problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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values | synthetic bool 1
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|---|---|---|---|---|---|---|---|---|
$$
\begin{array}{l}
\text { 2 For a row consisting of } n \text { A's, } n \text { B's, and } n \text { C's, define a new row below it (one letter shorter) such that if the two letters above it are different, write the third letter, and if they are the same, write that letter. Repeat this operation on the newly obtaine... | Solution: There is only one solution: $n=1$.
Method 1: When $n=1$, there are only the following 6 cases:
\begin{tabular}{cccccc}
$\mathrm{ABC}$ & $\mathrm{ACB}$ & $\mathrm{BAC}$ & $\mathrm{BCA}$ & $\mathrm{CAB}$ & $\mathrm{CBA}$ \\
$\mathrm{CA}$ & $\mathrm{BA}$ & $\mathrm{CB}$ & $\mathrm{AB}$ & $\mathrm{BC}$ & $\mathr... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
13.25 Given 4 coins, one of which may be counterfeit, each genuine coin weighs 10 grams, and the counterfeit coin weighs 9 grams. Now there is a balance scale with one pan that can measure the total weight of the objects on the pan. To identify whether each coin is genuine or counterfeit, what is the minimum number of ... | [Solution] Let the weights of the 4 coins be denoted as $a, b, c, d$. Weigh them 3 times: the first time, weigh $S_{1}=a+b+c$, the second time, weigh $S_{2}=a+b+d$, and the third time, weigh $S_{3}=a+c+d$. Thus, we have
$$
S=S_{1}+S_{2}+S_{3}=3 a+2 b+2 c+2 d .
$$
From the parity of $S$, we can determine whether $a$ is... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. Color each cell of a 2018 by 4 grid either red or blue, such that the number of red cells equals the number of blue cells in each row, and the number of red cells equals the number of blue cells in each column. Let $m$ be the total number of coloring methods that satisfy the above requirements. Find the remainder wh... | 7. A coloring method that satisfies the conditions is called a "pattern", and all patterns form a set $M$.
Use $(i, j)$ to denote the cell in the $i$-th row and $j$-th column, and consider the row number $i$ modulo 2018, i.e., the 2018-th row can be seen as the 0-th row. For a pattern $A$, an operation $T$ can be perf... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
There are $n$ people, and it is known that any two of them make at most one phone call. The total number of phone calls made among any $n-2$ of them is equal, and it is $3^{m}$ times, where $m$ is a natural number. Find all possible values of $n$.
The text above is translated into English, preserving the original text... | Obviously, $n \geqslant 5$. Let the $n$ people be several points $A_{1}, A_{2}, \cdots, A_{n}$. If $A_{i}, A_{j}$ make a phone call, then connect $\left(A_{i}, A_{j}\right)$. Therefore, among these $n$ points, there must be an edge, let's say $\left(A_{1}, A_{2}\right)$.
Suppose $A_{1}$ and $A_{3}$ are not connected b... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (3 points) Calculate: $\left[2 \frac{3}{5}-\left(8.5-2 \frac{2}{3}\right) \div 3.5\right] \times 7 \frac{1}{2}=$ | 1. (3 points) Calculate: $\left[2 \frac{3}{5}-\left(8.5-2 \frac{2}{3}\right) \div 3.5\right] \times 7 \frac{1}{2}=7$.
【Solution】Solution: $\left[2 \frac{3}{5}-\left(8.5-2 \frac{2}{3}\right) \div 3.5\right] \times 7 \frac{1}{2}=$
$$
\begin{array}{l}
=\left[\frac{13}{5}-\left(\frac{17}{2}-\frac{8}{3}\right) \div \frac{7}... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 2 $x$ is a positive real number, find the minimum value of the function $y=x^{2}+x+\frac{3}{x}$. | By substituting some simple values of $x$ such as $\frac{1}{2}, 1, 3$, one can find that the given function is not monotonic, which often makes it difficult to approach at first glance. However, with some problem-solving experience, one would realize that such problems can be easily solved by appropriately transforming... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8-38 If $(8 y-1)^{2}+|x-16 y|=0$, then the value of $\log _{2} y^{x}$ is
(A) -6.
(B) 0.
(C) $\frac{1}{64}$.
(D) $\frac{1}{8}$.
(Shanghai Junior High School Mathematics Competition, 1983) | [Solution] From the given conditions and $(8 y-1)^{2} \geqslant 0,|x-16 y| \geqslant 0$, we have $\left\{\begin{array}{l}8 y-1=0, \\ x-16 y=0 .\end{array}\right.$ Solving these, we get $\left\{\begin{array}{l}x=2, \\ y=\frac{1}{8} .\end{array}\right.$ $\therefore \quad \log _{2} y^{x}=2 \log _{2} \frac{1}{8}=-6$. There... | -6 | Algebra | MCQ | Yes | Yes | olympiads | false |
6 If the inequality $a \sin ^{2} x+\cos x \geqslant a^{2}-1$ holds for any $x \in \mathbf{R}$, then the range of real number $a$ is $\qquad$. | 6 $a=0$ Hint: Let $\cos x=-1, \sin x=0, a \sin ^{2} x+\cos x \geqslant$ $a^{2}-1$, we get $a^{2} \leqslant 0$. And when $a=0$, the original inequality obviously always holds. Therefore, $a=0$. | 0 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
39. Balabala Supermarket has received many Magic Eggs, which are packaged in large and small baskets. Each large basket holds 150 Magic Eggs, and each small basket holds 80 Magic Eggs. If all the Magic Eggs are sold at the set price, the total revenue would be 1366.40 yuan. If the price of each Magic Egg is increased b... | $12$ | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Example 3.6.6] Find all positive integers $n>1$, such that $\frac{2^{n}+1}{n^{2}}$ is an integer. | 3 is the only solution.
We will prove it in three steps.
Step 1: Prove that $n$ is a multiple of 3, thus $n=3^{m} \cdot c$.
Step 2: Prove that $m=1$, i.e., $n=3 c$.
Step 3: Prove that $c=1$, i.e., $n=3$.
Below is the proof step by step.
Step 1: Prove that $3 \mid n$.
Let $p$ be the smallest prime factor of $n$. Since $... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
15.9 If $\{a\} \subset(A \cup B) \subset\{a, b, c, d\}$, and $a \in B, A \cap B=\varnothing$, then the number of sets $A$ that satisfy the above conditions is
(A) 5.
(B) 6.
(C) 7.
(D) 8.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991) | [Solution] $\because a \in B$, and $A \cap B=\varnothing$,
$\therefore \quad a \notin A$.
Also, $\because \quad\{a\} \subset(A \cup B) \subset\{a, b, c, d\}$,
$\therefore A$ could be $\varnothing ;\{b\} ;\{c\} ;\{d\} ;\{b, c\} ;\{b, d\} ;\{c, d\} ;\{b, c, d\}$, a total of 8 possibilities.
Therefore, the answer is $(D)$... | 8 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
5. Find all positive integers $x, y$ such that $\frac{x^{3}+y^{3}-x^{2} y^{2}}{(x+y)^{2}}$ is a non-negative integer. | 5. (1) When $x=y$, the original expression $=\frac{2 x-x^{2}}{4} \geqslant 0 \Rightarrow 0 \leqslant x \leqslant 2 \Rightarrow x=y=2$.
(2) By symmetry, assume $x \geqslant y$, when $x>y$, according to the problem, the original expression $=\frac{x^{2}-x y+y^{2}}{x+y}-\frac{x^{2} y^{2}}{(x+y)^{2}}=x+y-\frac{3 x y}{x+y}-... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Write 2004 numbers on the blackboard: $1,2, \cdots, 2004$, in each step, erase some numbers from the blackboard, and rewrite the remainder of their sum divided by 167. After several steps, two numbers remain on the blackboard, one of which is 999. What is the second remaining number? | 1. Let the second number remaining on the board be $x$. Since $999>167$, 999 is not a number that is rewritten on the board after the operation, thus $0 \leqslant x \leqslant 166$.
