
5.20.
Let there be $x$ white blocks, then the white blocks have a total of $6x$ edges, and the black blocks have a total of 60 edges. Since each white block has 3 edges connected to black blocks and 3 edges connected to other white blocks, then $6x \div 2 = 60, \therefore x = 20$.
|
20
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (This question is worth 25 points) Prove that $1997 \times 1998 \times$ $1999 \times 2000+1$ is a perfect square of an integer, and find this integer.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
$$
\begin{array}{l}
\text { Original expression }=x(x+1)(x+2)(x+3)+1 . \\
\because x(x+1)(x+2)(x+3)+1 \\
=\left(x^{2}+3 x\right)\left(x^{2}+3 x+2\right)+1 \\
=\left[\left(x^{2}+3 x+1\right)-1\right]\left[\left(x^{2}+3 x+1\right)+1\right]+1 \\
=\left(x^{2}+3 x+1\right)^{2} \\
\because 1997 \times 1998 \times 1999 \times 2000 \\
=\left(1997^{2}+3 \times 1997+1\right)^{2},
\end{array}
$$
$1997^{2}, 3 \times 1997,1$ are all integers, $\therefore 1997 \times 1998 \times 1999 \times 2000$ is the square of an integer.
$$
\begin{array}{l}
1997^{2}-3 \times 1997+1=1997 \times 2000+1 \\
=3994001 .
\end{array}
$$
$\therefore$ This integer is 3994001.
|
3994001
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $S=\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+$ $\sqrt{1+\frac{1}{3^{2}}+\frac{1}{4^{2}}}+\cdots+\sqrt{1+\frac{1}{1997^{2}}+\frac{1}{1998^{2}}}$. Then the integer closest to $S$ is ( ).
(A) 1997
(B) 1908
(C) 1009
(D) 2000
|
5. (B).
When $n$ is an integer, we have
$$
\begin{array}{l}
\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\left(1+\frac{1}{n}\right)^{2}-\frac{2}{n}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\left(\frac{n+1}{n}\right)^{2}-2 \cdot \frac{n+1}{n} \cdot \frac{1}{n+1}+\left(\frac{1}{n+1}\right)^{2}} \\
=\sqrt{\left(\frac{n+1}{n}-\frac{1}{n+1}\right)^{2}}=\frac{n+1}{n}-\frac{1}{n+1} \\
=1+\frac{1}{n}-\frac{1}{n+1} . \\
\begin{array}{l}
\therefore S=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right) \\
\quad+\cdots+\left(1+\frac{1}{1997}-\frac{1}{1998}\right) \\
= 1997+\left(1-\frac{1}{1998}\right) \\
=1998-\frac{1}{1998} .
\end{array}
\end{array}
$$
Therefore, the integer closest to $S$ is 1998.
|
1998
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $n$ be a positive integer. $0<x \leqslant 1$. In $\triangle A B C$, if $A B=n+x, B C=n+2 x, C A=n+3 x, B C$ has a height $A D=n$. Then, the number of such triangles is ( ).
(A) 10
(B) 11
(C) 12
(D) infinitely many
|
6. (C).
Let $\triangle A B C$ be the condition; then $c=n+1, a=n+2 x$, $b=n+3 x, p=\frac{1}{2}(a+b+c)=\frac{1}{2}(3 n+6 x)$. By Heron's formula, we get
$$
\begin{array}{l}
\sqrt{\frac{3 n+6 x}{2} \cdot \frac{n+4 x}{2} \cdot \frac{n+2 x}{2} \cdot \frac{n}{2}}=\frac{n(n+2 x)}{2} . \\
\therefore \frac{3 n(n+2 x)^{2}(n+4 x)}{16}=\frac{n^{2}(n+2 x)^{2}}{4},
\end{array}
$$
which simplifies to
$$
\begin{array}{l}
\frac{3(n+4 x)}{4}=n . \\
\therefore n=12 x . \\
\because 0<x \leqslant 1, n \text { is a positive integer. } \\
\therefore(x, n)=\left(\frac{1}{12}, 1\right),\left(\frac{2}{12}, 2\right), \cdots,\left(\frac{12}{12}, 12\right) .
\end{array}
$$
Therefore, there are 12 such triangles.
|
12
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $a, b$ are positive integers, and satisfy $\frac{a+b}{a^{2}}=\frac{4}{40}$. Then the value of $a+b$ is $\qquad$
|
Let $a+b=4k$ ($k$ is a positive integer), then $a^{2}+ab+b^{2}:=49k$, i.e., $(a+b)^{2}-ab=49k$.
$$
\therefore ab=16k^{2}-49k \text{. }
$$
It is easy to know that $a, b$ are the two positive integer roots of the equation about $x$
$$
x^{2}-4kx+(16k^{2}-49k)=0
$$
By $\Delta=16k^{2}-4(16k^{2}-49k) \geqslant 0$, we get $0 \leqslant k \leqslant \frac{49}{12}$.
$\because k$ is a positive integer, $\therefore k=1,2,3,4$.
It is easy to verify that when $k=1,2,3$, equation (1) has no positive integer roots; when $k=4$, equation (1) is $x^{2}-16x+60=0$, the solutions are $x=10$ or $x=6$. Therefore, $a+b=4k=16$.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Among the lines connecting the eight vertices of a square, the number of pairs of skew lines is ( ).
(A) 114
(B) 138
(C) 174
(D) 228
|
3. (C).
The vertex connections include 12 edges, 12 face diagonals, and 4 body diagonals. Skew lines can be categorized as follows:
(1) Edge to edge $\frac{12 \times 4}{2}=24$ pairs.
(2) Edge to face diagonal $12 \times 6=72$ pairs.
(3) Edge to body diagonal $12 \times 2=24$ pairs.
(4) Face diagonal to face diagonal $\frac{12 \times 5}{2}=30$ pairs.
(5) Face diagonal to body diagonal $12 \times 2=24$ pairs.
(G) Body diagonal to body diagonal (six at the center) 0 pairs.
Total: 174 pairs.
|
174
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let $[x]$ denote the greatest integer not greater than the real number $x$. The number of real roots of the equation $\lg ^{2} x-[\lg x]-2=0$ is $\qquad$
(1995, National High School Mathematics Competition)
|
Analysis: The difficulty of this problem lies in the uncertainty of $[\lg x]$. Since $[\lg x] \leqslant \lg x$, the original problem is first transformed into finding the values of $x$ that satisfy the inequality $\lg ^{2} x-\lg x-2 \leqslant 0$.
Solving this, we get $-1 \leqslant \lg x \leqslant 2$.
Therefore, the number of roots of the equation can be found by solving for $[\lg x]$ taking the values -1, 0, 1, 2, respectively, yielding the three roots as $x_{1}=\frac{1}{10}, x_{2}=10^{\sqrt{3}}, x_{3}=100$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Seven, (Full marks 12 points) A
home appliance manufacturing company, based on market research and analysis, has decided to adjust its production plan. They plan to produce a total of 360 units of air conditioners, color TVs, and refrigerators per week (calculated at 120 labor hours per week), with at least 60 refrigerators to be produced. The labor hours required per unit and the output value per unit of these home appliances are as follows:
How many units of air conditioners, color TVs, and refrigerators should be produced each week to maximize output value? What is the highest output value (in thousand yuan)?
保留源文本的换行和格式，直接输出翻译结果如下：
Seven, (Full marks 12 points) A
home appliance manufacturing company, based on market research and analysis, has decided to adjust its production plan. They plan to produce a total of 360 units of air conditioners, color TVs, and refrigerators per week (calculated at 120 labor hours per week), with at least 60 refrigerators to be produced. The labor hours required per unit and the output value per unit of these home appliances are as follows:
How many units of air conditioners, color TVs, and refrigerators should be produced each week to maximize output value? What is the highest output value (in thousand yuan)?
|
$$
\begin{array}{l}
\because x+y+z=360 \text {, then } z \geqslant 60 \text {, } \\
\therefore x+y \leqslant 300 \text {. } \\
\text { Total labor hours }=\frac{1}{2} x+\frac{1}{3} y+\frac{1}{4} z \\
=\frac{1}{12}(6 x+4 y+3 z) \\
=\frac{1}{4}(x+y+z)+\frac{1}{12}(3 x+y) \\
=\frac{1}{4} \times 360+\frac{1}{12}(3 x+y) \\
=90+\frac{1}{12}(3 x+y) \text {. } \\
\text { Total output value }=4 x+3 y+2 z \\
=2(x+y+z)+(2 x+y) \\
=2 \times 360+(2 x+y) \\
=720+(2 x+y) \text {. } \\
\left\{\begin{array}{l}
x+y \leqslant 300, \\
90+\frac{1}{12}(3 x+y)=120 \Rightarrow 3 x+y=360
\end{array}\right\} \Rightarrow\left\{\begin{array}{l}
x \geqslant 30, \\
3 x+y=360
\end{array}\right\} \\
\Rightarrow A=1080-x \leqslant 1050 \text {, } \\
\end{array}
$$
When the total output value $A$ reaches its maximum value of 1050, $x=30, y=270$, $z=60$.
Answer: Each week, 30 air conditioners, 270 televisions, and 60 refrigerators should be produced to maximize the output value. The maximum output value is 1050 thousand yuan.
|
1050
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (Full marks 12 points) A store sells a product that costs 10 yuan each at 18 yuan each, and can sell 60 units per day. After conducting a market survey, the store manager found that if the selling price of this product (based on 18 yuan each) is increased by 1 yuan, the daily sales volume will decrease by 5 units; if the selling price (based on 18 yuan each) is decreased by 1 yuan, the daily sales volume will increase by 10 units. To maximize daily profit, what should the selling price of this product be set to in yuan per unit?
|
Three, let the selling price of each item be $x$ yuan, and the daily profit be $s$ yuan. When $x \geqslant 18$, we have
$$
\begin{aligned}
s & =[60-5(x-18)](x-10) \\
& =-5(x-20)^{2}+500 .
\end{aligned}
$$
That is, when the price of the item is increased, the daily profit $s$ is maximized when $x=20$, and the maximum daily profit is 500 yuan. When $x \leqslant 18$, we have
$$
\begin{aligned}
s & =[60+10(18-x)](x-10) \\
& =-10(x-17)^{2}+490 .
\end{aligned}
$$
That is, when the price of the item is decreased, the daily profit is maximized when $x=17$, and the maximum daily profit is 490 yuan.
In summary, the selling price of this item should be set at 20 yuan per item.
Note: (1) This question tests the students' mastery and application of the method of completing the square. If students use the knowledge of quadratic functions to solve the problem correctly, full marks will also be given.
(2) This question must be solved by considering two different scenarios. If only one scenario is explained, or if only one scenario is correct, a maximum of 7 points will be awarded.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. (Full score 12 points) How many prime numbers are there among all integers that start and end with 1 and alternate between 1 and 0 (such as 101, 10101, 1010101, ...)? Why? And find all the prime numbers.
|
13. The following proves that $10101, 1010101, \cdots \cdots$, are not prime numbers.
For $n \geqslant 2$, then
$$
\begin{aligned}
A & =\underbrace{1010101 \cdots 01}_{2 n} \\
& =10^{2 n}+10^{2 n} 2+\cdots+10^{2}+1 . \\
& =\frac{10^{2 n+2}-1}{10^{2}-1}=\frac{\left(10^{n+1}+1\right)\left(10^{n+1}-1\right)}{99} .
\end{aligned}
$$
If $n$ is odd, i.e., $n=2 m+1$, then
$$
\frac{10^{n+1}-1}{99}=\frac{10^{2 n+2}-1}{99}=\underbrace{1010 \cdots 101}_{2 m} \text {. }
$$
Thus, $A=\underbrace{10101 \cdots 101}_{2 m \text { terms }} \times\left(10^{n+1}+1\right)$.
Hence, $A$ is a composite number.
If $n$ is even, then $9\left|\left(10^{n+1}-1\right), 11\right|\left(10^{n+1}+1\right)$, so $A$ is also a composite number.
