problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
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, r... | 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 s... | 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... | 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... | 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 ... | 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 ... | 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
$$
\begin{array}{l}
(40-15)^{\prime}=25^{\prime} . \\
\therefore \frac{15}{60}(44+v)=\frac{25}{... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $a$ is an integer, the two real roots of the equation $x^{2}+(2 a-1) x+$ $a^{2}=0$ are $x_{1}$ and $x_{2}$. Then $\left|\sqrt{x_{1}}-\sqrt{x_{2}}\right|=$ | 4. 1.
From the problem, the discriminant $\Delta=-4 a+1 \geqslant 0$, so $a \leqslant 0$.
$$
\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_... | 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 $16-5+5 \times 4=31$ ... | 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 vertic... | 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 ... | 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 work... | 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
$$
t(x)=\max \left\{\frac{6000}{5 x}, \frac{2000}{3(214-x)}\right\},(x \in N... | 137 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 The mathematical model of the population dynamics of four types of chicks, where the young can become mature within the same year, estimates the number of a female population after 5 years.
(1) Each female breeds once a year;
(2) Each female breeds only once in the first year of development, and does not bree... | Solution: Let $a_{n}$ represent the number of female birds in the $n$th generation.
(1) According to the problem, $a_{0}=1, a_{n}=2 a_{n-1}, n \geqslant 1$.
It is easy to get $a_{n}=2^{n} a_{0}=2^{n}$.
Therefore, $a_{5}=2^{5}=32$.
(2) According to the problem, $a_{0}=1, a_{n}=a_{n-1}+1$.
Thus, $a_{5}=6$.
(3) Let $b_{... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. A stationery store that operates both wholesale and retail has the following rule: If you buy 51 pencils or more (including 51), you pay the wholesale price; if you buy 50 pencils or fewer (including 50), you pay the retail price. The wholesale price for 60 pencils is 1 yuan cheaper than the retail price for 60 penc... | (Tip: Let the whole class have $x$ students, the equation is $\frac{60 m}{x}-\frac{60 m}{x+10}$ $=1(40<x \leqslant 50)$, solving it yields $x=50)$ | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 If $k$ is a positive integer, and the two roots of the linear equation $(k-1) x^{2}-p x+k=0$ are both positive integers, then the value of $k^{p k}\left(p^{p}+k^{k}\right)+(p+k)$ is $\qquad$ . | Solution: From $x_{1} x_{2}=\frac{k}{k-1}$, we know $\frac{k}{k-1}$ is a positive integer.
$$
\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}
$$
From $x_{1}+x_{2}=\frac{p}{k-1}$, we get $p=3$.
Therefore, the original expression $=2^{6}\l... | 1989 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. There are two flour mills supplying flour to three residential areas. The first flour mill produces 60 tons per month, and the second flour mill produces 100 tons per month. The first residential area requires 45 tons per month, the second residential area requires 75 tons per month, and the third residential area r... | (1960, Shanghai City Mathematics Recontest)(Hint: Let the flour transported by the first flour mill to the first and second residential areas be $x, y$ tons, respectively. Thus, $s=15 x+6 y+840$, where $0 \leqslant x \leqslant 45,0 \leqslant y \leqslant 60 . x+y$ $\geqslant 20$. When $x=20, y=0$, $S_{\text {min }}=960$... | 960 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2. Rearrange $1,2,3,4,5,6,7,8$.
(1) such that the even numbers are in their original positions, and the odd numbers are not in their original positions, how many different arrangements are there?
(2) such that the prime numbers are in the odd positions, and the composite numbers are in the even positions, how m... | Solution: (1) In essence, it is the Bernoulli-Euler letter misplacement problem for $n=4$, the number of different arrangements is
$$
u_{1}=9 \text { (ways). }
$$
(2) The different arrangements of even numbers in odd positions and odd numbers in even positions are 4! ways each. By the multiplication principle, the numb... | 225 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 For any $a>1, b>1$,
we have $\frac{a^{2}}{b-1}+\frac{b^{2}}{a-1} \geqslant 8$.