Notice that, each operation does not change the remainder of the sum of all numbers on the board when divided by 167.
$$
1+2+\cdots+2004=\f... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
33.68 If the octal representation of a perfect square is $a b 3 c$ (i.e., $n^{2}=$ $\left.(a b 3 c)_{8}=8^{3} a+8^{2} b+8 \cdot 3+c\right)$, where $a \neq 0$, then $c$ is
(A) 0.
(B) 1.
(C) 3.
(D) 4.
(E) cannot be uniquely determined.
(33rd American High School Mathematics Examination, 1982) | [Solution 1] If $n^{2}=(a b 3 c)_{8}$, let $n=(d e)_{8}$, then
$$
n^{2}=(8 d+e)^{2}=64 d^{2}+8(2 d e)+e^{2} \text {. }
$$
Therefore, 3 is the sum of the second digit (counting from the right, same below) of $e^{2}$ in octal and the first digit of $2 d e$ (the sum in octal sense). The latter is an even number, and 3 is... | 1 | Number Theory | MCQ | Yes | Yes | olympiads | false |
5.103 Suppose there are 100 mutually hostile countries on Mars. To maintain peace, it is decided to form several alliances, with each alliance including at most 50 countries, and any two countries must belong to at least one alliance. Try to answer the following questions:
(1) What is the minimum number of alliances ne... | [Solution] Since each alliance has at most 50 countries, in this alliance, each country has at most 49 allies. However, after forming alliances, each country has 99 allies, so each country must join at least 3 alliances. Therefore, the number of alliances formed by 100 countries is no less than $3 \times 100 \div 50 = ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(5) Let $A=\{x \mid 1 \leqslant x \leqslant 9, x \in \mathbf{Z}\}, B=\{(a, b) \mid a, b \in A\}$, define the mapping $f:(a, b) \rightarrow a b-a-b$ from $B$ to $\mathbf{Z}$, then the number of ordered pairs $(a, b)$ that satisfy $(a, b) \xrightarrow{f} 11$ is ( ).
(A) 4 pairs
(B) 6 pairs
(C) 8 pairs
(D) 12 pairs | 5 Because
$$
f(a, b)=a b-a-b=11,
$$
that is
$$
(a-1)(b-1)=2 \times 6=3 \times 4=1 \times 12 .
$$
then
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ a - 1 = 2 , } \\
{ b - 1 = 6 , }
\end{array} \left\{\begin{array} { l }
{ a - 1 = 6 , } \\
{ b - 1 = 2 , }
\end{array} \left\{\begin{array}{l}
a-1=3, \\
b-1=4,
\end{... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
1. The number of points on the circle $(x-3)^{2}+(y-3)^{2}=9$ that are 1 unit away from the line $3 x+4 y-11=0$ is
A. 1
B. 2
C. 3
D. 4 | 1. C Hint: It is evident that the circle $(x-3)^{2}+(y-3)^{2}=9$ intersects with the line $3 x+4 y-11=0$. The distance from the center of the circle to the line $d=2$, and the radius of the circle $r=3$. Since the radius $r=$ distance from the center to the line +1. Therefore, there are 3 such points. | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
Example 9 (2007 High School Mathematics League Additional Test Question) As shown in the figure, in a $7 \times 8$ rectangular chessboard, a chess piece is placed at the center of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be conne... | Solution: At least 11 chess pieces must be taken out to possibly meet the requirement. The reason is as follows: If a square is in the $i$-th row and the $j$-th column, then this square is denoted as $(i, j)$.
The first step is to prove that if any 10 chess pieces are taken, then among the remaining pieces, there must... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the country of Ouguo, everyone hates odd numbers so much that they don't even want to see any odd number digits. Therefore, they use marks to replace odd number digits when communicating. For example: when people in Ouguo write “$3 \times 4=12$”, they write it as “ふ $\times 4=ふ 2$”.
Convert the subtraction equation ... | $6$ | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Let the first term and common difference of an arithmetic sequence be non-negative integers, the number of terms be no less than 3, and the sum of all terms be $97^{2}$. Then the number of such sequences is:
(A) 2;
(B) 3;
(C) 4;
(D) 5. | 3. (C)
Let the first term of the arithmetic sequence be $a$, and the common difference be $d$. Then, according to the problem, we have
$$
n a+\frac{n(n-1)}{2} d=97^{2},
$$
which can be rewritten as
$$
[2 a+(n-1) d] n=2 \times 97^{2} .
$$
Since $n$ is a natural number not less than 3, and 97 is a prime number, the va... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
37. There are two-digit numbers where the sum of the tens digit and the units digit is 8. There are $\qquad$ such two-digit numbers. | answer: 8 | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
28.2.3 In a rectangle $A B C D$ with an area of 1 (including the boundary), there are 5 points, no three of which are collinear. Find the minimum number of triangles, with these 5 points as vertices, that have an area not greater than $\frac{1}{4}$. | The proof of this problem requires the following common conclusion, which we will use as a lemma: The area of any triangle within a rectangle does not exceed half the area of the rectangle.
In rectangle \(ABCD\), if a triangle formed by any three points has an area not greater than \(\frac{1}{4}\), it is called a good... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
16. (15 points) Observe the following operations:
If $\overline{\mathrm{abc}}$ is a three-digit number, since $\overline{\mathrm{abc}}=100 a+10 b+c=99 a+9 b+(a+b+c)$,
Therefore, if $a+b+c$ is divisible by 9, $\overline{\mathrm{abc}}$ is divisible by 9.
This conclusion can be extended to any number of digits.
Using the ... | 【Solution】Solution: (1) $2011 \times 2=4022$;
$$
4022 \div 9=446 \cdots 8 \text {, }
$$
Therefore, when $N$ is divided by 9, the remainder is 8;
(2) The sum of the digits of the natural number $N$ is $7n$; since $n \div 9$ has a remainder of 3, we can assume $n=9k+3$, then $7n=7(9k+3)=63k+21=(63k+18)+3=9(7k+2)+3$;
Th... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11. (This sub-question is worth 15 points) Divide the set $M=\{1,2, \cdots, 12\}$ of the first 12 positive integers into four three-element subsets $M_{1}, M_{2}, M_{3}, M_{4}$, such that in each three-element subset, one number is equal to the sum of the other two. Find the number of different ways to do this. | Let the four subsets be $M_{i}=\left(a_{i}, b_{i}, c_{i}\right), i=1,2,3,4$, where $a_{i}=b_{i}+c_{i}, b_{i}>c_{i}, i=1,2,3,4$. Given $a_{1}27$, we have $10 \leq a_{3} \leq 11$. If $a_{3}=10$, then from $a_{1}+a_{2}=17, a_{2}a_{1}+a_{2}=17$, we get $a_{2}=9, a_{1}=8$, which gives $\left(a_{1}, a_{2}, a_{3}, a_{4}\right... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$4 \cdot 43$ Find the smallest natural number $n$, such that in any two-coloring of $K_{n}$, there always exist two monochromatic triangles, each sharing exactly 1 vertex. | [Solution] Color the 8-vertex complete graph $K_{8}$ as shown in the figure, where blue lines are drawn, and all undrawn lines are red. It is easy to see that there are 8 blue triangles and no red triangles in the graph, and any two blue triangles either share a common edge or have no common vertex. Therefore, the smal... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$4 \cdot 74$ If $\alpha, \beta, \gamma$ are the roots of the equation $x^{3}-x-1=0$, find
$$
\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}
$$
the value. | ［Solution］From the given, we have
$$
x^{3}-x-1=(x-\alpha)(x-\beta)(x-\gamma) \text {, }
$$
Therefore,
$$
\left\{\begin{array}{l}
\alpha+\beta+\gamma=0, \\
\alpha \beta+\beta \gamma+\gamma \alpha=-1, \\
\alpha \beta \gamma=1 .