Since 101 is a prime number, the only prime number in the sequence is 101.
|
101
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let $f(x)=\frac{4^{x}}{4^{x}+2}$. Then, $\sum_{k=1}^{1000} f\left(\frac{k}{1001}\right)$ equals $\qquad$ -
|
Sampling: From the characteristics of the variable's values, we can deduce that $f(x)$ satisfies the following symmetric property:
If $x+y=1$, then $f(x)+f(y)=1$. Therefore, $f\left(\frac{k}{1001}\right)+f\left(\frac{1001-k}{1001}\right)=1$,
$(k=1,2, \cdots, 1000)$
$$
\therefore \sum_{k=1}^{1000} f\left(\frac{k}{1001}\right)=500 .
$$
|
500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. For all real numbers $x, y$, if the function $f$ satisfies:
$$
f(x y)=f(x) \cdot f(y)
$$
and $f(0) \neq 0$, then $f(1998)=$ $\qquad$ .
|
$\approx 、 1.1$.
Let $f(x y)=f(x) \cdot f(y)$, set $y=0$, we have $f(x \cdot 0)=$ $f(x) \cdot f(0)$, which means $f(0)=f(x) \cdot f(0)$, and since $f(0) \neq 0$, it follows that $f(x)=1$, thus $f(1998)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Quadrilateral $ABCD$ is inscribed in a circle, $BC=CD=4$, $AC$ and $BD$ intersect at $E$, $AE=6$, and the lengths of $BE$ and $DE$ are both integers. Then the length of $BD$ is $\qquad$
|
4.7.
From $\overparen{B C}=\overparen{C D}, \angle B D C=\angle C A D, \angle A C D=\angle A C D$, we know $\triangle D C E \backsim \triangle A C D$, so $\frac{C E}{C D}=\frac{C D}{A C}$, which means $C E \cdot A C=C D^{2}$. Therefore, $C E(C E+6)=16$, solving this gives $C E=2$. Also, from $B E \cdot D E=A E \cdot C E=12$, and $B E+C E>B C$, then $B E>$ 2. Similarly, $D E>2$. Since $B E$ and $D E$ are integers, then $B E=3, D E=4$ (or $B E=4, D E=3$). Hence, $B D=B E+D E=7$.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (Full marks 25 points) As shown in the figure, $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at $M$, and the angle between their two external common tangents is $60^{\circ}$. The line connecting the centers intersects $\odot O_{1}$ and $\odot O_{2}$ at $A$ and $B$ (different from $M$), respectively. A line through $B$ intersects $\odot O_{1}$ at points $C$ and $D$. Find the value of $\operatorname{ctg} \angle B A C \cdot \operatorname{ctg} \angle B A D$.
|
From the given conditions, we have $O_{1} E \perp P E, O_{1} F \perp P F, O_{1} E= O_{1} F$, thus $O_{1}$ lies on the bisector of $\angle E P F$. Similarly, $O_{2}$ lies on the bisector of $\angle E P F$. Therefore, $P A$ is the bisector of $\angle E P F$. Since $O_{2} Q / / P E$, we have $\angle Q O_{2} O_{1}=\angle E P O_{1}=30^{\circ}$, hence $O_{1} Q=\frac{1}{2} O_{1} O_{2}$.
Let the radii of the larger and smaller circles be $R$ and $r$. Then $O_{1} Q=R-r, O_{1} O_{2}=R+r$, so $R-r=\frac{1}{2}(R+r)$, thus $R=3 r$.
$\because \angle A C M=\angle A D M=90^{\circ}$,
$\therefore \operatorname{ctg} \angle B A C \cdot \operatorname{ctg} \angle B A D=\frac{A C}{C M} \cdot \frac{A D}{D M}$.
From $\triangle B A C \sim \triangle B D M, \triangle B A D \sim \triangle B C M$ we get
$$
\frac{A C}{D M}=\frac{A B}{D B}, \frac{A D}{C M}=\frac{D B}{M B} \text {. }
$$
Therefore, $\operatorname{ctg} \angle B A C \cdot \operatorname{ctg} \angle B A D=\frac{A B}{D B} \cdot \frac{D B}{M B}=\frac{8 r}{2 r}=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
iii. (Full marks 25 points) A mall installs an escalator between the first and second floors, which moves upwards at a uniform speed. A boy and a girl start walking up the escalator to the second floor at the same time (the escalator itself is also moving). If the boy and the girl are both considered to be moving at a uniform speed, and the boy walks twice as many steps per minute as the girl, it is known that the boy walked 27 steps to reach the top of the escalator, while the girl walked 18 steps to reach the top (assuming the boy and girl each step up one step at a time).
(i) How many steps are visible on the escalator?
(ii) If there is a staircase near the escalator that leads from the second floor to the first floor, with the same number of steps as the escalator, and the two children each walk down the stairs at their original speeds, then reach the bottom and ride the escalator again (ignoring the distance between the escalator and the stairs), how many steps will the boy have walked when he catches up to the girl for the first time?
|
(i) Let the girl's speed be $x$ levels/min, the escalator's speed be $y$ levels/min, and the stairs have $s$ levels. Then the boy's speed is $2x$ levels/min. According to the problem, we have:
$$
\left\{\begin{array}{l}
\frac{27}{2 x}=\frac{-27}{y}, \\
\frac{18}{x}=\frac{s-18}{y} ;
\end{array}\right.
$$
Solving, we get $\frac{3}{4}=\frac{s-27}{s-18}$.
Solving for $s$, we get $s=54$ (levels). That is, the escalator has 54 levels exposed.
(ii) Let the boy pass the escalator $m$ times and the stairs $n$ times when he first catches up with the girl, then the girl has passed the escalator $(m-1)$ times and the stairs $(n-1)$ times.
From the system of equations in (i), we can solve for $y=2x$, so the boy ascends the escalator at $4x$ levels/min, and the girl ascends the escalator at $3x$ levels/min.
$$
\frac{54 m}{4 x}+\frac{54 n}{2 x}=\frac{54(m-1)}{3 x}+\frac{54(n-1)}{x},
$$
$\frac{m}{4}+\frac{n}{2}=\frac{m-1}{3}+\frac{n-1}{1}$, which simplifies to $6 n+m=16$.
And $m, n$ must be positive integers, and $0 \leqslant m-n \leqslant$
1.
\begin{tabular}{c|ccccc}
$m$ & 1 & 2 & 3 & 4 & 5 \\
\hline$n=\frac{16-m}{6}$ & $2 \frac{1}{2}$ & $2 \frac{1}{3}$ & $2 \frac{1}{6}$ & 2 & $1 \frac{5}{6}$
\end{tabular}
\begin{tabular}{c|cc}
$n$ & 1 & 2 \\
\hline$m=16-6 n$ & 10 & 4
\end{tabular}
Clearly, it can only be $m=3, n=2 \frac{1}{6}$.
Thus, the number of levels the boy has walked
$$
=3 \times 27+2 \frac{1}{6} \times 54=198 \text { (levels). }
$$
|
198
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $f(x)=x^{10}+2 x^{9}-2 x^{8}-2 x^{7}+x^{6}$ $+3 x^{2}+6 x+1$, then $f(\sqrt{2}-1)=$
|
$=1.4$.
Let $x=\sqrt{2}-1$, then $x+1=\sqrt{2} \Rightarrow (x+1)^{2}=2 \Rightarrow x^{2}+2x-1=0$. That is, $x=\sqrt{2}-1$ is a root of $x^{2}+2x-1=0$. But $f(x)=(x^{8}-x^{6}+3)(x^{2}+2x-1)+4$, so $f(\sqrt{2}-1)=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If
$$
\dot{z}=\frac{(1+i)^{2000}(6+2 i)-(1-i)^{1998}(3-i)}{(1+i)^{1996}(23-7 i)+(1-i)^{1994}(10+2 i)} \text {, }
$$
then $|z|=$ . $\qquad$
|
3.1.
$$
\begin{array}{l}
\text { When }(1+i)^{2}=2 i, (1-i)^{2}=-2 i \text { and } X \in \mathbb{Z}, \\
i^{4 k+1}=i, i^{4 k+2}=-1, i^{4 k+3}=-i, i^{4 k}=1. \\
\text { Therefore, }(1+i)^{2000}=(2 i)^{1000}=2^{1000}, \\
(1-i)^{1998}=(-2 i)^{999}=2^{999} \cdot i, \\
(1+i)^{1996}=(2 i)^{998}=-2^{998}, \\
(1-i)^{1994}=(-2 i)^{997}=-2^{997} i. \\
\text { Then } z=\frac{2^{1000}(6+2 i)-2^{999} i(3-i)}{-2^{998}(23-7 i)-2^{997} i(10+2 i)} \\
=\frac{8(6+2 i)-4 i(3-i)}{-2(23-7 i)-i(10+2 i)} \\
=\frac{48+16 i-12 i-4}{-46+14 i-10 i+2} \\
=\frac{44+4 i}{-44+4 i}=-\frac{11+i}{11-i}. \\
\text { Hence }|z|=\frac{|11+i|}{|11-i|}=1. \\
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-3 x-\right.$ $3)^{2001}$ is expanded and like terms are combined. The sum of the coefficients of the odd powers of $x$ in the resulting expression is $\qquad$.
|
4. -1 .
$$
\text { Let } \begin{aligned}
f(x)= & \left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-3 x-3\right)^{2001} \\
= & A_{0}+A_{1} x+A_{2} x^{2}+\cdots+A_{4001} x^{4001} \\
& +A_{4002} x^{4002} .
\end{aligned}
$$
$$
\begin{array}{l}
\text { Then } A_{0}+A_{1}+A_{2}+\cdots+A_{4001}+A_{4002} \\
\quad=f(1)=0, \\
A_{0}-A_{1}+A_{2}-\cdots-A_{4001}+A_{4002} \\
=f(-1)=2 .
\end{array}
$$
Subtracting the two equations gives $2\left(A_{1}+A_{3}+\cdots+A_{4001}\right)=-2$.
$$
\text { Therefore, } A_{1}+A_{3}+\cdots+A_{4001}=-1 \text {. }
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. On the coordinate plane, the area of the plane region bounded by the conditions $\left\{\begin{array}{l}y \geqslant-|x|-1, \\ y \leqslant-2|x|+3\end{array}\right.$ is
|
6. 16 .
For $\left\{\begin{array}{l}y \geqslant-|x|-1, \\ y \leqslant-2|x|+3\end{array}\right.$, discuss for $x \geqslant 0, x<0$.
When $x \geqslant 0$, it is $\left\{\begin{array}{l}x \geqslant 0, \\ y \geqslant-x-1, \\ y \leqslant-2 x+3 .\end{array}\right.$
When $x<0$, it is $\left\{\begin{array}{l}x<0, \\ y \geqslant x-1, \\ y \leqslant 2 x+3 .\end{array}\right.$
On the coordinate plane, draw the lines $y=-x-1, y=-2 x+3$ and $x=0$, intersecting to form $\triangle A B C, A(4,-5), B(0,-1)$, $C(0,3)$. Then draw the lines $y=x-1, y=2 x+3, y=0$, intersecting to form $\triangle A^{\prime} B C$, where $A^{\prime}(-4,-5)$. Clearly,
$\triangle A B C \simeq \triangle A^{\prime} B C$.
The region defined by $\left\{\begin{array}{l}y \geqslant-|x|-1, \\ y \leqslant-2|x|+3\end{array}\right.$ is shown as the shaded area in the figure.
Its area is $2 S_{\triangle A B C}=2 \times 8=16$.
|
16
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (Full marks 20 points) An arithmetic sequence with a common difference of 4 and a finite number of terms, the square of its first term plus the sum of the rest of the terms does not exceed 100. Please answer, how many terms can this arithmetic sequence have at most?
保留源文本的换行和格式，所以翻译结果如下：
Four, (Full marks 20 points) An arithmetic sequence with a common difference of 4 and a finite number of terms, the square of its first term plus the sum of the rest of the terms does not exceed 100. Please answer, how many terms can this arithmetic sequence have at most?
|
Let the arithmetic sequence be $a_{1}, a_{2}, \cdots, a_{n}$, with common difference $d=4$. Then
$$
\begin{array}{l}
a_{1}^{2}+a_{2}+\cdots+a_{n} \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+\frac{2 a_{1}+4 n}{2}(n-1) \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+(n-1) a_{1}+\left(2 n^{2}-2 n-100\right) \leqslant 0 .
