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Example 8 (2003 Hunan Province Competition Question) Let $x, y, z$ all take positive real numbers, and $x+y+z=1$. Find the minimum value of the ternary function $f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}$, and provide a proof. | Since the three fractional functions in the ternary function are independent variable expressions, and both the numerator and denominator contain 2nd degree variables, we perform a reduction by lowering the degree. For this, we first examine the function $g(t)=\frac{t}{1+t^{2}}$, and we know that $g(t)$ is an odd funct... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
4. The Little Fairy has the same number of magic balls and magic wands. If the number of magic balls doubles, and the number of magic wands is halved, the magic balls will be 9 more than the magic wands. The Little Fairy originally had $\qquad$ magic balls. | $6$ | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. By folding a polygonal paper, a line segment can be drawn on the paper and the paper can be folded along it. As shown in Figure 1, for the given shape on the paper, cutting along the boundary of the shaded part will yield a polygonal paper. For this polygonal paper, a rectangular paper can be obtained with at most f... | 1. There are different ways to fold the given paper into a rectangle according to the problem requirements. As shown in Figure 3 (a～d), the folding methods are given in sequence, requiring only 4 folds.
Note, in the third and fourth folds, the crease lines can be slightly moved up or down, and all crease lines must be... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (5 points) A number, when reduced by 5 times, and then increased by 20 times, becomes 40. This number is $\qquad$ | 【Solution】Solution: $40 \div 20 \times 5$
$$
\begin{array}{l}
=2 \times 5 \\
=10
\end{array}
$$
Therefore, the answer is: 10. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Let $x, y$ be real numbers. Then the minimum value of $(x y-1)^{2}+(x+y)^{2}$ is ( ).
(A) 0
(B) $\frac{1}{4}$
(C) $\frac{1}{2}$
(D) 1
(E) 2 | 9. D.
$$
\begin{array}{l}
(x y-1)^{2}+(x+y)^{2} \\
=x^{2} y^{2}-2 x y+1+x^{2}+2 x y+y^{2} \\
=x^{2} y^{2}+x^{2}+y^{2}+1 \\
\geqslant 1,
\end{array}
$$
The equality holds if and only if $x=y=0$. | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. There is a rectangular iron sheet of size $80 \times 50$. Now, a square of the same size is to be cut from each corner, and then it is to be made into an open box. What should be the side length of the square to be cut so that the volume of this open box is maximized? | 7. Let the side length of the cut-out square be $x$, then the volume of the made box is $V=x(80-2 x)(50-2 x)$. Introducing parameters $\alpha, \beta>0$, we have
$$
\begin{aligned}
V & =\frac{1}{\alpha \beta}(\alpha x)[\beta(50-2 x)](80-2 x) \\
& \leqslant \frac{1}{\alpha \beta}\left[\frac{(50 \beta+80)+(\alpha-2 \beta-... | 10 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
$7 \cdot 108$ For any real numbers $x, y$, the function $f(x)$ satisfies the relation
$$
f(x+y)=f\left(x^{2}\right)+f(2 y) .
$$
Then the value of $f(1985)$ is
(A) 1985 .
(B) $\sqrt{1985}$.
(C) 3990 .
(D) None of the above.
(China Beijing High School Mathematics Competition, 1985) | [Solution] Let $y=x$, we get $f(2 x)=f\left(x^{2}\right)+f(2 x)$,
i.e., $\square$
$$
f\left(x^{2}\right)=0 \text {. }
$$
Let $x=\sqrt{1985}$, substituting in we get $f(1985)=0$.
Clearly, (A), (B), and (C) are not true.
Therefore, the answer is (D). | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
9. $A, B, C, D$ four people have some number of cards that are all different.
$A$ says: “I have 16 more cards than $C$.”
$B$ says: “$D$ has 6 more cards than $C$.”
$C$ says: “$A$ has 9 more cards than $D$.”
$D$ says: “If $A$ gives me 2 more cards, I will have 3 times as many cards as $C$.”
It is known that among these ... | 【Answer】 10
【Solution】First, translate what each person said:
$A$ means: $A-C=16$;
$B$ means: $D-C=6$;
$C$ means: $A-D=9$;
$D$ means: $D+2=3 C$.
Since only one person is lying, and both $A$ and $C$ claim that $A$ is not the least, therefore, $A$ is telling the truth. From $B$ and $D$'s statements, we can infer that $D... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9. (5 points) A division operation, the dividend is 10, the divisor is less than 10, then the sum of all possible different remainders is . $\qquad$ | 【Analysis】The divisor is less than 10, so we can divide 10 by $1 \sim 9$, and among the remainders obtained, there are 2 that are 0, i.e., the remainder is 0 when dividing by $1$ and $5$. The different remainders are $1, 2, 3, 4$. We can then find the sum.