\end{array}\right.
$$
Thus, we have
$$
\begin{aligned}
& \frac{1+\alpha}{1-\alpha}+\frac{1+\... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For any integer $n(n \geqslant 2)$, the positive numbers $a$ and $b$ that satisfy $a^{n}=a+1, b^{2 n}=b+3 a$ have the size relationship
(A) $a>b>1$
(B) $b>a>1$
(C) $a>1.01$ | 5. (A).
First, $a>1, b>1$. Otherwise, if $0< a \leq 1$, then $a^{2n} \leq 1 < 3a$, which implies $a>1$, a contradiction; if $0< b \leq 1$, then $b^{2n} \leq 1 < 3a$, which implies $b>1$, a contradiction.
Therefore, both $a$ and $b$ are greater than 1.
On one hand,
$$
a^{2n} - b^{2n} = (a+1)^2 - (b+3a) = a^2 - a - b + ... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
$31 \cdot 19$ The number of integer solutions to the equation $x^{2}-y^{2}=12$ is
(A) 2 pairs.
(B) 4 pairs.
(C) 8 pairs.
(D) 12 pairs.
(3rd "Jinyun Cup" Junior High School Mathematics Invitational, 1986) | [Solution] From the given, we have $(x+y)(x-y)=1 \cdot 12=3 \cdot 4=2 \cdot 6$. Also, $x+y$ and $x-y$ have the same parity, so we have
$$
\left\{\begin{array}{l}
x + y = 2, \\
x - y = 6;
\end{array} \text { or } \quad \left\{\begin{array}{l}
x+y=-2, \\
x-y=-6 ;
\end{array}\right.\right.
$$
or $\left\{\begin{array}{l}x... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. Niu Niu runs 5 kilometers from Monday to Friday every day, and 8 kilometers on Saturday and Sunday, but she doesn't run on rainy days. May 1, 2020, was a Friday, and Niu Niu ran a total of 170 kilometers in May 2020. So, how many rainy days were there this month? | $3$ | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Find the smallest positive integer $t$ such that: If each cell of a $100 \times 100$ grid is colored with some color, and the number of cells of each color does not exceed 104, then there exists a $1 \times t$ or $t \times 1$ rectangle containing at least three different colors in its $t$ cells. | Divide the grid paper into 100 $10 \times 10$ squares, each of the 100 small squares in each square is dyed the same color, and different squares are dyed different colors. This coloring method satisfies the problem's conditions, and it is easy to see that any $1 \times 11$ or $11 \times 1$ rectangle contains at most t... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) Given $1 ※ 2=1+2=3, 2 ※ 3=2+3+4=9, 5 ※ 4=5+6+7+8=26$, if $a ※ 15=165$, then $a=$ $\qquad$ | 【Solution】Solution: According to the operation rules, we have
165 is the sum of an arithmetic sequence, 15 is the number of terms in the arithmetic sequence, the common difference is $1, a$ is the first term,
$$
\begin{array}{l}
a=\left[165-\frac{15 \times(15-1) \times 1}{2}\right] \div 15 \\
=[165-105] \div 15 \\
=60 ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. (10 points) In a math test at Sheep Village Elementary School's fourth grade, the test consists of 15 questions. If Xiaoxixi, Xiaobeibe, Xiaomeimei, and Xiaolan all answered 11, 12, 13, and 14 questions correctly, respectively, then the minimum number of questions they all answered correctly is $\qquad$ questions. | 【Answer】Solution: The four people got 4, 3, 2, and 1 questions wrong, respectively. The total number of questions they got wrong is: $4+3+2+1=10$ (questions). If the questions that each of the four people got wrong are all different, then the number of questions they all got right is the least. Therefore, the minimum n... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 5 Find the real solution of the equation $\sqrt{1+\sqrt{1+x}}=\sqrt[3]{x}$.
$(2015$, Finnish High School Mathematics Competition) | 【Analysis】Substitution can make the structure of the equation more elegant, facilitating subsequent observation and processing. Sometimes substitution can also reduce the degree of the equation.
Let $\sqrt{1+x}=t$. Then
$$
\sqrt{1+t}=\sqrt[3]{t^{2}-1} \Rightarrow(1+t)^{3}=\left(t^{2}-1\right)^{2} \text {. }
$$
Since $... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
16. As shown in Figure $5, A, B, \cdots, F$ represent six different numbers 1,2 , $\cdots, 6$. Each of the five lines passes through some of these points. By adding the numbers corresponding to the points on each line, five sums can be obtained, and the sum of these five numbers is 47. Then the number corresponding to ... | 16. E.
From the problem, we know that these five sums are $A+B+C$,
$$
A+E+F, \ C+D+E, \ B+D, \ B+F \text{.}
$$
Adding them together, we get
$$
\begin{array}{l}
2 A+3 B+2 C+2 D+2 E+2 F=47 \\
\Rightarrow 2(A+B+C+D+E+F)+B=47.
\end{array}
$$
Since $A$ to $F$ are six different numbers from 1 to 6, then
$$
A+B+C+D+E+F=1+2... | 5 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
4. Let $a \triangle b$ and $a \nabla b$ represent the minimum and maximum of the two numbers $a$ and $b$, respectively. For example, $3 \triangle 4=3, 3 \nabla 4=4$. For different natural numbers $x$, the number of possible values for $6 \triangle[4 \nabla(x \triangle 5)]$ is $\qquad$. | 【Analysis】First, classify $x$ and derive 2 results respectively, and solve accordingly.
【Solution】Solution: When $x \geqslant 5$, $x \triangle 5=5$, $4 \nabla(x \triangle 5)=5$, $6 \triangle[4 \nabla(x \triangle 5)]=5$; when $x<4$, $x \triangle 5=x$, $4 \nabla(x \triangle 5)=4$, $6 \triangle[4 \nabla(x \triangle 5)]=4$... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Example 7 If real numbers $x, y$ satisfy $x^{2}+y^{2}-2 x+4 y=0$, find the maximum value of $x-2 y$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | The original equation can be transformed into $(x-1)^{2}+(y+2)^{2}=(\sqrt{5})^{2}$, which represents a circle.
We can set $\left\{\begin{array}{l}x=1+\sqrt{5} \cos \theta, \\ y=-2+\sqrt{5} \sin \theta,\end{array} \quad \theta\right.$ as the parameter.
$$
x-2 y=5+\sqrt{5} \cos \theta-2 \sqrt{5} \sin \theta=5+5 \cos (\th... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$21 \cdot 3$ On a plane, there are two congruent equilateral triangles, one of which has a vertex at the center of the other. Then the ratio of the maximum area of the overlapping part to the minimum area is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 . | [Solution] As shown in Figure (2), the maximum overlapping area is $\frac{2}{9} S$ (refer to the measurement in Figure (1)), where $S$ is the area of the original equilateral triangle.
The minimum overlapping area, as shown in Figure (3), is $\frac{1}{9} S$.
Thus, the ratio of the two areas is $2: 1$.
Therefore, the an... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
1.5.13 $\star \star$ Find the largest real number $k$, such that for any positive real numbers $a, b, c$, we have
$$
\begin{aligned}
& \frac{(b-c)^{2}(b+c)}{a}+\frac{(c-a)^{2}(c+a)}{b}+\frac{(a-b)^{2}(a+b)}{c} \\
\geqslant & k\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)
\end{aligned}
$$ | Parse: In (1), let $a=b=1$, we get $2(1-c)^{2}(1+c) \geqslant k\left(1-c^{2}\right)$. Let $c \rightarrow 0^{+}$, we can get $k \leqslant 2$. Below, we prove that when $k=2$, inequality (1) holds, that is, to prove
$$
\sum \frac{(b-c)^{2}(b+c)}{a} \geqslant 2\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=\sum(b-c)^{2} .
$$
... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. Given $a, b>0, a \neq 1$, and $a^{b}=\log _{a} b$, then the value of $a^{a^{b}}-\log _{a} \log _{a} b^{a}$ is | 1. -1 Explanation: From $a^{b}=\log _{a} b$ we know: $b=a^{a^{b}}, b=\log _{a} \log _{a} b$,
$$
a^{a^{b}}-\log _{a} \log _{a} b^{a}=b-\log _{a}\left(a \log _{a} b\right)=b-\log _{a} a-\log _{a} \log _{a} b=b-1-b=-1
$$ | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.79 Determine all pairs of positive integers $(m, n)$ such that
$$
1+x^{n}+x^{2 n}+\cdots+x^{m n}
$$
can be divided by $1+x+x^{2}+\cdots+x^{m}$.