\end{array}
$$
It holds for at least one value of $a_{1}$, so the discriminant is non-negative. That is,
$$
\begin{array}{l}
(n-1)^{2}-4\left(2 n^{2}-2 n-100\right) \geqslant 0 \\
\Leftrightarrow 7 n^{2}-6 n-401 \leqslant 0 \\
\Leftrightarrow-\frac{\sqrt{2} 816}{7} \leqslant n \leqslant \frac{3+\sqrt{2816}}{7} .
\end{array}
$$
But $n \in \mathbb{N}$, it is easy to see that $n \leqslant 8$.
In fact, the arithmetic sequence with a common difference of 4:
$-4,0,4,8,12,16,20,24$ has 8 terms, and
$$
(-4)^{2}+0+4+8+12+16+20+24=100 \text {. }
$$
Therefore, the maximum value of $n$ is 8. That is, such an arithmetic sequence can have at most 8 terms.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (Full marks 50 points) There are 21 points on a circle. Prove: Among all the arcs formed with these points as endpoints, there are no fewer than 100 arcs that do not exceed $120^{\circ}$.
|
Three points on a circle divide the circle into three arcs, at least one of which is greater than $120^{\circ}$. Connecting the endpoints of the arc that does not exceed $120^{\circ}$ with a chord, we can see that among any three points on the circle, at least two points are connected by a chord (referred to as an "edge"). Since such "edges" correspond one-to-one with arcs not exceeding $120^{\circ}$, it is sufficient to prove that the number of "edges" connecting 21 points on the circle is no less than 100.
Let $A_{1}$ be the vertex with the fewest "edges" connected, and let $A_{1} A_{2}$, $A_{1} A_{3}, \cdots A_{1} A_{k}$ be the $k-1$ "edges" drawn from $A_{1}$. Since each point $A_{i}(i=1,2, \cdots, k)$ has no fewer than $k-1$ "edges" connected, the total number of these "edges" is no less than $\frac{k(k-1)}{2}$. The remaining $21-k$ points should not have "edges" connected to $A_{1}$. However, among any three points, at least two points must have an "edge" connected, so every two points among these $21-k$ points must have an "edge" connected. Thus, we get no fewer than $\frac{(21-k)(20-k)}{2}$ "edges". Let $M$ represent the total number of "edges" connected among these 21 points, then
$$
\begin{aligned}
M & \geqslant \frac{k(k-1)}{2}+\frac{(21-k)(20-k)}{2} \\
& =k^{2}-21 k+210=f(k) .
\end{aligned}
$$
Given $k \in \mathbb{N}$, the minimum value of $k^{2}-21 k+210$ occurs at $k=\frac{21}{2}$, with the nearest integers being $k=10$ and $k=11$.
In $f(k)=k^{2}-21 k+210$, $f(10)=f(11)=$ 100,
$$
\therefore M \geqslant f(k) \geqslant f(10)=100 \text {. }
$$
This minimum value can be achieved. Draw a diameter $A B$ of the circle. Take 10 points near point $A$ and 11 points near point $B$ on the circle. This configuration meets the requirement. The 21 points are connected by
$$
C_{10}^{2}+C_{11}^{2}=45+55=100 \text { "edges". }
$$
|
100
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Initial 65. Given a real-coefficient polynomial function $y=a x^{2}+b x+c$, for any $|x| \leqslant 1$, it is known that $|y| \leqslant 1$. Try to find the maximum value of $|a|+|b|+|c|$.
|
Proof: First, we prove the following auxiliary proposition.
Proposition: Let real numbers $A, B$ satisfy $|A| \leqslant 2, |B| \leqslant 2$. Then $|A+B| + |A-B| \leqslant 4$.
In fact, without loss of generality, let $|A| \geqslant |B|$.
From $A^2 \geqslant B^2$, we have $(A+B)(A-B) \geqslant 0$,
thus $|A+B| + |A-B| = |(A+B) + (A-B)| = |2A| \leqslant 4$.
Now, let's solve the original problem.
When $x=0$, $|y| \leqslant 1$,
i.e., $|c| \leqslant 1$.
For any $|x| \leqslant 1$, we have
\[
|a x^2 + b x| = |y - c| \leqslant |y| + |c| \leqslant 2.
\]
Taking $x=1$, we get $|a + b| \leqslant 2$,
taking $x=-1$, we get $|a - b| \leqslant 2$.
According to the auxiliary proposition, we have
\[
\begin{aligned}
|a| + |b| &= \frac{1}{2}(|2a| + |2b|) \\
&= \frac{1}{2}(|(a+b) + (a-b)| + |(a+b) - (a-b)|) \\
&\leqslant \frac{1}{2} \times 4 = 2.
\end{aligned}
\]
From (1) and (2), we know $|a| + |b| + |c| \leqslant 3$.
Next, we show that $|a| + |b| + |c|$ can achieve the maximum value of 3.
Take $a=2, b=0, c=-1$. In this case, $y = 2x^2 - 1$. It is easy to verify that when $|x| \leqslant 1$, $|y| \leqslant 1$.
Therefore, the maximum value of $|a| + |b| + |c|$ is 3.
(Chen Kuanhong, Nengshi High School, Yueyang County, Hunan Province, 414113)
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 If any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>$ 0 , to make $\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant$ $k \log _{\frac{x_{0}}{x_{3}}} 1993$ always hold, then the maximum value of $k$ is $\qquad$
(1993, National High School Mathematics Competition)
|
Analysis: The inequality can be transformed into
$$
\begin{array}{l}
k \leqslant-\frac{\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993}{\log _{\frac{x_{0}}{x_{3}}} 1993} \\
=f\left(x_{0}, x_{1}, x_{2}, x_{3}\right) \text {. } \\
\end{array}
$$
Thus,
$$
\{k\}_{\text {max }}=\left\{f\left(x_{0}, x_{1}, x_{2}, x_{3}\right)\right\}_{\text {min }} .
$$
By the harmonic mean inequality, we get
$$
\begin{array}{l}
f\left(x_{0}, x_{1}, x_{2}, x_{3}\right) \geqslant \\
\frac{3}{\left(\log _{1993} \frac{x_{0}}{x_{1}}+\log _{1993} \frac{x_{1}}{x_{2}}+\log _{1993} \frac{x_{2}}{x_{3}}\right) \log _{\frac{x_{0}}{x_{3}}} 1993} \\
\end{array}
$$
$$
=3 \text {. }
$$
Therefore, $k_{\max }=3$.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12 Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in R$, and $x^{3}+\sin x-2 a=0,4 y^{3}+\sin y \cos y+a=$ 0. Then $\cos (x+2 y)=$ $\qquad$
(1994, National High School Mathematics Competition)
|
Analysis: Since $2 a=x^{3}+\sin x=(-2 y)^{3}+$ $\sin (-2 y)$, if we let $f(t)=t^{3}+\sin t$, then we have $f(x)$ $=f(-2 y)$.
And $f(t)$ is strictly increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so by monotonicity we have
$$
x=-2 y \text {, hence } x+2 y=0 \text {. }
$$
Thus $\cos (x+2 y)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given $f(x)=\frac{x^{2}}{1+x^{2}}$. Then the sum
$$
\begin{array}{l}
f\left(\frac{1}{1}\right)+f\left(\frac{2}{1}\right)+\cdots+f\left(\frac{100}{1}\right)+f\left(\frac{1}{2}\right) \\
+f\left(\frac{2}{2}\right)+\cdots+f\left(\frac{100}{2}\right)+\cdots f\left(\frac{1}{100}\right)+ \\
f\left(\frac{2}{100}\right)+\cdots+f\left(\frac{100}{100}\right)
\end{array}
$$
is equal to ( ).
(A) 10000
(B) 5000
(C) 1000
(D) 100
(1989, Suzhou City High School Mathematics Competition)
|
List the square number table as shown in the figure, and add the two number tables after rotating around the main diagonal by $180^{\circ}$,
$$
\begin{array}{ccccc}
f\left(\frac{1}{1}\right) & f\left(\frac{2}{1}\right) & f\left(\frac{3}{1}\right) & \cdots & f\left(\frac{100}{1}\right) \\
f\left(\frac{1}{2}\right) & f\left(\frac{2}{2}\right) & f\left(\frac{3}{2}\right) & \cdots & f\left(\frac{100}{2}\right) \\
f\left(\frac{1}{3}\right) & f\left(\frac{2}{3}\right) & f\left(\frac{3}{3}\right) & \cdots & f\left(\frac{100}{3}\right) \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
f\left(\frac{1}{100}\right) & f\left(\frac{2}{100}\right) & f\left(\frac{3}{100}\right) & \cdots & f\left(\frac{100}{100}\right)
\end{array}
$$
Then the sum of the two numbers at each position is
$$
f\left(\frac{j}{i}\right)+f\left(\frac{i}{j}\right)=1(i, j=1,2, \cdots,
$$
100 ).
Thus $2 S=100^{2}$, i.e., the sum $S=5000$, option $B$ should be selected.
|
5000
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2. Let $D=\{1,2, \cdots, 10\}, f$ be a one-to-one mapping from $D$ to $D$. Let $f_{1}(x)=f(x), f_{n+1}(x)=$ $f\left[f_{n}(x)\right]$. Prove: There exists a permutation $x_{1}, x_{2}$, $\cdots, x_{n}$ of $D$, such that the following equation holds:
$$
\sum_{i=1}^{10} x_{i} f_{2520}\left(x_{i}\right)=220 \text {. }
$$
(Adapted from the 1996 Jiangsu Province High School Mathematics Competition)
|
Proof: Since $f$ is a one-to-one mapping from $D$ to $D$, by Proposition 1, for any $i \in D$, there exists $m_{i}\left(1 \leqslant m_{i} \leqslant 10\right)$, such that $f_{m_{i}}(i)=i$.
Considering $2520=2^{3} \cdot 3^{2} \cdot 5 \cdot 7$, which is the least common multiple of $1,2, \cdots, 10$, it follows from the conclusion of the previous problem that $f_{2520}(i)=i$ for all $i \in D$. Therefore, we have
$$
\sum_{i=1}^{10} x_{i} f_{2520}\left(x_{i}\right)=\sum_{i=1}^{10} x_{i} \cdot i.
$$
By the rearrangement inequality, we know
$$
\begin{aligned}
\sum_{i=1}^{10} i \cdot x_{i} & \geqslant 1 \times 10+2 \times 9+\cdots+10 \times 1 \\
& =220.
\end{aligned}
$$
Equality holds if and only if $x_{1}, x_{2}, \cdots, x_{10}$ is the reverse permutation of $1,2, \cdots, 10$. Therefore, there exists a permutation of $D$, $10,9, \cdots, 1$, such that the given equation holds.
|
220
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Let the set $A=\{1,2, \cdots, 10\},$ and the mapping $f$ from $A$ to $A$ satisfies the following two conditions:
(1) For any $x \in A, f_{30}(x)=x$;
(2) For each $k \in \mathbb{Z}^{+}, 1 \leqslant k \leqslant 29$, there exists at least one $a \in A$ such that $f_{k}(a) \neq a$.
Find the total number of such mappings.
(1992, Japan Mathematical Olympiad Preliminary)
|
Solution: Notice that $10=5+3+2,30=5 \times 3 \times 2$. This suggests dividing $A$ into three disjoint subsets
$$
A=\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\} \cup\left\{b_{1}, b_{2}, b_{3}\right\}
$$
$\cup\left\{c_{1}, c_{2}\right\}$.
Since $f$ satisfies conditions (1) and (2), $f$ is a one-to-one mapping from $A$ to $A$, and from the proof process of Proposition 1, we know that there exists a mapping cycle in $A$. Therefore, define the mapping
$$
\begin{array}{l}
f: f\left(a_{1}\right)=a_{2}, f\left(a_{2}\right)=a_{3}, f\left(a_{3}\right)=a_{4}, \\
f\left(a_{4}\right)=a_{5}, f\left(a_{5}\right)=a_{1} ; f\left(b_{1}\right)=b_{2}, f\left(b_{2}\right) \\
=b_{3}, f\left(b_{3}\right)=b_{1} ; f\left(c_{1}\right)=c_{2}, f\left(c_{2}\right)=c_{1} .
\end{array}
$$
Since 30 is the least common multiple of $5,3,2$, by Proposition 1 and Proposition 2, we know that $f$ is the unique class of mappings that satisfy conditions (1) and (2) of the problem.