【Solution】According to the analysis, $10 \div 6=1 \cdots 4$;
$... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The integer part of the result of the expression $0.81 \times 0.6+0.83 \times 0.6+\cdots+0.97 \times 0.6+0.99 \times 0.6$ is
$\qquad$ . | answer: 5 | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Let $a$ be the sum of the digits of $4568^{777}$, $b$ be the sum of the digits of $a$, and $c$ be the sum of the digits of $b$. Find $c$. | 6. Solution: $\because 4568^{7777}<10000^{777}=10^{4 \times 777}$ has a total of $4 \times 7777+1=31109$ digits,
$$
\therefore a<9 \times 31109=279981, b<2+5 \times 9=
$$
$47, c<4+9=13$.
$$
\begin{array}{l}
\because 4568^{7777} \equiv 5^{7777} \equiv 5^{3 \times 2592+1} \equiv 5(-1)^{2592} \equiv \\
5(\bmod 9), \\
\the... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
50. One day, from 8 a.m. to 6 p.m., the hour hand and the minute hand overlap $\qquad$ times. | Answer: 9 | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
14.4.13 ** Find all integers \( k \geqslant 1 \), such that the sequence \( k, k+1, k+2, \cdots, k+99 \) contains the maximum number of prime numbers. | Parse $2 \times 3 \times 5 \times 7 \times 11=2310$. When $k=12,13, \cdots, 2310+11$, it is not difficult to verify that among $k, k+1, k+2, \cdots, k+99$ (1) at least 76 numbers are divisible by 2, 3, 5, 7, or 11. Therefore, when $k>11$, (1) contains at least 76 composite numbers and at most 24 prime numbers. When $k=... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1 -43 When $x-y=1$, the value of $x^{4}-x y^{3}-x^{3} y-3 x^{2} y+3 x y^{2}+y^{4}$ is
(A) -1 .
(B) 0 .
(C) 1 .
(D) 2 .
(China Beijing Junior High School Mathematics Competition, 1991) | $$
\text { [Solution] Original expression } \begin{aligned}
& =x\left(x^{3}-y^{3}\right)-y\left(x^{3}-y^{3}\right)-3 x y(x-y) \\
& =\left(x^{3}-y^{3}\right)(x-y)-3 x y(x-y),
\end{aligned}
$$
Since $x-y=1$, the expression can be simplified to
$$
\begin{aligned}
\text { Original expression } & =\left(x^{3}-y^{3}\right)-... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
3. (8 points) Place the numbers $1-8$ at the eight vertices of a cube, then write the average of the numbers at the two endpoints of each edge at the midpoint of that edge. If the numbers written at the midpoints of the four edges of the top face and the four edges of the bottom face are all integers, then among the nu... | 【Answer】Solution: For the odd-even problem, among the numbers $1 \sim 8$, there are 4 odd and 4 even numbers. To satisfy the condition that the sum of adjacent numbers is even in both the upper and lower groups of 4 numbers, the only scenario is that the numbers in each group are either all odd or all even. That is, th... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Example 3 Find the minimum value of the function $u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}$ with real numbers $x, y$ as variables.
(1991 "Hope Cup" Invitational Competition Question) | Solve $u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}=(x-y)^{2}+\left(\frac{9}{x}+\sqrt{2-y^{2}}\right)^{2}-2$. From the above equation, we can consider points $P_{1}\left(x, \frac{q}{x}\right), P_{2}\left(y,-\sqrt{2-y^{2}}\right)$ on the plane. When $x \in \mathbf{R}, x \neq 0$, the trajectory of $P_... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$36 \cdot 48$ In one night, the number of times the hour, minute, and second hands overlap is
(A) 2.
(B) 4.
(C) 6.
(D) 24.
(China Hubei Province Huanggang Junior High School Mathematics Competition, 1993) | [Solution] The three hands of a clock coincide if and only if it is exactly 12 o'clock. Thus, in a day and night, the hour, minute, and second hands coincide only twice.
Therefore, the answer is (A) | 2 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
$33 \cdot 4$ and the unit digit of $1988^{1989}+1989^{1988}$ is
(A) 9.
(B) 7.
(C) 5.
(D) 3.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1989) | [Solution] Let $\langle a\rangle$ denote the unit digit of $a$.
Then
$\left\langle 1988^{1989}\right\rangle=\left\langle 8^{1989}\right\rangle=\left\langle 8^{4 \cdot 497+1}\right\rangle=\left\langle 8^{1}\right\rangle=8$.
Also, $\left\langle 1989^{1988}\right\rangle=\left\langle 9^{1988}\right\rangle=\left\langle 9^{4... | 9 | Number Theory | MCQ | Yes | Yes | olympiads | false |
5. The number of common points of the curves $\rho=11 \cos \frac{\theta}{2}$ and $\rho=\cos \frac{\theta}{2}$ in the polar coordinate plane is ( ).
A. 0
B. 2
c. 3
D. 4 | 5. D
When $\rho=0$, the point $(0,0)$ lies on both curves. When $\rho \neq 0$, then from $1+\cos \frac{\theta}{2}=$ $\cos \frac{\theta+2 k \pi}{2} \rightarrow \cos \frac{\theta}{2}=-\frac{1}{2}, \rho=\frac{1}{2}$, which means the points $\left(\frac{1}{2}, \frac{4 \pi}{3}\right)\left(\frac{1}{2},-\frac{4 \pi}{3}\right... | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
5. (10 points) The grid in the figure is composed of 6 identical small squares. Color 4 of the small squares gray, requiring that each row and each column has a colored small square. Two colored grids are considered the same if they are identical after rotation. How many different coloring methods are there? $\qquad$ k... | 【Analysis】First, we can classify the discussion based on the number of colored cells in the first column, noting that rotations resulting in the same pattern are considered the same coloring method.