(6th United States of America Mathematical Olympiad, 1977) | [Solution] Since $1+x^{n}+x^{2 n}+\cdots+x^{m n}=\frac{x^{(m+1) n}-1}{x^{n}-1}$,
$$
1+x+x^{2}+\cdots+x^{m}=\frac{x^{m+1}-1}{x-1} .
$$
The polynomial $1+x^{n}+x^{2 n}+\cdots+x^{m n}$ can be divided by $1+x+x^{2}+\cdots+x^{m}$ if and only if
$$
\frac{x^{(m+1) n}-1}{x^{m+1}-1} \cdot \frac{x-1}{x^{n}-1}
$$
is an integer-... | +1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. If the set of real numbers
$$
A=\{2 x, 3 y\} \text { and } B=\{6, x y\}
$$
have exactly one common element, then the product of all elements in $A \cup B$ is $\qquad$ . | -1.0 .
Let the unique common element of sets $A$ and $B$ be denoted as $a$. If $a \neq 0$, then the other elements of sets $A$ and $B$ are both $\frac{6 x y}{a}$, which is a contradiction.
Therefore, the product of all elements in $A \cup B$ is 0. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Let real numbers $x, y$ be such that $x-y, x^{2}-y^{2}, x^{3}-y^{3}$ are all prime numbers, then the value of $x-y$ is | Let $x-y=p, x^{2}-y^{2}=q, x^{3}-y^{3}=r$ be prime numbers,
then $\left\{\begin{array}{l}x+y=\frac{q}{p}, \\ x-y=p\end{array} \Rightarrow\left\{\begin{array}{l}x=\frac{1}{2}\left(\frac{q}{p}+p\right), \\ y=\frac{1}{2}\left(\frac{q}{p}-p\right) .\end{array}\right.\right.$ Thus, $x^{3}-y^{3}=\frac{1}{8}\left(2 p^{3}+6 \c... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 4 Let $S=\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2011^{3}}$. Then the integer part of $4 S$ is ( ). ${ }^{[2]}$
(A) 4
(B) 5
(C) 6
(D) 7 | When $k=2,3, \cdots, 2011$,
$$
\frac{1}{k^{3}}<\frac{1}{k\left(k^{2}-1\right)}=\frac{1}{2}\left[\frac{1}{(k-1) k}-\frac{1}{k(k+1)}\right] \text {. }
$$
Then $1<S=1+\frac{1}{2^{3}}+\frac{1}{3^{3}}+\cdots+\frac{1}{2011^{3}}$
$$
<1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2011 \times 2012}\right)<\frac{5}{4} \text {. }
$$
... | 4 | Number Theory | MCQ | Yes | Yes | olympiads | false |
97 If $\cos A+\cos B+\cos C=0$, then the value of $\frac{\cos 3 A+\cos 3 B+\cos 3 C}{\cos A \cos B \cos C}$ is $\qquad$ | 97 12. Since
$$
\cos A+\cos B+\cos C=0,
$$
then
$$
\cos ^{3} A+\cos ^{3} B+\cos ^{3} C=3 \cos A \cos B \cos C .
$$
Therefore, we have
$$
\begin{aligned}
& \frac{\cos 3 A+\cos 3 B+\cos 3 C}{\cos A \cos B \cos C} \\
= & \frac{4\left(\cos ^{3} A+\cos ^{3} B+\cos ^{3} C\right)-3(\cos A+\cos B+\cos C)}{\cos A \cos B \cos ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Question 234: In the interval $[1,1000]$, take $\mathrm{n}$ different numbers $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{\mathrm{n}}$. There always exist two numbers $\mathrm{a}_{\mathrm{i}},$ $a_{j}$, such that $0<\left|a_{i}-a_{j}\right|<1+3 \sqrt[3]{a_{i} a_{j}}$. Find the minimum possible value of $n$. | Question 234, Solution: When $n \geq 11$, since $\sqrt[3]{a_{1}}, \sqrt[3]{a_{2}}, \ldots, \sqrt[3]{a_{n}}$ all fall within the interval $[1,10]$, according to the pigeonhole principle, there must exist two numbers $\sqrt[3]{a_{i}}, \sqrt[3]{a_{j}}$ such that $0<\left|\sqrt[3]{a_{i}}-\sqrt[3]{a_{j}}\right|<1$. Therefor... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (10 points) Two strips of paper, the longer one is $23 \mathrm{~cm}$, and the shorter one is $15 \mathrm{~cm}$. After cutting off the same length from both strips, the remaining longer strip should be at least twice as long as the remaining shorter strip. The minimum length that can be cut off is ( ) cm.
A. 6
B. 7
C... | 【Analysis】Let the length cut off be $x$ cm, then the longer remaining piece is $(23-x)$ cm, and the shorter remaining piece is $(15-x)$ cm. According to the requirement that "the longer of the two remaining strips must be at least twice the length of the shorter strip," we can set up the inequality: $23-x \geqslant 2(1... | 7 | Inequalities | MCQ | Yes | Yes | olympiads | false |
$33 \cdot 59$ number 10! (in decimal) when written in base 12, ends with exactly $k$ zeros. Then the value of $k$ is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(E) 5.
(21st American High School Mathematics Examination, 1970) | [Solution] In decimal,
$$
\begin{array}{l}
10!=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10=12^{4} \cdot 5^{2} \cdot 7 \text {. } \\
\text { and } \quad 5^{2} \cdot 7=175=1 \cdot 12^{2}+2 \cdot 12+7 \text {, } \\
\text { thus }\left(5^{2} \cdot 7 \cdot 12^{4}\right)_{10}=\left(127 \cdot 10... | 4 | Number Theory | MCQ | Yes | Yes | olympiads | false |
For example, let $8 \alpha, \beta, \gamma$ be the three interior angles of a given triangle. Prove that
$$
\csc ^{2} \frac{\alpha}{2}+\csc ^{2} \frac{\beta}{2}+\csc ^{2} \frac{\gamma}{2} \geqslant 12,
$$
and find the condition for equality. | Prove that by the Arithmetic-Geometric Mean Inequality, we have
$$
\csc ^{2} \frac{\alpha}{2}+\csc ^{2} \frac{\beta}{2}+\csc ^{2} \frac{\gamma}{2} \geqslant 3\left(\csc \frac{\alpha}{2} \cdot \csc \frac{\beta}{2} \cdot \csc \frac{\gamma}{2}\right)^{\frac{2}{3}},
$$
where equality holds if and only if $\alpha=\beta=\ga... | 12 | Inequalities | proof | Yes | Yes | olympiads | false |
9. (10 points) A small insect crawls from $A$ to $B$ along the route shown in the figure. It is stipulated: the edges marked with arrows can only be traveled in the direction of the arrow, and each edge can be passed through at most once in the same route. How many different routes are there for the insect to travel fr... | 【Analysis】The insect travels from $A$ to $B$. At the first hexagon's fork, there are 2 paths both above and below. The hexagon where $B$ is located also has 2 paths both above and below. Therefore, there are $2 \times 2 + 2 \times 2 = 8$ paths. There are 2 paths for the arrow pointing back in the middle, making a total... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. In the number table below, both the upper and lower rows of numbers are arithmetic sequences. For the corresponding numbers in the upper and lower rows, the difference between the larger number and the smaller number, the smallest is $\qquad$ _ | 【Analysis】The first sequence is $5 a(a=1,2,3, \ldots)$; the second sequence is $2020-7 a \quad(a=1,2,3, \ldots)$ they approach 12 each time, the initial difference is $2013-5=2008, \quad 2008 \div 12=168 \ldots \ldots .4$, so the smallest difference is 4 ( above is 840, below is 844 ). | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1・32 Given $2 x-3 y-z=0, \quad x+3 y-14 z=0, z \neq 0$, then the value of $\frac{x^{2}+3 x y}{y^{2}+z^{2}}$ is
(A) 7.