Therefore, the total number of $f$ is equivalent to selecting 5 elements from 10, then selecting 3 from the remaining 5, and finally selecting the last 2, and arranging them in circular permutations. It is equal to
$$
\left(C_{10}^{5} \cdot 4!\right)\left(C_{5}^{3} \cdot 2!\right)\left(C_{2}^{2} \cdot 1!\right)=120960
$$
|
120960
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Let the set $A=\{1,2,3,4,5,6\}$, and the mapping $f: A \rightarrow A$, such that its third composite mapping $f \cdot f \cdot f$ is the identity mapping. How many such $f$ are there?
(1996. Japan Mathematical Olympiad Preliminary)
|
Solution: Since the threefold composite mapping on set A is the identity mapping, it follows from Proposition 1 and Proposition 2 that there are three types of mappings $f$ that meet the conditions:
(1) $f$ is the identity mapping;
(2) There exists a three-element mapping cycle $a \rightarrow b \rightarrow c \rightarrow a (a, b, c$ are distinct), and the other three elements are fixed points;
(3) There exist two three-element mapping cycles $a \rightarrow b \rightarrow c \rightarrow a$ and $a' \rightarrow b' \rightarrow c' \rightarrow a' (a, b, c, a', b', c'$ are distinct).
There is only 1 $f$ of type (1).
For type (2), first choose 3 elements $a, b, c$ from 6 elements, which can be done in $C_{6}^{3}=20$ ways. Then, the arrangement of $a, b, c$ in a circle has (3-1)! $=2$ ways, so there are $20 \times 2=40$ such $f$ (s).
For type (3), first divide 6 elements into two groups, which can be done in $C_{6}^{3} \div 2=10$ ways. Each group can be arranged in a circle in (3-1)!(3-1)! =4 ways, so there are $10 \times 4=40$ such $f$ (s).
In summary, the total number of $f$ is
$$
1+40+40=81 \text { (s). }
$$
|
81
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, let integers $a, b, c$ satisfy $1 \leqslant a<b<c \leqslant 5, a^{3}$, $b^{3}, c^{3}$ have unit digits $x, y, z$ respectively. When $(x-y)(y-z)(z-x)$ is the smallest, find the maximum value of the product $a b c$.
Let integers $a, b, c$ satisfy $1 \leqslant a<b<c \leqslant 5, a^{3}$, $b^{3}, c^{3}$ have unit digits $x, y, z$ respectively. When $(x-y)(y-z)(z-x)$ is the smallest, find the maximum value of the product $a b c$.
|
$$
\text { Three, } \because 1^{3}=1,2^{3}=8,3^{3}=27,4^{3}=64,5^{3}=125 \text {. }
$$
$\therefore(x, y, z)$ has the following 10 possibilities:
(1) $(1,8,7) ;(2)(1,8,4) ;(3)(1,8,5)$;
(4) $(1,7,4) ;(5)(1,7,5) ;(6)(1,4,5)$;
(7) $(8,7,4) ;(8)(8,7,5) ;(9)(8,4,5)$;
$(10)(7,4,5)$.
Then the values of $(x-y)(y-z)(z-x)$ are
$$
-42,-84,-84,-54,-48,12,-12,-6 \text {, }
$$
12,6 .
Therefore, the minimum value of $(x-y)(y-z)(z-x)$ is -84.
At this time, $(x, y, z)=(1,8,4)$ or $(1,8,5)$.
The corresponding $a b c=1 \cdot 2 \cdot 4=8$ or $a b c=1 \cdot 2 \cdot 5=10$. Hence the maximum value of $a b c$ is 10.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, on a circular road, there are 4 middle schools arranged clockwise: $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$. They have 15, 8, 5, and 12 color TVs, respectively. To make the number of color TVs in each school the same, some schools are allowed to transfer color TVs to adjacent schools. How should the TVs be transferred to minimize the total number of TVs transferred? And what is the minimum total number of TVs transferred?
|
Let $A_{1}$ High School transfer $x_{1}$ color TVs to $A_{2}$ High School (if $x_{1}$ is negative, it means $A_{2}$ High School transfers $|x_{1}|$ color TVs to $A_{1}$ High School. The same applies below), $A_{2}$ High School transfer $x_{2}$ color TVs to $A_{3}$ High School, $A_{3}$ High School transfer $x_{3}$ color TVs to $A_{4}$ High School, and $A_{4}$ High School transfer $x_{4}$ color TVs to $A_{1}$ High School.
Since there are 40 color TVs in total, with an average of 10 per school, therefore,
$$
\begin{array}{l}
15-x_{1}+x_{4}=10,8-x_{2}+x_{1}=10, \\
5-x_{3}+x_{2}=10,12-x_{4}+x_{3}=10 . \\
\text { Then }=x_{1}-5, x_{1}=x_{2}+2, x_{2}=x_{3}+5, x_{3}= \\
x_{3}-2, x_{3}=\left(x_{1}-5\right)-2=x_{1}-7, x_{2}=\left(x_{1}-7\right)+5 \\
=x_{:}-2 .
\end{array}
$$
Then $x_{4}=x_{1}-5, x_{1}:=x_{2}+2, \tilde{x}_{2}=x_{3}+5, x_{3}=$ $=x:-\hat{2}$.
The problem requires $y=|x_{1}|+|x_{2}|+|x_{3}|+|x_{4}|$
$$
=|x_{1}|+|x_{1}-2|+|x_{1}-7|+|x_{1}
$$
$-5|$ to be minimized,
where $x_{1}$ is an integer satisfying $-8 \leqslant x_{1} \leqslant 15$.
Let $x_{1}=x$, and consider the function defined on $-8 \leqslant x \leqslant 15$
$$
y=|x|+|x-2|+|x-7|+|x-5| \text {, }
$$
$|x|+|x-7|$ represents the sum of the distances from $x$ to 0 and 7. When $0 \leqslant x \leqslant 7$, $|x|+|x-7|$ takes the minimum value 7; similarly, when $2 \leqslant x \leqslant 5$, $|x-2|+|x-5|$ takes the minimum value 3.
Therefore, when $2 \leqslant x \leqslant 5$, $y$ takes the minimum value 10. In other words, when $x_{1}=2,3,4,5$, $|x_{1}|+|x_{1}-2|+|x_{1}-7|+|x_{1}-5|$ takes the minimum value 10.
Thus, the minimum total number of color TVs transferred is 10, and there are four allocation schemes:
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1: Xiao Zhang is riding a bicycle on a road next to a double-track railway. He notices that every 12 minutes, a train catches up with him from behind, and every 4 minutes, a train comes towards him from the opposite direction. If the intervals between each train are constant, the speeds are the same, and both the train and bicycle speeds are uniform, find the interval in minutes at which trains depart from the stations in front of and behind Xiao Zhang.
(1990, Xianyang City Junior High School Mathematics Competition Selection)
Analysis: Let the interval be $x$ minutes at which trains depart from the stations in front of and behind Xiao Zhang. Let Xiao Zhang's cycling speed be $v_{1}$, and the train speed be $v_{2}$. Let $AB$ be the distance between two adjacent trains in the same direction, then $|AB| = x v_{2}$.
(1) First, consider the trains coming towards Xiao Zhang from the opposite direction. Assume Xiao Zhang meets the first train at point $A$, then when he meets the next train, both the train and Xiao Zhang have traveled the distance $|AB|$, so we have $4(v_{1} + v_{2}) = |AB|$. As shown in Figure 1.
(2) Similarly,
consider the trains
catching up with
Xiao Zhang. At point
$B$, he meets the
first train, and the next train is at point $A$ at that moment. To catch up with Xiao Zhang, the next train travels an additional distance of $|AB|$, so we have $12(v_{2} - v_{1}) = |AB|$. As shown in Figure 2.
|
Solution: Let the trains depart from the station ahead and behind Xiao Zhang every $x$ minutes, Xiao Zhang's cycling speed be $v_{1}$, and the train speed be $v_{2}$, then
$$
\left\{\begin{array}{l}
4\left(v_{1}+v_{2}\right)=x v_{2}, \\
12\left(v_{2}-v_{1}\right)=x v_{2} .
\end{array}\right.
$$
Solving, we get $x=6$ (minutes).
Answer: Omitted.
(2) Time equivalence relationship
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2: Person A and Person B start from points $A$ and $B$ respectively at the same time and walk towards each other. They meet at point $C$, which is 10 kilometers away from $A$. After meeting, both continue at the same speed, reach $B$ and $A$ respectively, and immediately return. They meet again at point $D$, which is 3 kilometers away from $B$. Find the distance between $A$ and $B$.
Analysis: As shown in Figure 3, the time it takes for A to travel from $A$ to $C$ is equal to the time it takes for B to travel from $B$ to $C$; the time it takes for A to travel from $A$ to $B$ and then to $D$ is equal to the time it takes for B to travel from $B$ to $A$ and then to $D$.
|
Solution: Let the distance between $A$ and $B$ be $x$ kilometers, the speed of person A be $v_{1}$ kilometers/hour, and the speed of person B be $v_{2}$ kilometers/hour, then
$$
\left\{\begin{array}{l}
\frac{10}{v_{1}}=\frac{x-10}{v_{2}}, \\
\frac{x+3}{v_{1}}=\frac{2 x-3}{v_{2}} .
\end{array}\right.
$$
Eliminating the parameter $v_{1} \backslash v_{2}$, we get
$$
\frac{x-10}{10}=\frac{2 x-3}{x+3} \text {. }
$$
Solving for $x$ yields $x=27$ (kilometers).
Answer: Omitted.
(3) Speed equivalence relationship
|
27
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 A person walks from place A to place B, and there are regular buses running between A and B, with equal intervals for departures from both places. He notices that a bus going to A passes by every 6 minutes, and a bus going to B passes by every 12 minutes. How often do the buses depart from their respective starting stations? (Assume that each bus travels at the same speed)
|
Analysis: Let the distance between two consecutive buses in the same direction be $s$, and the time be $t$ minutes, then the speed of the bus is $\frac{s}{t}$.
(1) A person's speed relative to the bus going to location A is $\frac{s}{6}$, so the person's speed is $\frac{s}{6}-\frac{s}{t}$.
(2) A person's speed relative to the bus going to location B is $\frac{s}{12}$, so the person's speed is $\frac{s}{t}-\frac{s}{12}$.
$$
\therefore \frac{s}{6}-\frac{s}{t}=\frac{s}{t}-\frac{s}{12} \text {. }
$$
Solving for $t$ gives $t=8$ (minutes).
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 In the border desert area, patrol vehicles travel 200 kilometers per day, and each patrol vehicle can carry enough gasoline to travel for 14 days. There are 5 patrol vehicles that set out from base $A$ simultaneously, complete their mission, and then return along the original route to the base. To allow three of them to patrol as far as possible (and then return together), vehicles Jia and Yi travel to a certain point $B$ along the way, leaving only enough gasoline for their return to the base, and leaving the extra gasoline for the other three vehicles to use. How far can the other three vehicles travel at most?
(1905, Hebei Province Junior High School Mathematics Joint Competition)
|
Analysis: The key is where point $B$ is most appropriate. Suppose car A and car B travel for $x$ days to reach point $B$. After reaching point $B$, considering the fuel capacity of the patrol car, we should have
$$
\frac{2(14-2 x)}{3}+(14-x) \leqslant 14 \text {. }
$$
Thus, $x \geqslant 4$.
For the other three cars, the fuel consumption for each car throughout the process is
$$
\frac{2(14-2 x)}{3}+14 \text {. }
$$
Therefore, the distance these three cars leave from base $A$ is
$$
\begin{aligned}
f(x) & =\frac{\frac{2(14-2 x)}{3}+14}{2} \times 200 \\
& =\frac{100}{3}(70-4 x) .
\end{aligned}
$$
This is a decreasing function of $x$.
Thus, $f(x)_{\max }=1800$ (km).
Note: Establishing indeterminate equations and inequalities based on the equal distance relationship can solve the problem of the maximum travel distance.
(2) Inequality relationship
|
1800
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
A function in itself, and for any $s$ and $t$ in $N$, it satisfies $f\left(t^{2} f(s)\right)=s(f(t))^{2}$. Determine the minimum value that $f(1998)$ can achieve among all functions $f$.
untranslated part:
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。
(As requested, the untranslated part is not included in the translation.)
|
Solution: For every function $f$ that satisfies the conditions of the problem,
it is known that $f\left(t^{2} f(s)\right)=s(f(t))^{2}$.