(1992, 26th Commonwealth Mathematics Olympiad (10th grade)) | To prove for $a:=1+t_{1}, b:=1+t_{2}$, where $t_{1}, t_{2}>0$. The inequality to be proven becomes
$$
\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)}{\... | 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 the... | 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 ... | 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... | 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 colo... | 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
$$
\left\{\begin{array}{l}
1+2+\cdots+k=\frac{k(k+1)}{2}<1997, \\
1+2+\cdots+(k+1)=\frac{(k+1)(k+2)}{2} \geqslant 1997 .
\end{array}\right.
$$
... | 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 .
$$
\begin{array}{l}
\text { Let } \omega=\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7} \text {. Then } \omega^{7}=1 . \\
\omega^{k}=\cos \frac{2 k \pi}{7}+i \sin \frac{2 k \pi}{7}, \\
\omega^{-k}=\cos \frac{2 k \pi}{7}-i \sin \frac{2 k \pi}{7}, \\
\therefore \omega^{k}+\omega^{-k}=2 \cos \frac{2 k \pi}{7} . \\
\tex... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$\begin{array}{l}\text { 6. Given } P(x)=x^{5}+a_{1} x^{4}+a_{2} x^{3}+ \\ a_{3} x^{2}+a_{4} x+a_{5} \text {, and when } k=1,2,3,4 \text {, } P(k) \\ =k \cdot 1997 \text {. Then } P(10)-P(-5)=\end{array}$ | 6.75315 .
Let $Q(x)=P(x)-1997 x$. Then for $k=1,2,3,4$, $Q(k)=P(k)-1997 k=0$. Hence, $1,2,3,4$ are roots of $Q(x)=0$. Since $Q(x)$ is a fifth-degree polynomial, we can assume
$$
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)+... | 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 dis... | $$
\begin{array}{l}
a=5, b \\
=4, c=3 . \text { Let } \angle X F P_{1} \\
=\alpha \cdot, \angle P_{1} F P_{2}= \\
\angle P_{2} F P_{3}=\cdots= \\
\angle P_{16} F P_{1}=\frac{\pi}{8} . \\
F M=d:-P_{i} F \cos \left[(i-1) \frac{\pi}{8}+\alpha\right] \\
\quad=\frac{a^{2}}{c}-c=\frac{16}{3} .
\end{array}
$$
By the definiti... | 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
$$
\begin{array}{l}
x_{n}=10000^{x_{n-1}}=10^{4 x_{n-1}} . \\
\therefore a_{1}<x_{1997}=10^{4 x_{1996}}, \\
a_{2} \leqslant 9 \times 4 x_{1996}<100 x_{1996} \\
=100 \cdot 10^{4 x 1995}=10^{4 x_{1995}+2}, \... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $m=\sqrt{5}+1$. Then the integer part of $m+\frac{1}{m}$ is $\qquad$ . | \begin{array}{l}\text { 1. } m=\sqrt{5}+1, \frac{1}{m}=\frac{1}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{4}, \\ \therefore m+\frac{1}{m}=\frac{5}{4} \sqrt{5}+\frac{3}{4},\left[m+\frac{1}{m}\right]=3 .\end{array} | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $x^{2}-x-1=0$. Then, the value of the algebraic expression $x^{3}$ $-2 x+1$ is $\qquad$ . | $\begin{array}{l} \text { 3. } x^{3}-2 x+1 \\ \quad=\left(x^{3}-x^{2}-x\right)+\left(x^{2}-x-1\right)+2 \\ =x\left(x^{2}-x-1\right)+\left(x^{2}-x-1\right)+2=2\end{array}$
The translation is as follows:
$\begin{array}{l} \text { 3. } x^{3}-2 x+1 \\ \quad=\left(x^{3}-x^{2}-x\right)+\left(x^{2}-x-1\right)+2 \\ =x\left(x... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $m$ and $n$ are rational numbers, and the equation $x^{2}+$ $m x+n=0$ has a root $\sqrt{5}-2$. Then the value of $m+n$ is $\qquad$ . | 4. Since $m, n$ are rational, the other root is $-\sqrt{5}-2$, thus by Vieta's formulas,
$$
\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}
$$
When $a+\frac{b-1}{2}=0, b-1=0$,
i.e., $a=0, b=1$, the equality in the abo... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Every book has an international book number:
ABCDEFGHIJ
where $A B C D E F G H I$ are composed of nine digits, and $J$ is the check digit.
$$
\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}
$$
$r$ is the remainder when $S$ is divided by 11. If $r$ is not 0 or 1, then $J... | $$
\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}
$$
$\therefore S$ the remainder when divided by 11 is equal to the remainder when $7 y+1$ is divided by 11.