(B) 2.
(C) 0.
(D) $-\frac{20}{17}$.
(E) -2.
(15th American High School Mathematics Examination, 1964) | [Solution] From the given, we have $2 x-3 y=z, x+3 y=14 z$,
solving these gives $\quad x=5 z, y=3 z$.
$$
\therefore \quad \frac{x^{2}+3 x y}{y^{2}+z^{2}}=\frac{x(x+3 y)}{y^{2}+z^{2}}=\frac{5 z \cdot 14 z}{9 z^{2}+z^{2}}=\frac{70 z^{2}}{10 z^{2}}=7 \text {. }
$$
Therefore, the answer is (A). | 7 | Algebra | MCQ | Yes | Yes | olympiads | false |
$10 \cdot 22$ The sum of the first $n$ terms of an arithmetic sequence is 153, the common difference is 2, the first term is an integer, and $n>1$, then the number of possible values for $n$ is
(A) 2.
(B) 3.
(C) 4.
(D) 5.
(E) 6.
(15th American High School Mathematics Examination, 1964) | ［Solution］Let $a$ be the first term of the arithmetic sequence. According to the problem, we have
$$
\frac{n}{2}\{a+[a+2(n-1)]\}=153 \text {, }
$$
or
$$
n^{2}+(a-1) n-153=0 \text {. }
$$
The product of the roots of this quadratic equation is -153, so the possible values of $n$ among the factors greater than 1 are: $3... | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
$8, Class 3 (1) 21 students made a total of 69 paper airplanes. Each girl made 2, and each boy made 5. So, there are ( ) boys and ( ) girls.
$ | 【Analysis】Assuming all are girls, they can make 42 paper airplanes in total, which is 27 less than the actual 69 paper airplanes. For every girl replaced by a boy, 3 more paper airplanes can be made, so there are $27 \div 3=9$ boys, and 12 girls. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.21 In a simple graph with 8 vertices, what is the maximum number of edges in a graph without a quadrilateral? (A simple graph is one where no vertex is connected to itself, and there is at most one edge between any two vertices.)
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。 | [Solution 1] In the graph shown in the right figure, there are 8 vertices and 11 edges, and there is no quadrilateral. From this, we know that the maximum number of edges sought is no less than 11.
Next, we will prove that if the graph has 12 edges, then it must contain a quadrilateral.
First, we point out two obvious... | 11 | Logic and Puzzles | other | Yes | Yes | olympiads | false |
4. Calculate: $2017 \times 2016 + 2016 \times 2014 - 2015 \times 2016 - 2015 \times 2017$. | Reference answer: 1 | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. As shown in the figure, 10 identical regular pentagons are joined edge-to-edge, forming a regular decagon in the middle. 3 identical regular $n$-sided polygons are joined edge-to-edge, forming a regular triangle in the middle, then the value of $n$ is $\qquad$ . | $12$ | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. If the function $f(x)=\frac{(\sqrt{1008} x+\sqrt{1009})^{2}+\sin 2018 x}{2016 x^{2}+2018}$ has a maximum value of $M$ and a minimum value of $m$, then
$M+m=$ $\qquad$ | Solve
$$
f(x)=\frac{1008 x^{2}+1009+2 \sqrt{1008} \sqrt{1009} x+\sin 2018 x}{2\left(1008 x^{2}+1009\right)}=\frac{1}{2}+\frac{2 \sqrt{1008} \sqrt{1009} x+\sin 2018 x}{2\left(1008 x^{2}+1009\right)}
$$
Since $g(x)=\frac{2 \sqrt{1008} \sqrt{1009} x+\sin 2018 x}{2\left(1008 x^{2}+1009\right)}$ is an odd function,
thus $M... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Set $A=\left\{1, \frac{x+y}{2}-1\right\}$, Set $B=\{-\ln (x y), x\}$, if $A=B$ and $0<y<2$, then $\left(x^{2}-\frac{1}{y^{2}}\right)+\left(x^{4}-\right.$ $\left.\frac{1}{y^{4}}\right)+\cdots+\left(x^{2022}-\frac{1}{y^{2022}}\right)$ is $\qquad$ | Answer: 0
Solution: Given $A=B$, if $1=x$, substituting into $\frac{x+y}{2}-1=-\ln (x y)$, it is easy to get $y=1$, at this point the result of the required expression is 0; if $\frac{x+y}{2}-1=x$, we get $y(y-2)=\frac{1}{e}$, but $0<y<2, y(y-2)<0$, which is a contradiction. In summary, the required value is 0. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Given that $z_{1}, z_{2}, \ldots, z_{7}$ are the seventh roots of $2021+i$, and their principal arguments in ascending order are $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{7}$, find the value of $\tan \alpha_{1} \tan \alpha_{3}+\tan \alpha_{2} \tan \alpha_{4}+\cdots+\tan \alpha_{7} \tan \alpha_{2}$ is $\qquad$ | Answer: -7
Solution: It is evident that $\left\{a_{n}\right\}$ forms an arithmetic sequence with a common difference of $\frac{2 \pi}{7}$. Moreover, we have $1+\tan a_{1} \tan a_{3}=\frac{\tan a_{3}-\tan a_{1}}{\tan \frac{4 \pi}{7}}, 1+$ $\tan \alpha_{2} \tan \alpha_{4}=\frac{\tan \alpha_{4}-\tan a_{2}}{\tan \frac{4 \p... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
B4. Numbers $a, b$ and $c$ form an arithmetic sequence if $b-a=c-b$. Let $a, b, c$ be positive integers forming an arithmetic sequence with $a<b<c$. Let $f(x)=a x^{2}+b x+c$. Two distinct real numbers $r$ and $s$ satisfy $f(r)=s$ and $f(s)=r$. If $r s=2017$, determine the smallest possible value of $a$. | Correct answer: 9 .
Solution 1: Note that
$$
\begin{array}{l}
a r^{2}+b r+c=s \\
a s^{2}+b s+c=r
\end{array}
$$
Subtracting the second equation from the first yields
$$
\begin{aligned}
a\left(r^{2}-s^{2}\right)+b(r-s) & =-(r-s) \Rightarrow a(r+s)(r-s)+(b+1)(r-s)=0 \\
\Rightarrow & (a(r+s)+b+1)(r-s)=0 .
\end{aligned}
$... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.12 The shape of the game board is a rhombus containing a $60^{\circ}$ angle. Each side is divided into 9 equal parts, and through each division point, two lines are drawn parallel to the sides and the shorter diagonal, dividing the rhombus into many small equilateral triangles. If a piece is placed in one of these sm... | [Solution] When 1 chess piece is placed in each of the 6 small squares shaded in the figure on the right, all the squares on the entire game board are controlled, so the minimum value we are looking for is no more than 6.