Substituting $t=1$ yields $f(f(s))=s(f(1))^{2}$,
and substituting $s=1$ into (1) gives
$f\left(t^{2} f(1)\right)=(f(t))^{2}$.
Now, let $f(1)=k$, then the above two equations can be rewritten as
$f(f(s))=k^{2} s$,
$f\left(k t^{2}\right)=(f(t))^{2}$.
Substituting $s=1$ into (4) gives
$f(k)=k^{2}$.
Substituting $s=f(k)$ into (1) gives
$f\left(t^{2} k^{3}\right)=k^{2}(f(t))^{2}$.
On the other hand, from (5)
$f\left(t^{2} k^{3}\right)=f\left(k(k t)^{2}\right)=(f(k t))^{2}$.
Comparing the two equations, we get
$$
(f(k t))^{2}=k^{2}(f(t))^{2} \Rightarrow f(k t)=k f(t) \text {. }
$$
Substituting (6) into (5) gives
$k f\left(t^{2}\right)=(f(t))^{2}$.
Similarly, we can write
$k f\left(t^{4}\right)=\left(f\left(t^{2}\right)\right)^{2}$,
$k f\left(t^{2^{m}}\right)=\left(f\left(t^{2^{m-1}}\right)\right)^{2}$.
Squaring the first equation and comparing it with the second gives $k^{3} f\left(t^{4}\right)=(f(t))^{4}$.
Squaring again and comparing with the third equation gives
$k^{7} f\left(t^{8}\right)=(f(t))^{8}$.
Continuing this process, we finally get
$$
(f(t))^{2^{m}}=k^{2^{m}-1} f\left(t^{2^{m}}\right) \text {. }
$$
Here $m$ is any positive integer.
Taking any prime factor $p$ of $k$, we have $p^{a} \| k$, and $p^{g} \| f(t)$.
From (7), we get $\beta \cdot 2^{m} \geqslant \alpha\left(2^{m}-1\right)$,
which implies $\frac{\beta}{\alpha} \geqslant \frac{2^{m}-1}{2^{m}}$.
Taking the limit on both sides gives
$$
\frac{\beta}{\alpha} \geqslant \lim _{m \rightarrow \infty} \frac{2^{m}-1}{2^{m}}=1 \Rightarrow \beta \geqslant \alpha \text {. }
$$
This, along with the arbitrariness of $p$, implies $k \mid f(t)$.
Now, let $g(t)=\frac{f(t)}{k}$. Then $g(t)$ is also a function from $N \rightarrow N$.
From (6), we get
$f\left(i^{2} \hat{j}(s)\right)=f\left(i_{i}^{2} \cdot k g(s)\right)=k f\left(t^{2} \cdot g(s)\right)$
$$
=b^{2} \cdot g\left(t^{2} \cdot g(s)\right)
$$
and $s(f(t))^{2}=k^{2} s(g(t))^{2}$.
From (1), we get
$$
g\left(t^{2} \cdot g(s)\right)=s(g(t))^{2} .
$$
This shows that $g$ is also a function that satisfies the conditions of the problem, and $g(1)=1$. Thus, all the results obtained for $f$ apply to $g$, with $k$ replaced by 1. Therefore, from (4) and (5), we get
$$
\begin{array}{l}
g(g(s))=s, \\
g\left(t^{2}\right)=(g(t))^{2} \text {. } \\
\text { Substituting } g(s) \text { for } s \text { in }(8) \text { gives } \\
g\left(t^{2} s\right)=g(s) \cdot(g(t))^{2} . \\
\text { Applying }(10) \text { and }(11) \text { gives } \\
(g(a b))^{2}=g\left(a^{2} b^{2}\right)=g\left(a^{2}\right)(g(b))^{2} \\
=(g(a) \cdot g(b))^{2} \\
\Rightarrow g(a b)=g(a) g(b) .
\end{array}
$$
From (9), we see that $g$ is injective, so for all $a>1$, we have $g(a)>1$. Furthermore, from (11), the function values of composite numbers are also composite, so for prime $p$, $g(g(p))=p \Rightarrow g(p)$ is prime.
Thus, we have
$$
\begin{array}{l}
f(1998)=k g(1998)=k g(2) \cdot g(3)^{3} \cdot g(37) \\
\geqslant\left(2^{3} \cdot 3 \cdot 5\right) k=120 k \geqslant 120 .
\end{array}
$$
On the other hand, consider a function $f$ defined as follows:
$$
\left\{\begin{array}{l}
f(1)=1, \\
f(2)=3, \\
f(3)=2, \\
f(5)=37, \\
f(37)=5, \\
f(p)=p(p \text { is any other prime }), \\
f(n)=f\left(p_{1}\right)^{\alpha_{1}} f\left(p_{2}\right)^{\alpha_{2} \cdots} f\left(p_{k}\right)^{a_{k}} \\
\left(n=p_{1}^{\alpha_{1}} \cdots p_{k}^{\alpha_{k}} \text { is the prime factorization of the composite number } n\right. \text {) }
\end{array}\right.
$$
It is easy to verify that it satisfies the conditions of the problem, and $f(1998)=120$. Therefore, the minimum value of $f(1998)$ is indeed 120.
|
120
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, let $M=\{2,3,4, \cdots, 1000\}$. Find the smallest natural number $n$ such that in any $n$-element subset of $M$, there exist 3 pairwise disjoint 4-element subsets $S, T, U$ satisfying the following three conditions:
(1) For any two elements in $S$, the larger number is a multiple of the smaller number, and the same property holds for $T$ and $U$;
(2) For any $s \in S$ and $t \in T$, $(s, t)=1$;
(3) For any $s \in S$ and $u \in U$, $(s, u)>1$.
|
Solution: Note that $999=37 \times 27$, let $A=\{3,5, \cdots, 37\}$, $B=M-A$, then $,|A|=18,|B|=981$.
Below we prove that the subset $B$ of $M$ cannot simultaneously satisfy conditions (1) $\sim$ (3). If not, let $S=\left\{s_{1}, s_{2} ; s_{3}, s_{4}\right\}, T=\left\{t_{1}, t_{2}, t_{3}, t_{4}\right\}$ and there is $s_{1}1000$. Contradiction. This indicates that the smallest natural number $n \geqslant 982$.
On the other hand, let
$\left\{\begin{array}{l}S_{1}=\{3,9,27,81,243,729\}, \\ T_{1}=\{2,4,8,16,32,64\}, \\ U_{1}=\{6,12,24,48,96,192\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{2}=\{5,15,45,135,405\}, \\ T_{2}=\{41,82,164,328,656\}, \\ U_{2}=\{10,20,40,80,160\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{3}=\{7,21,63,189,567\}, \\ T_{3}=\{43,86,172,344,688\}, \\ U_{3}=\{14,28,56,112,224\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{4}=\{11,33,99,297,891\}, \\ T_{4}=\{47,94,188,376,752\}, \\ U_{4}=\{22,44,88,176,352\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{5}=\{13,39,117,351\}, \\ T_{5}=\{53,106,212,424\}, \\ U_{5}=\{26,52,104,208\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{6}=\{17,51,153,459\}, \\ T_{6}=\{59,118,236,472\}, \\ U_{6}=\{34,68,136,272\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{7}=\{19,57,171,513\}, \\ T_{7}=\{61,122,244,488\}, \\ U_{7}=\{38,76,152,304\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{8}=\{23,69,207,621\}, \\ T_{8}=\{67,134,268,536\}, \\ U_{8}=\{46,92,184,368\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{9}=\{25,75,225,675\}, \\ T_{9}=\{71,142,284,568\}, \\ U_{9}=\{50,100,200,400\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{10}=\{29,87,261,783\}, \\ T_{10}=\{73,146,292,584\}, \\ U_{10}=\{58,116,232,464\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{11}=\{31,93,279,837\}, \\ T_{11}=\{79,158,316,632\}, \\ U_{11}=\{62,124,248,496\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{12}=\{35,105,315,945\}, \\ T_{12}=\{83,166,332,664\}, \\ \left.U_{12}=\mid 70,140,280,560\right\} ;\end{array}\right.$
$\left\{\begin{array}{l}S_{13}=\{37,111,333,999\}, \\ T_{13}=\{89,178,356,712\}, \\ U_{13}=\{74,148,296,592\}\end{array}\right.$
Let $S_{i}, T_{i}, U_{i}$ be the 3-tuples formed by the 3 numbers in each group, a total of 57 such 3-tuples can be obtained. For any 982-element subset $B$ of $M$, only 17 numbers in $M$ are not in $B$, so at least 40 of the 57 3-tuples are contained in $B$. These 40 3-tuples belong to the 13 groups mentioned above. By the pigeonhole principle, at least 4 of these 3-tuples must belong to the same group among the 13 groups. Writing these 4 3-tuples as a $3 \times 4$ number table, the 3 rows of numbers are $S, T, U$ which satisfy the requirements of the problem.
In summary, the smallest natural number $n=982$.
|
982
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, $A, B, C=$ Three countries are holding a Go broadcast tournament, each team with 9 members. The rules are as follows: Each match involves one person from each of two teams competing, the winner stays to defend, the loser is eliminated, and the other team sends one person to challenge. The first match starts with one person from each of $A$ and $B$ teams, and then proceeds in sequence. If any team's 9 members are all eliminated, the remaining two teams continue to compete until another team is completely eliminated. The team of the winner of the last match is the champion team. Answer the following questions and provide reasons:
(1) What is the minimum number of wins for the champion team?
(2) If the champion team wins 11 matches at the end of the tournament, what is the minimum number of matches that could have been played in the entire tournament?
|
Solution: (1) When the champion team finally wins, the other two teams' 18 people have all been eliminated. Since team $C$ plays later, when $C$ team wins the championship, it can win one less game. To minimize the number of wins for team $C$, teams $A$ and $B$ need to eliminate as many of their own members as possible through mutual matches.
According to the match procedure, for every two adjacent members of teams $A$ and $B$ who are eliminated, there must be one member of team $C$ who is also eliminated. Since team $C$ can have at most 8 members eliminated, the number of members eliminated from teams $A$ and $B$ through mutual matches is at most 9. Therefore, team $C$ must win at least 9 games.
On the other hand, if $A_{1}$ consecutively defeats $B_{1}, C_{1}, B_{2}, C_{2}$, ..., $B_{8}, C_{8}, B_{9}$, and then $C_{9}$ defeats all 9 members of team $A$, team $C$ will win the championship with exactly 9 wins.
In summary, the champion team must win at least 9 games.
(2) If the champion team wins a total of 11 games, 11 of the 18 people from teams $A$ and $B$ lose to members of the champion team, while the remaining 7 are eliminated through mutual matches between teams $A$ and $B$. Therefore, 6 members of team $C$ are eliminated. At least 24 games are played in total.
On the other hand, if $A_{1}$ consecutively defeats $B_{1}, C_{1}, B_{2}, C_{2}$, ..., $B_{6}, C_{6}, B_{7}$, and then $C_{7}$ consecutively defeats $A_{1}, B_{8}, A_{2}, B_{9}$, $A_{3}$, ..., $A_{9}$, team $C$ will win a total of 11 games and win the championship, with a total of 24 games played.
In summary, the entire competition must have at least 24 games.
|
24
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 A shipping company has a ship leaving Harvard for New York every noon, and at the same time every day, a ship also leaves New York for Harvard. It takes seven days and seven nights for the ships to complete their journeys in both directions, and they all sail on the same route. How many ships of the same company will the ship leaving Harvard at noon today encounter on its way to New York?
|
Solution 1: Establishing Equal Relationships of Distance and Time.
Analysis: If this ship meets another ship opposite to it at point $A$, then the next ship opposite to it is at point $B$. By the equal relationship of distance, it is easy to know that they meet at the midpoint of $A B$, and the interval is exactly half a day. Therefore, throughout the process, it should meet 15 ships coming from the opposite direction.
Solution 2: Establishing Indeterminate Equation Method.