From the check digit, we know that the remainder when... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given that the line $y=-2 x+3$ intersects the parabola $y=$ $x^{2}$ at points $A$ and $B$, and $O$ is the origin. Then, the area of $\triangle O A B$ is $\qquad$ . | 7.6 .
As shown in the figure, the line $y=$
$-2 x+3$ intersects the parabola $y$ $=x^{2}$ at points $A(1,1), B(-3,9)$.
Construct $A A_{1}, B B_{1}$ perpendicular to the $x$-axis, with feet of the perpendiculars at
$$
\begin{array}{l}
A_{1} 、 B_{1} . \\
\therefore S_{\triangle O A B}=S_{\text {trapezoid } A A_{1} B_{1}... | 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... | 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 co... | 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. | $$
\begin{array}{l}
\text { Original expression }=x(x+1)(x+2)(x+3)+1 . \\
\because x(x+1)(x+2)(x+3)+1 \\
=\left(x^{2}+3 x\right)\left(x^{2}+3 x+2\right)+1 \\
=\left[\left(x^{2}+3 x+1\right)-1\right]\left[\left(x^{2}+3 x+1\right)+1\right]+1 \\
=\left(x^{2}+3 x+1\right)^{2} \\
\because 1997 \times 1998 \times 1999 \times... | 3994001 | Number Theory | proof | Yes | Yes | cn_contest | false |
5. Let $S=\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+$ $\sqrt{1+\frac{1}{3^{2}}+\frac{1}{4^{2}}}+\cdots+\sqrt{1+\frac{1}{1997^{2}}+\frac{1}{1998^{2}}}$. Then the integer closest to $S$ is ( ).
(A) 1997
(B) 1908
(C) 1009
(D) 2000 | 5. (B).
When $n$ is an integer, we have
$$
\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... | 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
$$
\begin{array}{l}
\sqrt{\frac{3 n+6 x}{2} \cdot \frac{n+4 x}{2} \cdot \frac{n+2 x}{2} \cdot \frac{n}{2}}=\frac{n(n+2 x)}{2} . \\
\therefore \frac{3 n(n+2 x)^{2}(n+4 ... | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false |
1. Given $a, b$ are positive integers, and satisfy $\frac{a+b}{a^{2}}=\frac{4}{40}$. Then the value of $a+b$ is $\qquad$ | Let $a+b=4k$ ($k$ is a positive integer), then $a^{2}+ab+b^{2}:=49k$, i.e., $(a+b)^{2}-ab=49k$.
$$
\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
$$
By $\Delta=16k^{2}-4(16k^{2}-49k) \geqslant 0$, we get $... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Among the lines connecting the eight vertices of a square, the number of pairs of skew lines is ( ).
(A) 114
(B) 138
(C) 174
(D) 228 | 3. (C).
The vertex connections include 12 edges, 12 face diagonals, and 4 body diagonals. Skew lines can be categorized as follows:
(1) Edge to edge $\frac{12 \times 4}{2}=24$ pairs.
(2) Edge to face diagonal $12 \times 6=72$ pairs.
(3) Edge to body diagonal $12 \times 2=24$ pairs.
(4) Face diagonal to face diagonal $... | 174 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
Example 2 Let $[x]$ denote the greatest integer not greater than the real number $x$. The number of real roots of the equation $\lg ^{2} x-[\lg x]-2=0$ is $\qquad$
(1995, National High School Mathematics Competition) | Analysis: The difficulty of this problem lies in the uncertainty of $[\lg x]$. Since $[\lg x] \leqslant \lg x$, the original problem is first transformed into finding the values of $x$ that satisfy the inequality $\lg ^{2} x-\lg x-2 \leqslant 0$.
Solving this, we get $-1 \leqslant \lg x \leqslant 2$.