On the other hand, suppose there are at most 5 chess pieces placed on the game board, and let's a... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Given $y z \neq 0$, and the set $\{2 x, 3 z, x y\}$ can also be represented as $\left\{y, 2 x^{2}, 3 x z\right\}$, then $x=$ | If $2 x=y \Rightarrow\left\{\begin{array}{l}2 x^{2}=x y, \\ 3 x z=3 z\end{array} \Rightarrow\left\{\begin{array}{l}x=1, \\ y=2,\end{array}\right.\right.$ satisfies the conditions;
If $2 x=2 x^{2} \Rightarrow x=1 \Rightarrow\left\{\begin{array}{l}x y=y, \\ 3 z=3 x z,\end{array}\right.$ satisfies the conditions;
If $2 x=... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
20. As shown in the figure, the line $y=-x+3 \sqrt{2}$ intersects the $y$-axis and the $x$-axis at points $A$ and $B$, respectively. The line $AP$ intersects the $x$-axis at $P$. An isosceles right triangle $APC$ is constructed with $AP$ as one of its sides. The line $CB$ intersects the $y$-axis at point $D$. Then the ... | $6$ | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
52 The number of integer solutions to the equation $x^{1988}+y^{1988}+z^{1988}=7^{1990}$ is
A. 0
B. 1
C. $2^{3} \cdot P_{3}^{3}$
D. Not less than 49 | 52 A. Suppose the original equation has an integer solution $(x, y, z)=(m, n, r)$, then
$$
m^{1988}+n^{1988}+r^{1988}=7^{1990} .
$$
On one hand, since $7^{1990}=7^{1 \times 497+2}$, the last digit of $7^{1990}$ is 9; on the other hand, for any $t \in \mathbf{Z}$, the last digit of $t^{4}$ can only be one of $0,1,5,6$.... | 0 | Number Theory | MCQ | Yes | Yes | olympiads | false |
1. The number of equilateral triangles that can be formed by three vertices of a cube is
(A) 4 ;
(B) 8 ;
(C) 12 ;
(D) 24 . | 1. (B)
In a cube, there is a vertex where three faces, each perpendicular to the other two, meet. On these three faces, there are three diagonals that together form a unique equilateral triangle. This establishes a one-to-one correspondence between the equilateral triangles and the vertices. Since a cube has 8 vertice... | 8 | Geometry | MCQ | Yes | Yes | olympiads | false |
1. Given $A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}$, find the integer part of $A$ | Reference answer: 1 | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$36 \cdot 31$ A student observes during a $e$-day vacation:
(1) It rains seven times, either in the morning or in the afternoon;
(2) When it rains in the afternoon, the morning is clear;
(3) There are five clear afternoons in total;
(4) There are six clear mornings in total.
Then $e$ equals
(A) 7.
(B) 9.
(C) 10.
(D) 11... | ［Solution］Let the number of days corresponding to the weather conditions be as in the table below:
\begin{tabular}{c|c|c}
\hline & Rainy in the morning & Sunny in the morning \\
\hline Rainy in the afternoon & $a$ days & $b$ days \\
\hline Sunny in the afternoon & $c$ days & $d$ days \\
\hline
\end{tabular}
From the g... | 9 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
10. (9th Central American and Caribbean Mathematical Olympiad) Let $S$ be a finite set of integers. Suppose that for any two distinct elements $p, q \in S$, there exist three elements $a, b, c \in S$ (not necessarily distinct, and $a \neq 0$) such that the polynomial $F(x) = a x^{2} + b x + c$ satisfies $F(p) = F(q) = ... | 10. It is easy to verify that $S=\{-1,0,1\}$ satisfies the condition.
Below is the proof: $|S|_{\max }=3$.
(1) At least one of $1, -1$ belongs to $S$.
Otherwise, there exist $a_{1}, a_{2} \in S$, and
$$
\left|a_{1}\right| \cdot\left|a_{2}\right| \geqslant 2\left(\left|a_{2}\right| \geqslant\left|a_{1}\right|\right) \... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In the figure below, there are three scales. By observing the first two scales, we can find that 5 " $\boldsymbol{\Delta}$ " and 3 " $\bigcirc$ " weigh the same, and 1 " " weighs the same as 1 " $\boldsymbol{\Delta}$ " plus 2 " $\boldsymbol{\square}$ ". Therefore, 1 " $\boldsymbol{\Delta}$ " plus 1 " 0 " weighs the ... | Analysis: [Equivalent Substitution] From the problem, we know: $5 \boldsymbol{\Delta}=3 \boldsymbol{\bullet}, 1 \boldsymbol{=}=1 \boldsymbol{\Delta}+2 \boldsymbol{\square}$. Combining the two equations, we get $5 \boldsymbol{\Delta}=3 \boldsymbol{\Delta}+6 \boldsymbol{\square}$, which simplifies to $2 \boldsymbol{\Delt... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. If the union of two sets $A$ and $B$ has two elements, and $f(A)$ denotes the number of elements in $A$, $f(B)$ denotes the number of elements in $B$, then the number of such pairs $(f(A), f(B))$ is $\qquad$ pairs. | 7. 6 .
Let $A \cup B=\left\{a_{1}, a_{2}\right\}$, then when $A=\varnothing$, $B=\left\{a_{1}, a_{2}\right\}$; when $A=\left\{a_{1}\right\}$, $B=\left\{a_{2}\right\}$ or $\left\{a_{1}, a_{2}\right\}$; when $A=\left\{a_{2}\right\}$, $B=\left\{a_{1}\right\}$ or $\left\{a_{1}, a_{2}\right\}$; when $A=\left\{a_{1}, a_{2}\... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
12. (15th All-Russian Mathematical Olympiad, 1989) Insert “+” or “-” signs between the numbers $1, 2, 3, \cdots, 1989$. What is the smallest non-negative number that can be obtained from the resulting sum? | 12. Except for 995, all numbers $1,2,3, \cdots, 1989$ can be divided into 994 pairs: $(1,1989),(2,1988), \cdots,(994,996)$. Since the parity of the two numbers in each pair is the same, the result of the operation is always even regardless of how the “+” or “-” signs are placed before each pair. Since 995 is odd, the s... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
25. As shown in the figure, the sum of the numbers on the opposite faces of the cube is 7. If it is stipulated that the outside of side 2 is the front, and the cube is first flipped backward 15 times, then flipped to the right 30 times, each time flipping $90^{\circ}$, then the number on the top of the cube at this tim... | answer: 2 | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
38. Using 80 squares with a side length of 2 cm, you can form $\qquad$ types of rectangles with an area of 320 square cm. | answer: 5 | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
18.105 The sum of the two base angles of a trapezoid is $90^{\circ}$, the upper base is 5, and the lower base is 11. Then the length of the line segment connecting the midpoints of the two bases is
(A) 3.
(B) 4.
(C) 5.
(D) 6.
(Anhui Junior High School Mathematics Competition, 1996) | [Solution] Draw $C F / / A D$ intersecting $A B$ at $F$. According to the problem, $\angle F C B=90^{\circ}$.
Since $C E$ is the median of the hypotenuse of the right triangle $\mathrm{Rt} \triangle F C B$, then
$$
C E=\frac{1}{2} F B=\frac{1}{2} \cdot(11-5)=3 \text {. }
$$
Let $G$ and $H$ be the midpoints of the uppe... | 3 | Geometry | MCQ | Yes | Yes | olympiads | false |
7.35 Prove that for any $n \in N$ greater than a certain number $n_{0}$, the plane can be divided into $n$ regions using some straight lines, and there must be intersecting lines among them, and find the smallest $n_{0}$. | [Solution] According to the given information, at least two of the used straight lines intersect, dividing the plane into 4 regions. Adding any 1 more line will increase the number of regions by at least 2, so it is impossible to have exactly 5 regions. Therefore, $n_{0} \geqslant 5$.
On the other hand, any division i... | 5 | Geometry | proof | Yes | Yes | olympiads | false |
9. In the known sequence $1,4,8,10,16,19,21,25,30,43$, the number of arrays where the sum of adjacent numbers is divisible by 11 is $\qquad$ . | Given the sequence $\left\{x_{n}\right\}$, the remainders when each term is divided by 11 are $1,4,-3,-1,5,-3,-1,3,-3,-1$.