Let the ship that departs from Harvard at noon today meet a ship that departed $x$ days ago, then
\[
\begin{array}{l}
x+y=7(\hat{u} \leqslant x \leqslant 7), \\
x-y=n(n \in Z). \\
\therefore 2 x=7+n. \\
\text { Thus, }-7 \leqslant n \leqslant 7.
\end{array}
\]
Therefore, $n$ can take $-7,-6, \cdots,-1,0,1, \cdots, 6,7$. There are a total of 15 ships.
|
15
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (Full marks 20 points) Given that $x$, $y$, $z$ are positive integers, and satisfy $x^{3}-y^{3}-z^{3}=3 x y z$, $x^{2}=2(y+z)$. Find the value of $x y+y z+z x$.
---
The translation is provided as requested, maintaining the original format and line breaks.
|
$$
\begin{aligned}
- & \because x^{3}-y^{3}-z^{3}-3 x y z \\
= & x^{3}-(y+z)^{3}-3 x y z+3 y^{2} z+3 y z^{2} \\
= & (x-y-z)\left(x^{2}+x y+x z+y^{2}+2 y z+z^{2}\right) \\
& -3 y z(x-y-z) \\
= & (x-y-z)\left(x^{2}+y^{2}+z^{2}-x y-y z+x z\right),
\end{aligned}
$$
X. $x^{3}-y^{3}-z^{3}=3 x y z$,
$$
\begin{array}{l}
\therefore(x-y-z)\left(x^{2}+y^{2}+z^{2}+x y-y z+x z\right) \\
\quad=0 .
\end{array}
$$
Since $x, y, z$ are all natural numbers,
$$
\begin{array}{l}
\text { and } x^{2}+y^{2}+z^{2}+x y-y z+x z \\
=\frac{1}{2}\left[(x+y)^{2}+(y-z)^{2}+(z+x)^{2} \neq 0,\right. \\
\therefore x-y-z=0,
\end{array}
$$
i.e., $x=y+z$.
Substituting (1) into $x^{2}=2(y+z)$, we get $x^{2}=2 x$.
$$
\therefore x=2, y+z=2 \text {. }
$$
Given $y \geqslant 1$ and $z \geqslant 1$, we know $y=z=1$.
$$
\therefore x y+y z+z x=5 \text {. }
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (Full marks 25 points) On the blackboard, all natural numbers from 1 to 1997 are written. Students $A$ and $B$ take turns to perform the following operations: Student $A$ subtracts the same natural number from each number on the blackboard (the number subtracted can be different in different operations); Student $B$ erases two numbers from the blackboard and writes down their sum. Student $A$ goes first, and the operations continue until only one number remains on the blackboard. If this number is non-negative, find this number.
|
Three, because after each operation by student $A$ and student $B$, the number of numbers written on the blackboard decreases by 1. Since students $A$ and $B$ take turns operating, when $B$ completes the last operation, only one number remains on the blackboard, and both have performed 1996 operations. Let $d_{k} (k=1,2, \cdots, 1996)$ be the natural number removed by student $A$ during the $k$-th operation. Since in the $k$-th operation, the sum of the numbers on the blackboard decreases by $(1998-k) d_{k}$, and student $B$'s operation does not change this sum, after 1996 alternating operations, the number written on the blackboard is
\[
\begin{aligned}
x= & (1+2+\cdots+1997)-1997 d_{1}-1996 d_{2}-\cdots \\
& -2 d_{1996} \\
= & 1997\left(1-d_{1}\right)+1996\left(1-d_{2}\right)+\cdots+(1998 \\
& -k)\left(1-d_{k}\right)+\cdots+2\left(1-d_{1996}\right)+1 .
\end{aligned}
\]
For all $k=1,2, \cdots, 1996$, the number $1-d_{k}$ is non-positive.
If for some $k$, $d_{k} \geqslant 2$, then
\[
(1998-k)\left(d_{k}-1\right) \geqslant 2 \text{. }
\]
Thus, $x \leqslant(1998-k)\left(1-d_{k}\right)+1 \leqslant-1$.
This contradicts the conditions given in the problem. Therefore, for all $k=1,2, \cdots, 1996$, $d_{k}=1$. This means $x=1$.
(243000, Anhui Province, Maguan Second Middle School, Cheng Lihu)
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $M$ be a subset of the set $S=\{1,2,3, \cdots, 1998\}$, and each natural number (element) in $M$ contains exactly one 0. Then the maximum number of elements in the set $M$ is ( ).
(A) 324
(B) 243
(C) 495
(D) 414
|
6. (D).
Divide the set $S$ into four subsets
$$
\begin{array}{l}
S_{1}=\{1,2, \cdots, 9\}, \\
S_{2}=\{10,11, \cdots, 99\}, \\
S_{3}=\{100,101, \cdots, 999\}, \\
S_{4}=\{1000,1001, \cdots, 1998\} .
\end{array}
$$
Let $M_{i}^{r}$ represent the set formed by all elements in $S_{i}(i=1,2,3,4)$ that contain exactly one digit 0.
Obviously, none of the elements in $S_{1}$ contain 0, so $\left|M_{1}\right|=0$;
In $S_{2}$, there are 9 elements where the unit digit is 0, so $\left|M_{2}\right|=9$;
In $S_{3}$, there are $9^{2}$ elements where only the unit digit or the tens digit is 0, so $\left|M_{3}\right|=2 \times 9^{2}=162$;
In $S_{4}$, there are $9^{2}$ elements where only the unit digit, the tens digit, or the hundreds digit is 0, so $\left|M_{4}\right|=3 \times 9^{2}=243$.
Therefore, $|M|_{\text {max }}=\left|M_{1}\right|+\left|M_{2}\right|+\left|M_{3}\right|+\left|M_{4}\right|=414$.
|
414
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. From the center of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, two perpendicular chords $A C$ and $B D$ are drawn. Connecting $A, B, C, D$ in sequence forms a quadrilateral. Then, the maximum value of the area $S$ of quadrilateral $A B C D$ is
|
3. 12.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Given $O A \perp O B$, we have
$$
\begin{array}{l}
x_{1} x_{2}+y_{1} y_{2}=0 . \\
\begin{aligned}
\therefore S & =4 S_{\triangle A O B}=2|O A| \cdot|O B| \\
& =2 \sqrt{\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)} \\
& =2 \sqrt{\left(x_{1} x_{2}+y_{1} y_{2}\right)^{2}+\left(x_{1} y_{2}-x_{2} y_{1}\right)^{2}} \\
& =2 \sqrt{\left(x_{1} y_{2}-x_{2} y_{1}\right)^{2}} \\
& =12 \sqrt{\left(\frac{x_{1}}{3} \cdot \frac{y_{2}}{2}-\frac{x_{2}}{3} \cdot \frac{y_{1}}{2}\right)^{2}} \\
& =12 \sqrt{\left(\frac{x_{1}^{2}}{9}+\frac{y_{1}^{2}}{4}\right)\left(\frac{x_{2}^{2}}{9}+\frac{y_{2}^{2}}{4}\right)-\left(\frac{x_{1} x_{2}}{9}+\frac{y_{1} y_{2}}{4}\right)^{2}} \\
& =12 \sqrt{1-\left(\frac{x_{1} x_{2}}{9}+\frac{y_{1} y_{2}}{4}\right)^{2}} \leqslant 12 .
\end{aligned}
\end{array}
$$
Equality holds if and only if $\frac{x_{1} x_{2}}{9}+\frac{y_{1} y_{2}}{4}=0$. Given $x_{1} x_{2}+$ $y_{1} y_{2}=0$, it follows that $x_{1} x_{2}=0, y_{1} y_{2}=0$. In this case, points $A$ and $B$ are the endpoints of the major and minor axes of the ellipse, respectively.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given a pyramid $S-ABCD$ inscribed in a sphere with the base being a rectangle $ABCD$, and $SA=4, SB=8, SD=7$, $\angle SAC=\angle SBC=\angle SDC$. Then the length of $BD$ is
|
5. $B D=9$.
As shown in the figure, $\because \angle S A C=\angle S B C=\angle S D C$,
$$
\therefore \frac{S C}{\sin \angle S A C}=\frac{S C}{\sin \angle S B C}=\frac{S C}{\sin \angle S D C} \text {. }
$$
Therefore, the circumcircles of $\triangle S A C$,
$\triangle S B C, \triangle S D C$ have equal diameters, meaning the circles formed by the intersection of the sphere with the planes $S A C$, plane $S B C$,
plane $S D C$ are equal.
The plane passing through the midpoint $O$ of $S C$ and perpendicular to $S C$ intersects $S A C, S B C, S D C$ at the midpoints of the arcs $A^{\prime} 、 B^{\prime} 、 D^{\prime}$, and $O A^{\prime}=O B^{\prime}=O D^{\prime}$, thus, $O$ is the circumcenter of $\triangle A^{\prime} B^{\prime} C^{\prime}$. Hence, $S C$ is the diameter of the sphere, $O$ is the center of the sphere, and $\angle \mathrm{SAC}=\angle \mathrm{SBC}=\angle \mathrm{SDC}=90^{\circ}$.
Let $B D=A C=x, A B=y, B C=z$. Then
$$
S C^{2}=x^{2}+4^{2}=y^{2}+7^{2}=z^{2}+8^{2} \text {, }
$$
which simplifies to $2\left(x^{2}+16\right)=\left(y^{2}+49\right)+\left(z^{2}+64\right)$
$$
=x^{2}+49+64 \text {. }
$$
Solving this, we get $x=9$, i.e., $B D=9$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given positive integers $m, n$ satisfying $m+n=19$. Then the maximum number of solutions to the equation $\cos m x=\cos n x$ in the interval $[0, \pi]$ is
|
6.18.
Since $m+n=19$ is odd, then $m \neq n$. Without loss of generality, assume $m>n$.
The solutions to the equation $\cos m x=\cos n x$ are
$x=\frac{2 k_{1} \pi}{m+n}, x=\frac{2 k_{2} \pi}{m-n}\left(k_{1}, k_{2} \in \mathbb{Z}\right)$.
From $0 \leqslant \frac{2 k_{1} \pi}{m+n} \leqslant \pi, 0 \leqslant \frac{2 k_{2} \pi}{m-n} \leqslant \pi$, we get
$k_{1}=0,1,2, \cdots, \frac{m+n-1}{2}$,
$k_{2}=0,1,2, \cdots, \frac{m-n-1}{2}$.
Therefore, the number of roots of the equation is no more than
$$
\left(\frac{m+n-1}{2}+1\right)+\left(\frac{m-n-1}{2}+1\right)=m+1 \text {. }
$$
Since $n \geqslant 1$, then $m \leqslant 18$, so the number of roots is no more than 19.
However, there cannot be 19 roots, because this would include the root when $k_{1}=k_{2}=0$, thus the number of roots is no more than 18.
Next, we prove that the number of roots of the equation is exactly 18. For this, we only need to prove that when $k_{1}, k_{2}$ are not equal to 0, $\frac{2 k_{1} \pi}{m+n} \neq \frac{2 k_{2} \pi}{m-n}$. If not, then $\frac{k_{1}}{19}=\frac{k_{2}}{m-n}$, i.e., $19 k_{2}=(m-n) k_{1}$, thus $k_{1}$ is a multiple of 19. But $k_{1} \leqslant \frac{m+n}{2}=\frac{19}{2}$, so this is impossible.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (Full marks 20 points) The complex numbers $z_{1}, z_{2}, z_{3}, z_{4}, z_{5}$ satisfy
$$
\left\{\begin{array}{l}
\left|z_{1}\right| \leqslant 1, \\
\left|z_{2}\right| \leqslant 1, \\
\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \leqslant\left|z_{1}-z_{2}\right|, \\
\left|2 z_{4}-\left(z_{1}+z_{2}\right)\right| \leqslant\left|z_{1}-z_{2}\right|, \\
\left|2 z_{5}-\left(z_{3}+z_{4}\right)\right| \leqslant\left|z_{3}-z_{4}\right| .