Therefore, the num... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Seven, (Full marks 12 points) A
home appliance manufacturing company, based on market research and analysis, has decided to adjust its production plan. They plan to produce a total of 360 units of air conditioners, color TVs, and refrigerators per week (calculated at 120 labor hours per week), with at least 60 refrige... | $$
\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) \\
... | 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 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 d... | 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
$$
\begin{aligned}
A & =\underbrace{1010101 \cdots 01}_{2 n} \\
& =10^{2 n}+10^{2 n} 2+\cdots+10^{2}+1 . \\
& =\frac{10^{2 n+2}-1}{10^{2}-1}=\frac{\left(10^{n+1}+1\right)\left(10^{n+1}-1\right)}{99} .
\end{al... | 101 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let $f(x)=\frac{4^{x}}{4^{x}+2}$. Then, $\sum_{k=1}^{1000} f\left(\frac{k}{1001}\right)$ equals $\qquad$ - | Sampling: From the characteristics of the variable's values, we can deduce that $f(x)$ satisfies the following symmetric property:
If $x+y=1$, then $f(x)+f(y)=1$. Therefore, $f\left(\frac{k}{1001}\right)+f\left(\frac{1001-k}{1001}\right)=1$,
$(k=1,2, \cdots, 1000)$
$$
\therefore \sum_{k=1}^{1000} f\left(\frac{k}{1001}\... | 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... | 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... | 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... | 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 ... | (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 ge... | 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.
$$
\begin{array}{l}
\text { When }(1+i)^{2}=2 i, (1-i)^{2}=-2 i \text { and } X \in \mathbb{Z}, \\
i^{4 k+1}=i, i^{4 k+2}=-1, i^{4 k+3}=-i, i^{4 k}=1. \\
\text { Therefore, }(1+i)^{2000}=(2 i)^{1000}=2^{1000}, \\
(1-i)^{1998}=(-2 i)^{999}=2^{999} \cdot i, \\
(1+i)^{1996}=(2 i)^{998}=-2^{998}, \\
(1-i)^{1994}=(-2 i... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-3 x-\right.$ $3)^{2001}$ is expanded and like terms are combined. The sum of the coefficients of the odd powers of $x$ in the resulting expression is $\qquad$. | 4. -1 .
$$
\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 | 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 \geqsla... | 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 po... | Let the arithmetic sequence be $a_{1}, a_{2}, \cdots, a_{n}$, with common difference $d=4$. Then
$$
\begin{array}{l}
a_{1}^{2}+a_{2}+\cdots+a_{n} \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+\frac{2 a_{1}+4 n}{2}(n-1) \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+(n-1) a_{1}+\left(2 n^{2}-2 n-100\right) \leqslant 0 .
\end{a... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Full marks 50 points) There are 21 points on a circle. Prove: Among all the arcs formed with these points as endpoints, there are no fewer than 100 arcs that do not exceed $120^{\circ}$. | Three points on a circle divide the circle into three arcs, at least one of which is greater than $120^{\circ}$. Connecting the endpoints of the arc that does not exceed $120^{\circ}$ with a chord, we can see that among any three points on the circle, at least two points are connected by a chord (referred to as an "edg... | 100 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Initial 65. Given a real-coefficient polynomial function $y=a x^{2}+b x+c$, for any $|x| \leqslant 1$, it is known that $|y| \leqslant 1$. Try to find the maximum value of $|a|+|b|+|c|$. | Proof: First, we prove the following auxiliary proposition.
Proposition: Let real numbers $A, B$ satisfy $|A| \leqslant 2, |B| \leqslant 2$. Then $|A+B| + |A-B| \leqslant 4$.
In fact, without loss of generality, let $|A| \geqslant |B|$.