Thus, starting from the first term, the remainders when the sum of any $k(k=1,2, \cdots, 10)$ consecutive terms is divided by 11 form the sequence $\left\{a_{n}\right\}$:
$1,5,2,1,6,3,2,5,2,1$. Tha... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 4-13 The 6 faces of a cube are to be colored using red and blue. How many different coloring schemes are there? | The rigid motion group that makes a regular hexahedron coincide has the following cases:
(1) Identity permutation, i.e., the unit element (1)(2)(3)(4)(5)(6), formatted as $(1)^{6}$.
(2) Rotation by $\pm 90^{\circ}$ around the axis $A B$ passing through the centers of faces (1) and (6) (as shown in Figure 4-15(a)). The ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Given a finite sequence that satisfies: the sum of any 3 consecutive terms is negative, and the sum of any 4 consecutive terms is positive, then the maximum number of terms in this sequence is
A. 7
B. 6
C. 5
D. 4 | Construct an array as shown below, with each row consisting of positive numbers, then the sum of 12 items is positive;
$$
\begin{array}{llll}
a_{1} & a_{2} & a_{3} & a_{4}
\end{array}
$$
and each column consisting of negative numbers, then the sum of 12 items is negative, which is a contradiction.
$$
a_{2} \quad a_{3}... | 5 | Inequalities | MCQ | Yes | Yes | olympiads | false |
1. Let the set $S=\left\{A_{0}, A_{1}, A_{2}, A_{3}\right\}$, and define the operation “ $\oplus$ ” on set $S$: $A_{i} \oplus A_{j}=A_{k}$, where $k$ is the remainder of $i+j$ divided by 4, $i, j \in\{0,1,2,3\}$. Then the number of $x(x \in S)$ that satisfies the relation $(x \oplus x) \oplus A_{2}=A_{0}$ is ( ).
(A) 1... | - 1. B.
From $(x \oplus x) \oplus A_{2}=A_{0}$, let $x \oplus x=A_{k}$, then
$$
\begin{array}{l}
A_{k} \oplus A_{2}=A_{0}, k=2 \\
\Rightarrow x \oplus x=A_{2} \\
\Rightarrow x=A_{1} \text { or } A_{3} .
\end{array}
$$ | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
[Example 4.5.6] Given that $p, q$ are both integers greater than 1, and $2 p-1$ is a multiple of $q$, $2 q-1$ is a multiple of $p$, find the value of $p+q$. | The problem does not provide specific numbers, yet it requires finding the value of $p+q$. This presents a certain difficulty. If we analyze the given conditions one by one, we almost get no useful information. However, if we look at the conditions as a whole, from the fact that $2 p-1$ is a multiple of $q$ and $2 q-1$... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 3: There are two red balls, one black ball, and one white ball. Question: (1) How many different ways are there to select the balls? Try to enumerate them respectively; (2) If 3 balls are randomly selected each time, how many different ways are there to select them? | Let $x_{1}, x_{2}, x_{3}$ represent red, black, and white balls respectively. The ways to choose 2 red balls correspond to $\left(x_{1} t\right)^{0},\left(x_{1} t\right)^{1},\left(x_{1} t\right)^{2}$. Thus, the possible ways to choose red balls are represented by $1+x_{1} t+\left(x_{1} t\right)^{2}$, where 1 indicates ... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
13. (15 points) Eight $2 \times 1$ small rectangles as shown in the figure can be used to form a $4 \times 4$ square. If a square formed by these rectangles is the same as another square after rotation, they are considered the same square. How many different squares can be formed where both diagonals are axes of symmet... | 【Analysis】Use the right figure
to replace the $2 \times 1$ rectangle in the problem, because the given small rectangle is asymmetrical up and down, so in the square figure formed by the small rectangles where both diagonals are its axes of symmetry, this way, we only need to consider different scenarios in the upper ha... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. In rectangle $A B C D$, $A B=3, A D=4, P$ is a point on the plane of rectangle $A B C D$, satisfying $P A=2$, $P C=\sqrt{21}$. Then $\overrightarrow{P B} \cdot \overrightarrow{P D}=$ $\qquad$ . | 11.0.
As shown in Figure 3, let $A C$ and $B D$ intersect at point $E$, and connect $P E$. Then $E$ is the midpoint of $A C$ and $B D$.
Notice that,
$$
\begin{array}{l}
\overrightarrow{P B} \cdot \overrightarrow{P D}=\frac{1}{4}\left[(\overrightarrow{P B}+\overrightarrow{P D})^{2}-(\overrightarrow{P B}-\overrightarrow... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. Zara collected 4 marbles, which are made of agate, tiger's eye, steel, and amber. She wants to display these 4 marbles in a row on a shelf, but she does not want the steel marble and the tiger's eye marble to be next to each other. How many arrangements are possible?
(A) 6
(B) 8
(C) 12
(D) 18
(E) 24 | 10. C.
Let $S$ and $T$ represent steel and tiger stone, respectively. First, arrange $S$ and $T$.
Since $S$ and $T$ cannot be adjacent, the only possible arrangements are:
$$
S \square T \square, S \square \square T, \square S \square T,
$$
and the cases where $S$ and $T$ are swapped, giving a total of $3 \times 2$ w... | 12 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
1. The number of equilateral triangles that can be formed by three vertices of a cube is
(A) 4
(B) 8
(C) 12
(D) 24 | (B)
I. Multiple Choice Question
1.【Analysis and Solution】In a cube, at a common vertex of three mutually perpendicular faces, there is a pair of diagonals that together form a unique equilateral triangle. This establishes a one-to-one correspondence between the equilateral triangle and the vertex. Since a cube has 8 ve... | 8 | Geometry | MCQ | Yes | Yes | olympiads | false |
2. Given that $z$ is an imaginary number, and $z^{2}=\bar{z}$, then $z^{3}=$ | $z^{2}=\bar{z} \Rightarrow|z|^{2}=|\bar{z}|=|z| \Rightarrow|z|=1$, so $z^{2}=\bar{z} \Rightarrow z^{3}=z \bar{z}=|z|^{2}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.65 The line $x=k$ intersects the curves $y=\log _{5} x$ and $y=\log _{5}(x+4)$ at two points with a distance of 0.5. If $k=a+\sqrt{b}$, where $a, b$ are integers, then $a+b=$
(A) 6.
(B) 7.
(C) 8.
(D) 9.
(E) 10.
(48th American High School Mathematics Examination, 1997) | [Solution] From the given, we know:
$$
\log _{5}(k+4)-\log _{5} k=0.5,
$$
which means $\frac{k+4}{k}=\sqrt{5}$, leading to $k=\sqrt{5}+1$, thus $a+b=6$. Therefore, the answer is $(A)$. | 6 | Algebra | MCQ | Yes | Yes | olympiads | false |
9. A gardener is preparing to plant a row of 20 trees, with two types of trees available: maple trees or sycamore trees. The number of trees between any two maple trees (not including these two maple trees) cannot be equal to 3. Therefore, the maximum number of maple trees among the 20 trees is $\qquad$ trees. | 【Answer】 12
【Solution】In any continuous eight trees, once a maple tree is planted, it means that another position can only plant a sycamore tree. We use the following diagram to illustrate, using — to represent a maple tree and ○ to represent a sycamore tree. Once the second position is planted with a maple tree, posit... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 20 (1) Let $n$ be a prime number greater than 3, find the value of $\left(1+2 \cos \frac{2 \pi}{n}\right)\left(1+2 \cos \frac{4 \pi}{n}\right) \cdots\left(1+2 \cos \frac{2 m \pi}{n}\right)$.
(2) Let $n$ be a natural number greater than 3, find the value of
$\left(1+2 \cos \frac{\pi}{n}\right)\left(1+2 \cos \fr... | (1) Let $\omega=\mathrm{e}^{\frac{2 \pi i}{n}}$, then $\omega^{n}=1, \omega^{-\frac{n}{2}}=\mathrm{e}^{\pi i}=-1, 2 \cos \frac{2 k \pi}{n}=\omega^{k}+\omega^{-k}$,
$$
\begin{aligned}
\prod_{k=1}^{n}\left(1+2 \cos \frac{2 k \pi}{n}\right) & =\prod_{k=1}^{n}\left(1+\omega^{k}+\omega^{-k}\right)=\prod_{k=1}^{n} \omega^{-k... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. If from any $\mathrm{n}$ numbers, one can always select 4 numbers such that their sum is a multiple of 4, then what is the minimum value of $n$? | 【Answer】7
【Explanation】Pigeonhole principle, residue classes.