\end{array}\right.
$$
Find the maximum value of $\left|z_{5}\right|$.
|
$$
\begin{array}{l}
\text { 3, } \because\left|z_{1}-z_{2}\right| \geqslant\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \\
\quad \geqslant|2| z_{3}|-| z_{1}+z_{2}||, \\
\therefore\left|z_{1}+z_{2}\right|-\left|z_{1}-z_{2}\right| \leqslant 2\left|z_{3}\right| \\
\quad \leqslant\left|z_{1}+z_{2}\right|+\left|z_{1}-z_{2}\right| . \\
\therefore\left|z_{3}\right| \leqslant \frac{\left|z_{1}+z_{2}\right|+\left|z_{1}-z_{2}\right|}{2} \\
\quad \leqslant \sqrt{\frac{\left|z_{1}+z_{2}\right|^{2}+\left|z_{1}-z_{2}\right|^{2}}{2}} \\
=\sqrt{\frac{2\left|z_{1}\right|^{2}+2\left|z_{2}\right|^{2}}{2}} \leqslant \sqrt{2} .
\end{array}
$$
Similarly, $\left|z_{4}\right| \leqslant \sqrt{2}$.
$$
\therefore\left|z_{5}\right| \leqslant \sqrt{\left|z_{3}\right|^{2}+\left|z_{4}\right|^{2}} \leqslant 2 \text {. }
$$
The equality in the above formula can sometimes be achieved. For example, when $z_{1}=1$, $z_{2}=$, $z_{3}=1+i$; and when $z_{1}=-1, z_{2}=i$, $z_{4}=-1+i$; when $z_{3}=1+i, z_{4}=-1+i$, $z_{5}=2i$, at this time $\left|z_{5}\right|=2$.
Therefore, $\left|z_{5}\right|_{\max }=2$.
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|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Polynomial $M=2 a^{2}-8 a b+17 b^{2}-$ $16 a-4 b+1998$. Then, what is the minimum value of $M$?
|
$$
\text { Sol: } \begin{aligned}
M= & 2\left(a^{2}+4 b^{2}+16-4 a b-8 a+\right. \\
& 16 b)+9\left(b^{2}-4 b+4\right)+1930 \\
= & 2(a-2 b-4)^{2}+9(b-2)^{2}+ \\
& 1930 .
\end{aligned}
$$
It is evident that for all real numbers $a, b$, we always have $M \geqslant 1930$. The minimum value of $M$ is 1930, which occurs only when $a-2 b-4=0$ and $b-2=0$, i.e., $a=8, b=2$.
2 Formula Method
When $a, b$ are both positive, $(\sqrt{a}-\sqrt{b})^{2} \geqslant 0$. Therefore, $a+b \geqslant 2 \sqrt{a b}(a>0, b>0$, equality holds only when $a=b$).
|
1930
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 13 Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ all be natural numbers, and $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} x_{4} x_{5}$. Try to find the maximum value of $x_{5}$.
|
Solution: Without loss of generality, let $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}$.
$$
\because x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} \text {. }
$$
$x_{4} x_{5}$,
$$
\begin{aligned}
\therefore 1= & \frac{1}{x_{2} x_{3} x_{4} x_{5}}+\frac{1}{x_{1} x_{3} x_{4} x_{5}} \\
& +\frac{1}{x_{1} x_{2} x_{4} x_{5}}+\frac{1}{x_{1} x_{2} x_{3} x_{5}} \\
& +\frac{1}{x_{1} x_{2} x_{3} x_{4}} \\
\leqslant & \frac{1}{x_{4} x_{5}}+\frac{1}{x_{4} x_{5}}+\frac{1}{x_{4} x_{5}}+\frac{1}{x_{5}}+\frac{1}{x_{4}} \\
& =\frac{3+x_{4}+x_{5}}{x_{4} x_{5}} .
\end{aligned}
$$
Thus, $x_{4} x_{5} \leqslant 3+x_{4}+x_{5}$.
Therefore, $\left(x_{4}-1\right)\left(x_{5}-1\right) \leqslant 4$.
If $x_{4}=1$, then $x_{1}=x_{2}=x_{3}=x_{4}=1$. From the given, we have $4+x_{5}=x_{5}$, which is a contradiction. Therefore, $x_{4} \geqslant 2$. Then $x_{5}-1 \leqslant\left(x_{4}-1\right)\left(x_{5}-1\right) \leqslant 4, x_{5} \leqslant 5$.
When $x_{5}=5$, there exist $x_{1}=x_{2}=x_{3}=1$, $x_{4}=2$ that satisfy the equation.
Therefore, the maximum value of $x_{5}$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 14 Try to find the maximum value of the following expression:
$$
P=|\cdots|\left|x_{1}-x_{2}\right|-x_{3}|-\cdots|-x_{1995} \mid,
$$
where $x_{1}, x_{2}, \cdots, x_{1995}$ are different natural numbers from 1 to 1995.
|
Solution: It is evident that when $x_{1} \geqslant 0, x_{2} \geqslant 0$, $\left|x_{1}-x_{2}\right|$ does not exceed the largest of $x_{1}$ and $x_{2}$; for $x_{1} \geqslant 0, x_{2} \geqslant$ $0, x_{3} \geqslant 0$, $\left|\left|x_{1}-x_{2}\right|-x_{3}\right|$ does not exceed the largest of $x_{1}$, $x_{2}$, and $x_{3}$. Therefore, $P \leqslant 1995$.
On the other hand, $P$ has the same parity as $x_{1}+x_{2}+\cdots+x_{1995}$, and $x_{1}+x_{2}+\cdots+x_{1995}=1+2+3+\cdots+1995$ is an even number, so $P \leqslant 1994$. In fact, $\left|\left|(4 k+1)-(4 k+3)\right|-(4 k+4)\right|-(4 k+2)=0$ holds for $k=0,1,2, \cdots$. Therefore, $1 \sim 1992$ can be processed according to the above method to yield a result of 0. And $\left|\left|1993-1994\right|-1995\right|=1994$. Hence, the maximum value of $P$ is 1994.
|
1994
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 If $x y=1$, then the minimum value of the algebraic expression $\frac{1}{4}+\frac{1}{4 y^{4}}$ is $\qquad$
|
Solution: From $a+b \geqslant 2 \sqrt{a b}$ we know $\frac{1}{x^{4}}+\frac{1}{4 y^{4}} \geqslant 2 \sqrt{\frac{1}{x^{4} \cdot 4 y^{4}}}=1$. Therefore, the minimum value of $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example: $3 A, B, C, D, E$ five people participate in an exam, with 7 questions, all of which are true or false questions. The scoring rule is: for each question, a correct answer earns 1 point, a wrong answer deducts 1 point, and no answer neither earns nor deducts points. Figure 1 records the answers of $A, B, C, D, E$ five people. It is known that $A, B, C, D$ each scored 2 points, how many points should $E$ get? What is the answer to each question?
保留源文本的换行和格式，直接输出翻译结果如下：
Example: $3 A, B, C, D, E$ five people participate in an exam, with 7 questions, all of which are true or false questions. The scoring rule is: for each question, a correct answer earns 1 point, a wrong answer deducts 1 point, and no answer neither earns nor deducts points. Figure 1 records the answers of $A, B, C, D, E$ five people. It is known that $A, B, C, D$ each scored 2 points, how many points should $E$ get? What is the answer to each question?
|
Let: Assign $k=1,2, \cdots, 7$. When the conclusion of the $k$-th question is correct, i.e., $x_{k}:=1$, if it is judged as correct (i.e., marked with the symbol “$\checkmark$”), then $x_{k}$ points are scored; if it is judged as incorrect (i.e., marked with the symbol “$X$”), then $-x_{k}$ points are scored. When the conclusion of the $k$-th question is incorrect, i.e., $x_{k}=-1$, if it is judged as correct, then $x_{k}$ points are scored; if it is judged as incorrect, then $-x_{k}$ points are scored. Since $A, B, C, D$ each scored 2 points, we can obtain the system of equations:
$$
\left\{\begin{array}{l}
x_{1}+0 \cdot x_{2}-x_{3}+x_{4}-x_{5}+x_{6}+x_{7}=2, \\
x_{1}-x_{2}+x_{3}+x_{4}-x_{5}-x_{6}+0 \cdot x_{7}=2, \\
0 \cdot x_{1}+x_{2}-x_{3}-x_{4}+x_{5}-x_{6}+x_{7}=2, \\
-x_{1}-x_{2}-x_{3}+x_{4}+x_{5}+0 \cdot x_{6}-x_{7}=2
\end{array}\right.
$$
Adding these four equations, we get
$$
x_{1}-x_{2}-2 x_{3}+2 x_{4}+0 \cdot x_{5}-x_{6}-x_{7}=8.
$$
Noting that $x_{i}= \pm 1(i=1,2, \cdots, 7)$, the left side of the above equation is $\leqslant 8$, and the right side is $=8$, hence $x_{1}=1, x_{2}=-1$, $x_{3}=-1, x_{4}=1, x_{6}=-1, x_{7}=1$. Substituting these results into the first equation of the system, we get $x_{5}=1$. Therefore, the 1st, 4th, 5th, and 7th questions are correct, and the 2nd, 3rd, and 6th questions are incorrect. Thus, according to the problem, $E$ scored 4 points.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Given that $x, y, z$ are all positive numbers, and $x y z \cdot (x+y+z)=1$. Then, the minimum value of $(x+y)(y+z)$ is . $\qquad$
|
Analysis: To find the minimum value of the original expression, that is, to find the minimum value of $x y+y^{2}+$ $x z+y z$, we can achieve this by appropriately combining terms so that the sum becomes the sum of two terms, and the product of these two terms should be a constant.
$$
\text { Solution: } \begin{array}{l}
\because(x+y)(y+z)=\left(x y+y^{2}+y z\right) \\
+x z \\
= y(x+y+z)+x z, \\
\text { and } x z>0, y(x+y+z)>0, \\
\therefore y(x+y+z)+x z \\
\geqslant 2 \sqrt{y(x+y+z) \cdot x z}=2, \\
\therefore(x+y)(y+z) \text { has a minimum value of } 2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. For a geometric sequence $\left\{a_{n}\right\}$ with all terms being real numbers, the sum of the first $n$ terms is denoted as $S_{n}$. If $S_{10}=10, S_{30}=70$, then $S_{40}$ equals ( ).
(A)150
(B) -200
(C)150 or -200
(D) 400 or -50
|
3. A.
Let $b_{1}=S_{10}, b_{2}=S_{20}-S_{10}, b_{3}=S_{30}-S_{20}, b_{4}=$ $S_{\infty 0}-S_{30}$. Suppose $q$ is the common ratio of $\left\{a_{n}\right\}$, then $b_{1}, b_{2}, b_{3}, b_{4}$ form a geometric sequence with the common ratio $r=q^{10}$. Therefore, $70=S_{30}=b_{1}+$
$$
\begin{array}{l}
b_{2}+b_{3}=b_{1}\left(1+r+r^{2}\right)=10\left(1+r+r^{2}\right) . \\
\quad \therefore r^{2}+r-6=0, r=2\left(\because r=q^{10}>0, \therefore r=-3\right.
\end{array}
$$
should be discarded).
Thus, $S_{40}=10\left(1+2+2^{2}+2^{3}\right)=150$.
|
150
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the complex number $z=\cos \theta+i \sin \theta\left(0^{\circ} \leqslant \theta \leqslant\right.$ $\left.180^{\circ}\right)$, and the complex numbers $z, (1+i)z, 2\bar{z}$ correspond to three points $P, Q, R$ in the complex plane. When $P, Q, R$ are not collinear, the fourth vertex of the parallelogram formed by segments $PQ, PR$ is $S$. Then the maximum distance from point $S$ to the origin is $\qquad$.
|
2.3.
Let the complex number $w$ correspond to point $S$. Since $Q P R S$ is a parallelogram, we have
$$
w+z=2 \bar{z}+(1+i) z \text{, i.e., } w=2 \bar{z}+i z \text{. }
$$
Therefore, $|w|^{2}=(2 \bar{z}+i z)(2 z-i \bar{z})$
$$
\begin{array}{l}
=4+1+2 i\left(z^{2}-\bar{z}^{2}\right) \\
=5-4 \sin 2 A<5 \div 4=9 .
\end{array}
$$
(When $\theta=135^{\circ}$, equality holds.)
Hence, $|w|_{\operatorname{max}}=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. From the 10 numbers $0,1,2,3,4,5,6,7,8,9$, choose 3 numbers such that their sum is an even number not less than 10.