From $A^2 \geqslant B^2$, we have $(A+B)(A-B) \geqslant 0$,
thus $|A+B| + |A-B| = |... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 If any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>$ 0 , to make $\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant$ $k \log _{\frac{x_{0}}{x_{3}}} 1993$ always hold, then the maximum value of $k$ is $\qquad$
(1993, National High School Mathematics Competi... | Analysis: The inequality can be transformed into
$$
\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 {m... | 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
$$
\begin{array}{l}
f\left(\frac{1}{1}\right)+f\left(\frac{2}{1}\right)+\cdots+f\left(\frac{100}{1}\right)+f\left(\frac{1}{2}\right) \\
+f\left(\frac{2}{2}\right)+\cdots+f\left(\frac{100}{2}\right)+\cdots f\left(\frac{1}{100}\right)+ \\
f\left(\frac{2}{100}\rig... | List the square number table as shown in the figure, and add the two number tables after rotating around the main diagonal by $180^{\circ}$,
$$
\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\l... | 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:
$$
\sum_{i=1}^{10} x_{i} f_{2520}\left(x_{i}\right)=220 \text {.... | Proof: Since $f$ is a one-to-one mapping from $D$ to $D$, by Proposition 1, for any $i \in D$, there exists $m_{i}\left(1 \leqslant m_{i} \leqslant 10\right)$, such that $f_{m_{i}}(i)=i$.
Considering $2520=2^{3} \cdot 3^{2} \cdot 5 \cdot 7$, which is the least common multiple of $1,2, \cdots, 10$, it follows from the ... | 220 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Example 3 Let the set $A=\{1,2, \cdots, 10\},$ and the mapping $f$ from $A$ to $A$ satisfies the following two conditions:
(1) For any $x \in A, f_{30}(x)=x$;
(2) For each $k \in \mathbb{Z}^{+}, 1 \leqslant k \leqslant 29$, there exists at least one $a \in A$ such that $f_{k}(a) \neq a$.
Find the total number of such m... | Solution: Notice that $10=5+3+2,30=5 \times 3 \times 2$. This suggests dividing $A$ into three disjoint subsets
$$
A=\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\} \cup\left\{b_{1}, b_{2}, b_{3}\right\}
$$
$\cup\left\{c_{1}, c_{2}\right\}$.
Since $f$ satisfies conditions (1) and (2), $f$ is a one-to-one mapping from ... | 120960 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let the set $A=\{1,2,3,4,5,6\}$, and the mapping $f: A \rightarrow A$, such that its third composite mapping $f \cdot f \cdot f$ is the identity mapping. How many such $f$ are there?
(1996. Japan Mathematical Olympiad Preliminary) | Solution: Since the threefold composite mapping on set A is the identity mapping, it follows from Proposition 1 and Proposition 2 that there are three types of mappings $f$ that meet the conditions:
(1) $f$ is the identity mapping;
(2) There exists a three-element mapping cycle $a \rightarrow b \rightarrow c \rightarro... | 81 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, let integers $a, b, c$ satisfy $1 \leqslant a<b<c \leqslant 5, a^{3}$, $b^{3}, c^{3}$ have unit digits $x, y, z$ respectively. When $(x-y)(y-z)(z-x)$ is the smallest, find the maximum value of the product $a b c$.
Let integers $a, b, c$ satisfy $1 \leqslant a<b<c \leqslant 5, a^{3}$, $b^{3}, c^{3}$ have unit di... | $$
\text { Three, } \because 1^{3}=1,2^{3}=8,3^{3}=27,4^{3}=64,5^{3}=125 \text {. }
$$
$\therefore(x, y, z)$ has the following 10 possibilities:
(1) $(1,8,7) ;(2)(1,8,4) ;(3)(1,8,5)$;
(4) $(1,7,4) ;(5)(1,7,5) ;(6)(1,4,5)$;
(7) $(8,7,4) ;(8)(8,7,5) ;(9)(8,4,5)$;
$(10)(7,4,5)$.
Then the values of $(x-y)(y-z)(z-x)$ are
$$... | 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 ... | 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... | 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 t... | 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 $x=6$ (... | 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 $D$, wh... | 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... | 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 st... | 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... | 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... | 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, t... | 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 rew... | 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 ... | 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 $... | 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 ... | 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... | 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 comp... | 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... | 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. $x^{3}-y^{3}-z^{3}=3 x y z$,
$$
\begin{array}{l}
\there... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Full marks 25 points) On the blackboard, all natural numbers from 1 to 1997 are written. Students $A$ and $B$ take turns to perform the following operations: Student $A$ subtracts the same natural number from each number on the blackboard (the number subtracted can be different in different operations); Student... | Three, because after each operation by student $A$ and student $B$, the number of numbers written on the blackboard decreases by 1. Since students $A$ and $B$ take turns operating, when $B$ completes the last operation, only one number remains on the blackboard, and both have performed 1996 operations. Let $d_{k} (k=1,... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $M$ be a subset of the set $S=\{1,2,3, \cdots, 1998\}$, and each natural number (element) in $M$ contains exactly one 0. Then the maximum number of elements in the set $M$ is ( ).