Natural numbers divided by 4 have 4 residue classes, with remainders of $0,1,2,3 \ldots$ First, if any class has 4 or more numbers, their sum will be a multiple of 4, so each class can have at most 3 numbers; and $1+3=2+2$, so if there are 2 numbers with a r... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In $\triangle A B C$, $a, b, c$ are the sides opposite to $\angle A, \angle B, \angle C$ respectively, satisfying $a^{2}+b^{2}=4-\cos ^{2} C, a b=2$. Then $S_{\triangle A B C}=$ $\qquad$ . | -1.1 .
From the problem, we have $(a-b)^{2}+\cos ^{2} C=0$.
Solving, we get $a=b=\sqrt{2}, \cos C=0$.
Therefore, $S_{\triangle A B C}=1$. | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. The integers $x_{0}, x_{1}, \cdots, x_{2004}$ satisfy the conditions, $x_{0}=0,\left|x_{1}\right|=\left|x_{0}+1\right|,\left|x_{2}\right|=\left|x_{1}+1\right|, \cdots$, $\left|x_{2004}\right|=\left|x_{2003}+1\right|$, find the minimum value of $\left|x_{1}+x_{2}+\cdots+x_{2004}\right|$. | 9. From the known, we have
$$
\left\{\begin{array}{l}
x^{2}=x_{0}^{2}+2 x_{0}+1, \\
x_{2}^{2}=x_{1}^{2}+2 x_{1}+1, \\
\cdots \cdots \\
x_{2004}^{2}=x_{2003}^{2}+2 x_{2003}+1 .
\end{array}\right.
$$
Thus, $x_{2004}^{2}=x^{2}+2\left(x_{0}+x_{1}+\cdots+x_{2003}\right)+2004$. Given $x_{0}=0$, then $2\left(x_{1}+x_{2}+\cdo... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. For a positive integer $n$, let the product of its digits be denoted as $a(n)$. Then the positive integer solution to the equation $n^{2}-17 n+56=a(n)$ is
$\qquad$ . | 7. 4 Detailed Explanation: First, prove $a(n) \leqslant n$. Let $n=\overline{b_{k} b_{k-1} \cdots b_{1} b_{0}}$, then $n \geqslant b_{k} \times 10^{k} \geqslant b_{k} \prod_{i=1}^{k-1} b_{i}=\prod_{i=1}^{k} b_{i}=a(n)$. $\therefore n^{2}-17 n+56 \leqslant n$, which means $n^{2}-18 n+56 \leqslant 0$, solving this yields... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 1 As shown in Figure 14-6, in $\triangle ABC$, $G$ is the centroid, and $P$ is a point inside the triangle. The line $PG$ intersects the lines $BC$, $CA$, and $AB$ at $A'$, $B'$, and $C'$, respectively. Prove that: $\frac{A'P}{A'G} + \frac{B'P}{B'G} + \frac{C'P}{C'G} = 3$. | Prove: Connect $B G, G C, P B, P C$, draw $G G^{\prime} \perp B C$ at $G^{\prime}$, draw $P P^{\prime} \perp B C$ at $P^{\prime}$, then $P P^{\prime} / / G G^{\prime}, \frac{P P^{\prime}}{G G^{\prime}}=\frac{A^{\prime} P}{A^{\prime} G}$.
Also, $\frac{S_{\triangle P B C}}{S_{\triangle G B C}}=\frac{P P^{\prime}}{G G^{\p... | 3 | Geometry | proof | Yes | Yes | olympiads | false |
10. As shown in the figure, ABCD is a trapezoid, the area of triangle ADE is 1, the area of triangle $\mathrm{ABF}$ is 9, and the area of triangle BCF is 27. What is the area of triangle $\mathrm{ACE}$? | 【Analysis】From the area of triangle $\mathrm{ABF}$ being 9, and the area of triangle $\mathrm{BCF}$ being 27, we can deduce that $\mathrm{AF}: \mathrm{FC}=1: 3$.
Since $\mathrm{ABCD}$ is a trapezoid, the area of triangle $\mathrm{CDF}$ is also 9.
Given $\mathrm{AF}: \mathrm{FC}=1: 3$, the area of triangle $\mathrm{ADF}... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. In an $8 \times 8$ square chessboard composed of 64 small squares, a circle with a radius of 4 is placed. If the sum of the areas inside the circle for all small squares that the circle's circumference passes through is denoted as $S_{1}$, and the sum of the areas outside the circle for all small squares that the ci... | 6. (B)
Solution According to the symmetry of the square, we only need to consider its $\frac{1}{4}$ part. Let the sum of the areas inside the circle of all small squares that the circumference passes through be $S_{1}^{\prime}$, and the sum of the areas outside the circle of all small squares that the circumference pa... | 1 | Geometry | MCQ | Yes | Yes | olympiads | false |
0.1 Given the Fibonacci sequence defined as follows: $F_{1}=1, F_{2}=1, F_{n+2}=F_{n+1}+F_{n}$ (n 1), find $\left(F_{2017}, F_{99} F_{101}+1\right)$ (40 points) | For any positive integers $\mathrm{m}, \mathrm{n}$ (let $\mathrm{m}>\mathrm{n}$), we have
$$
\begin{aligned}
\left(F_{m}, F_{n}\right) & =\left(F_{m-1} F_{2}+F_{m-2} F_{1}, F_{n}\right) \\
& =\left(F_{m-2} F_{2}+F_{m-3} F_{2}+F_{m-2} F_{1}, F_{n}\right) \\
& =\left(F_{m-2} F_{3}+F_{m-3} F_{2}, F_{n}\right) \\
& =\left(... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Let $x_{1}, x_{2}, x_{3}, \cdots, x_{9}$ be positive integers, and $x_{1}<x_{2}<\cdots<x_{9}, x_{1}+x_{2}+\cdots+x_{9}=220$. Then, when $x_{1}+x_{2}+\cdots+x_{5}$ is maximized, the minimum value of $x_{9}-x_{1}$ is
A. 8
B. 9
C. 10
D. 11 | 4. B $\quad x_{9} \geqslant x_{8}+1 \geqslant x_{7}+2 \geqslant \cdots \geqslant x_{1}+8$, thus $x_{9} \geqslant x_{1}+8, x_{9}-x_{1} \geqslant 8$, if the equality holds, $x_{1}$, $x_{2}, \cdots, x_{9}$ form an arithmetic sequence with a common difference of 1, and their sum is 220, i.e., $9 x_{5}=220, x_{5}$ is not an... | 9 | Number Theory | MCQ | Yes | Yes | olympiads | false |
2. Given $\log _{\sqrt{7}}(5 a-3)=\log _{\sqrt{a^{2}+1}} 5$, then the real number $a=$ | The left side of the equation is increasing with respect to $a$, while the right side is decreasing with respect to $a$, so there exists a unique real number $a=2$.
| 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the plane, there is a set of points $M$ and seven different circles $C_{1}, C_{2}, \cdots, C_{7}$, where circle $C_{7}$ passes through exactly 7 points in $M$, circle $C_{6}$ passes through exactly 6 points in $M$, $\cdots$, circle $C_{1}$ passes through exactly 1 point in $M$. The minimum number of points in $M$... | 5. (B)
To minimize the number of points in $M$, the circles should pass through as many common points in $M$ as possible. Since two distinct circles can have at most two common points, and $C_{7}$ passes through 7 points in $M$, $C_{6}$ must pass through at least four other points in $M$. Additionally, $C_{5}$, beside... | 12 | Geometry | MCQ | Yes | Yes | olympiads | false |
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