The number of different ways to choose them is $\qquad$ .
|
$3.5 i$
From these 10 numbers, the number of ways to choose 3 different even numbers is $C_{5}^{3}$; the number of ways to choose 1 even number and 2 different odd numbers is $C_{5}^{1} C_{5}^{2}$.
From these 10 numbers, the number of ways to choose 3 numbers such that their sum is an even number less than 10, there are the following 9 different ways:
$$
\begin{array}{l}
(0,1,3) ;(0,1,5),(0,2,4),(1,2,3) ;(0,1,7) ; \\
(0,2,6),(0,3,5),(1,2,5),(1,3,4) .
\end{array}
$$
Therefore, the number of different ways that meet the requirements of the problem is
$$
C_{5}^{3}+C_{5}^{1} C_{5}^{2}-9=51 \text { ways. }
$$
|
51
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In an arithmetic sequence with real number terms, the common difference is 4, and the square of the first term plus the sum of the remaining terms does not exceed 100. Such a sequence can have at most terms.
|
4.8 .
Let $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence with a common difference of 4, then
$$
\begin{aligned}
& a_{1}^{2}+a_{2}+a_{3}+\cdots+a_{n} \leqslant 100 \\
\Leftrightarrow & a_{1}^{2}+\frac{\left(a_{1}+4\right)+\left[a_{1}+4(n-1)\right]}{2} \cdot(n-1) \\
& \leqslant 100 \\
\Leftrightarrow & a_{1}^{2}+(n-1) a_{1}+\left(2 n^{2}-2 n-100\right) \leqslant 0 .
\end{aligned}
$$
There exists at least one real number $a_{1}$ satisfying inequality (1) if and only if $\Delta \dot{=}^{\prime}(n-1)^{2}-4\left(2 n^{2}-2 n-100\right) \geqslant 0$. Since $\Delta \geqslant 0 \Leftrightarrow 7 n^{2}-6 n-401 \leqslant 0$
$$
\Leftrightarrow n_{1} \leqslant n \leqslant n_{2} \text {, }
$$
where $n_{1}=\frac{3-\sqrt{2816}}{7}<0,8<n_{2}=\frac{3+\sqrt{2816}}{7}<9$.
Therefore, the maximum natural number $n$ satisfying inequality (2) is 8, meaning the sequence can have at most 8 terms.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 When $x$ varies, the minimum value of the fraction $\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$ is $\qquad$
(1993, National Junior High School Competition)
|
Let $y=\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$. Then,
$$
\left(3-\frac{1}{2} y\right) x^{2}+(6-y) x+(5-y)=0 \text {. }
$$
Since $x$ is a real number, $\Delta \geqslant 0$, so,
$$
y^{2}-10 y+24 \leqslant 0 \text {. }
$$
Thus, $4 \leqslant y \leqslant 6$.
When $y=4$, $x=1$.
Therefore, when $x=1$, the minimum value of the original expression is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, 50 points 5) For positive integers $a, n$, define $F_{a}(a)=q+r$, where $q, r$ are non-negative integers, $a=$ $q n+r$, and $0 \leqslant r<n$. Find the largest positive integer $A$ such that there exist positive integers $n_{1}, n_{2}, n_{3}, n_{4}, n_{5}, n_{6}$, for any positive integer $a \leqslant A$, we have
$$
F_{n_{6}}\left(F_{n_{5}}\left(F_{n_{4}}\left(F_{n_{3}}\left(F_{n_{2}}\left(F_{n_{1}}(a)\right)\right)\right)\right)\right)=1 .
$$
Prove your conclusion.
|
Three, the maximum positive integer $B$ that satisfies the condition "there exist positive integers $n_{1}, n_{2}, \ldots, n_{k}$, such that for any positive integer $a \leqslant B$,
$$
F_{n_{k}}\left(F_{\pi_{k-1}}\left(\cdots\left(F_{n_{1}}(a)\right) \cdots\right)\right) \doteq 1^{\prime}
$$
is denoted as $x_{k}$. Clearly, the maximum positive integer $A$ we are looking for is $x_{6}$.
(1) First, prove that $x_{1}=2$.
In fact, $F_{2}(1)=F_{2}(2)=1$, so $x_{1} \geq 2$.
Furthermore, when $n_{1} \geqslant 3$, $F_{n_{1}}(2)=2$, and $F_{2}(3)=F_{1}(2)=2$, so $x_{1}0, n_{1}>1$. Thus, $0<(q-1) n_{1}+n_{1}-1$ $=q n_{1}-1<x_{k+1}$. Therefore,
$$
F_{n_{1}}\left((q-1) n_{1} a_{s}-1\right)=q+n_{1}-2 \leqslant x_{k} \text {. }
$$
Hence, $q\left(n_{1}-1\right) \leqslant\left[\left(\frac{q+n_{1}-1}{2}\right)^{2}\right]$
$$
\leqslant\left[\left(\frac{x_{k}+1}{2}\right)^{2}\right]=\left[\frac{x_{k}^{2}}{4}+\frac{x_{k}}{2}+\frac{1}{4}\right] \text {. }
$$
Since $x_{k}$ is even, thus,
$q\left(n_{1}-1\right) \leqslant \frac{x_{k}^{2}}{4}+\frac{x_{k}}{2}$.
Since $x_{k} \geqslant 2$, so,
$$
x_{k}+\frac{x_{k}^{2}}{4}+\frac{x_{k}}{2} \geqslant x_{k}+2 \text {. }
$$
Therefore, we always have
$x_{k+1} \leqslant x_{k}+\frac{x_{k}^{2}}{4}+\frac{x_{k}}{2}=\frac{x_{k}\left(x_{k}+6\right)}{4}$.
On the other hand, if we take $n_{1}=\frac{x_{k}}{2}+2$, since $\frac{x_{k}\left(x_{k}+6\right)}{4}=$ $\frac{x_{k}}{2} \cdot n_{1}+\frac{x_{k}}{2}$, for each $a \leqslant \frac{x_{k}\left(x_{k}+6\right)}{4}$, let $a=q n_{1}+r$, then,
either $q=\frac{x_{k}}{2}, r \leqslant \frac{x_{k}}{2}$;
or $q \leqslant \frac{x_{k}}{2}-1, r \leqslant n_{1}-1=\frac{x_{k}}{2} \div 1$.
In both cases, we have $q+r \leqslant x_{k}$. Therefore,
$$
x_{k+1}=\frac{x_{k}\left(x_{k}+6\right)}{4} \text {. }
$$
Moreover, since $x_{k}$ is even, if $4 \mid x_{k}$, by $2 \mid\left(x_{k}+6\right)$ we get $8 \mid x_{k}\left(x_{k}+6\right)$; if $x_{k} \equiv 2(\bmod 4)$, by $x_{k}+6 \equiv 0$ $(\bmod 4)$ we also get $8 \mid x_{k}\left(x_{k}+6\right)$. Therefore, $x_{k+1}$ is also even.
Thus, the induction proof is completed: $x_{k+1}=\frac{x_{k}\left(x_{k}+6\right)}{4}$.
Starting from $x_{1}=2$, we successively obtain $x_{2}=4, x_{3}=10, x_{4}=40$, $x_{5}=460, x_{6}=\frac{460 \times 446}{4}=53590$.
Therefore, the maximum positive integer $A=53590$.
(Provided by 邆玉懒)
|
53590
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $\sqrt{x}+\sqrt{y}=35, \sqrt[3]{x}+\sqrt[3]{y}=13$. Then $x+y=$ $\qquad$
|
3.793 .
$$
\begin{aligned}
x+y= & (\sqrt[3]{x}+\sqrt[3]{y})\left[(\sqrt[3]{x}+\sqrt[3]{y})^{2}-3 \sqrt[3]{x y}\right] \\
= & 13\left[13^{2}-3 \sqrt[3]{x y}\right], \\
& (\sqrt{x}+\sqrt{y})^{2}=x+y+2 \sqrt{x y}=35^{2}, \\
& \therefore 35^{2}-2 \sqrt{x y}=13^{3}-39 \sqrt[3]{x y},
\end{aligned}
$$
i.e., $2 \sqrt{x y}-39 \sqrt[3]{x y}+972=0$.
Let $\sqrt[6]{x y}=t$, the above equation becomes
$$
2 t^{3}-39 t^{2}+972=0,
$$
i.e.,
$$
\begin{array}{l}
2 t^{3}-12 t^{2}-27 t^{2}+972=0, \\
2 t^{2}(t-6)-27(t-6)(t+6)=0, \\
(t-6)(t-18)(2 t+9)=0 . \\
t_{1}=6, t_{2}=18, t_{3}=-\frac{9}{2} \text { (discard). }
\end{array}
$$
When $t_{2}=18$, $\sqrt[3]{x y}=324$, at this time $x+y=13\left(13^{2}\right.$ $-3 \times 324)$ is a negative value, which does not meet the problem's requirements;
When $t_{1}=6$, $\sqrt[6]{x y}=6^{6}, \sqrt[3]{x y}=36$, $x+y=13\left(13^{2}-3 \times 36\right)=793$.
|
793
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Does there exist a four-digit number $\overline{a b c d}$, such that the last four digits of its square are also $\overline{a b c d}$? Does there exist a five-digit number $\overline{a b c d e}$, such that the last five digits of its square are also $\overline{a b c d e}$? If they exist, find all of them; if not, explain why.
|
Let $\overline{a b c d}=x$. Since $x^{2}$ and $x$ have the same last four digits, the last four digits of $x^{2}-x$ are 0000, i.e., $10000 \mid x^{2}-x$, or equivalently, $2^{4} \cdot 5^{4} \mid x(x-1)$.
(i) If $2^{4} \mid x$ and $5^{4} \mid x-1$, then $x-1$ is a four-digit odd number and a multiple of 625. Therefore, $x-1$ can only be 1875, 3125, 4375, 5625, 6875, 8125, 9375. Thus, $x$ is 1876, 3126, 4376, 5626, 6876, 8126, 9376. Among these 7 numbers, only 9376 is divisible by $2^{4}$. Verification shows that 9376 is the solution.
(ii) If $2^{4} \mid x-1$ and $5^{4} \mid x$, then $x$ is a four-digit odd number and a multiple of 625. Therefore, $x$ can only be 1875, 3125, 4375, 5625, 6875, 8125, 9375. Thus, $x-1$ is 1874, 3124, 4374, 5624, 6874, 8124, 9374. None of these 7 numbers are divisible by $2^{4}$. Hence, there is no solution in this case.
$\therefore \overline{a b c d}=9376$ is the solution. $\left(9376^{2}=87909376\right)$
Using the same method, we can find that $\overline{a b c d e}=90625$. $\left(90625^{2}=8212890625\right)$
|
9376
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Four cities each send 3 political advisors to participate in $k$ group inspection activities (each advisor can participate in several groups), with the rules: (1) advisors from the same city are not in the same group; (2) any two advisors from different cities exactly participate in one activity together. Then the minimum value of $k$ is $\qquad$ .
|
2.9 Group.
First, consider the CPPCC members from two cities, Jia and Yi. Let the members from Jia city be $A_{1}, A_{2}, A_{3}$, and the members from Yi city be $B_{1}, B_{2}, B_{3}$. They can form 9 pairs: $A_{1} B_{1}, A_{1} B_{2}, A_{1} B_{3}, A_{2} B_{1}, A_{2} B_{2}, A_{2} B_{3}$, $A_{3} B_{1}, A_{3} B_{2}, A_{3} B_{3}$. According to the problem, the number of groups $k$ is no less than 9. To minimize the number of groups, the number of people in each group should be as large as possible (up to 4 people), and the following nine groups can be arranged to meet the requirements:
$$
\begin{array}{ll}
\left(A_{1}, B_{1}, C_{1}, D_{1}\right) & \left(A_{1}, B_{2}, C_{2}, D_{2}\right) \\
\left(A_{1}, B_{3}, C_{3}, D_{3}\right) & \left(A_{2}, B_{1}, C_{2}, D_{3}\right) \\
\left(A_{2}, B_{2}, C_{3}, D_{1}\right) & \left(A_{2}, B_{3}, C_{1}, D_{2}\right) \\
\left(A_{3}, B_{1}, C_{3}, D_{2}\right) & \left(A_{3}, B_{2}, C_{1}, D_{3}\right) \\
\left(A_{3}, B_{3}, C_{2}, D_{1}\right) &
\end{array}
$$
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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