(A) 324
(B) 243
(C) 495
(D) 414 | 6. (D).
Divide the set $S$ into four subsets
$$
\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}
$$
Let $M_{i}^{r}$ represent the set formed by all elements in $S_{i}(i=1,2,3,4)$ that contain exactly one di... | 414 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
3. From the center of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, two perpendicular chords $A C$ and $B D$ are drawn. Connecting $A, B, C, D$ in sequence forms a quadrilateral. Then, the maximum value of the area $S$ of quadrilateral $A B C D$ is | 3. 12.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Given $O A \perp O B$, we have
$$
\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{\l... | 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, meani... | 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... | 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
$$
\left\{\begin{array}{l}
\left|z_{1}\right| \leqslant 1, \\
\left|z_{2}\right| \leqslant 1, \\
\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \leqslant\left|z_{1}-z_{2}\right|, \\
\left|2 z_{4}-\left(z_{1}+z_{2}\right)\right|... | $$
\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}... | 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 onl... | 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 $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}$.
$$
\because x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} \text {. }
$$
$x_{4} x_{5}$,
$$
\begin{aligned}
\therefore 1= & \frac{1}{x_{2} x_{3} x_{4} x_{5}}+\frac{1}{x_{1} x_{3} x_{4} x_{5}} \\
& +\frac{1... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 14 Try to find the maximum value of the following expression:
$$
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_... | 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, ... | 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 ... | 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... | 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, $70=S_{30}=b_{1}+$
$$
\begin{array}{l}
b_{2}+b_{3}=b_{1}\... | 150 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. Let the complex number $z=\cos \theta+i \sin \theta\left(0^{\circ} \leqslant \theta \leqslant\right.$ $\left.180^{\circ}\right)$, and the complex numbers $z, (1+i)z, 2\bar{z}$ correspond to three points $P, Q, R$ in the complex plane. When $P, Q, R$ are not collinear, the fourth vertex of the parallelogram formed by... | 2.3.
Let the complex number $w$ correspond to point $S$. Since $Q P R S$ is a parallelogram, we have
$$
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{a... | 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 a... | 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
$$
\begin{aligned}
& a_{1}^{2}+a_{2}+a_{3}+\cdots+a_{n} \leqslant 100 \\
\Leftrightarrow & a_{1}^{2}+\frac{\left(a_{1}+4\right)+\left[a_{1}+4(n-1)\right]}{2} \cdot(n-1) \\
& \leqslant 100 \\
\Leftrightarrow & a_{1}^{2... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 When $x$ varies, the minimum value of the fraction $\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$ is $\qquad$
(1993, National Junior High School Competition) | Let $y=\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$. Then,
$$
\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 {. }
$$
Thus, $4 \leqslant y \leqslant 6$.
When $y=4$, $x=1$.
Therefore, when $x=1$, the minimum value... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, 50 points 5) For positive integers $a, n$, define $F_{a}(a)=q+r$, where $q, r$ are non-negative integers, $a=$ $q n+r$, and $0 \leqslant r<n$. Find the largest positive integer $A$ such that there exist positive integers $n_{1}, n_{2}, n_{3}, n_{4}, n_{5}, n_{6}$, for any positive integer $a \leqslant A$, we hav... | Three, the maximum positive integer $B$ that satisfies the condition "there exist positive integers $n_{1}, n_{2}, \ldots, n_{k}$, such that for any positive integer $a \leqslant B$,
$$
F_{n_{k}}\left(F_{\pi_{k-1}}\left(\cdots\left(F_{n_{1}}(a)\right) \cdots\right)\right) \doteq 1^{\prime}
$$
is denoted as $x_{k}$. Cle... | 53590 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $\sqrt{x}+\sqrt{y}=35, \sqrt[3]{x}+\sqrt[3]{y}=13$. Then $x+y=$ $\qquad$ | 3.793 .
$$
\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 ... | 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 ... | 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, ... | 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 m... | 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... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.