【Solution】Solution: (1) When the first column has 3 colored cells, the coloring situations are as follows: there are 3 c... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. If $a$ is the positive root of the equation $x^{2}+3 x-2=0$, and $b$ is the root of the equation $x+\sqrt{x+1}=3$, then $a+b=$ $\qquad$
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 7. 2 .
Notice that, $b+1+\sqrt{b+1}=4$.
Let $t=\sqrt{b+1}-1$.
Then $(t+1)^{2}+(t+1)=4$, and $t>0$.
Thus, $a$ satisfies $(a+1)^{2}+(a+1)=4$.
Therefore, $a=t$.
Hence $b+1+a+1=4 \Rightarrow a+b=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A set of points $M$ in the plane satisfies the following conditions:
(a) No three points are collinear;
(b) Each point in $M$ can be colored either red or blue such that every triangle with vertices of the same color has at least one point of the other color inside it.
Determine the maximum possible number of elements... | Let $n(n \geqslant 3)$ red points form a convex hull $F$, among which $k$ red points are inside $F$. The convex $(n-k)$-gon $F$ is divided by diagonals into $n-k-2$ red triangles. Each internal red point further divides its triangle into 3, resulting in a total of $n-k-2+2k=n+k-2$ red triangles. According to the given ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. As shown in the figure, in $\triangle \mathrm{ABC}$, $\mathrm{M}$ is the midpoint of side $\mathrm{AB}$, $\mathrm{N}$ is a trisection point on side $\mathrm{AC}$, and $\mathrm{CM}$ intersects $\mathrm{BN}$ at point K. If the area of $\triangle B C K$ is 1, then the area of $\triangle A B C$ is $\qquad$ . | 【Analysis】Connect $\mathrm{AK}$, by the燕tail model (Note: This should be "by the tail model" or "by the cevian theorem", but "燕tail model" is kept as a literal translation)
Since $\mathrm{AM}=\mathrm{MB}$, we have $S_{\triangle A K C}=S_{\triangle B K C}=1$
Since $\mathrm{AN}=2 \mathrm{NC}$, we have $S_{\triangle A K B... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
15. Environmental volunteer Xiao Li goes to Ant Forest to plant trees. If he plants 60 trees every day, he can complete the task 10 days ahead of schedule; if he plants 50 trees every day, he can complete the task 8 days ahead of schedule; if he plants 40 trees every day, he can complete the task $\qquad$ days ahead of... | $5$ | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Given two quadratic trinomials $f(x)$ and $g(x)$ with leading coefficients of 1, each having two real roots, and $f(1)=g(2)$, $g(1)=f(2)$. Find the sum of the four roots of these two quadratic trinomials. | 1.6.
$$
\text { Let } f(x)=x^{2}+a x+b, g(x)=x^{2}+c x+d \text {. }
$$
According to the conditions in the problem, we have
$$
\begin{array}{l}
1+a+b=4+2 c+d, \\
4+2 a+b=1+c+d .
\end{array}
$$
(1) - (2) gives
$$
-3-a=3+c \Rightarrow-a-c=6 \text {. }
$$
According to Vieta's formulas, $-a$ is the sum of the roots of the... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. (15 points) Let $n$ be a positive integer. If from any $n$ non-negative integers, one can always find four distinct numbers $a, b, c, d$ such that $a+b-c-d$ is divisible by 20, what is the minimum value of $n$? | 【Analysis】First, explain that any 8 non-negative integers cannot satisfy the condition, because taking any 9 non-negative integers, and choosing any 7 from them, their pairwise sums result in 21 sums, and the remainders of these 21 sums when divided by 20 have 21 values. Since the remainders can have at most 20 differe... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$13 \cdot 3$ If $a, c, d$ are integers, $b$ is a positive integer, and they satisfy $a+b=c$, $b+c=d, \quad c+d=a$. Then, the maximum value of $a+b+c+d$ is
(A) -1 .
(B) -5 .
(C) 0 .
(D) 1 .
(China Junior High School Mathematics League, 1991) | [Sol] $a+b=c$ and $c+d=a$ are added, getting $a+b+c+d=a+c$.
So $b+d=0$, which means $b=-d$. Substituting $b=-d$ into $b+c=d$, we get $c=2d$.
Therefore, $a=c+d=2d+d=3d$.
Thus, $a+b+c+d=a+c=3d+2d=5d=-5b$.
Since $b$ is a positive integer, the minimum value of $b$ is 1.
Therefore, the maximum value of $a+b+c+d$ is -5.
Henc... | -5 | Algebra | MCQ | Yes | Yes | olympiads | false |
Example 5 Given that the function $y=f(x)$ defined on $\mathbf{R}$ satisfies $f(x)-f(2-$ $x)=0$ for any $x \in \mathbf{R}$, is its graph an axis-symmetric figure? If so, write down the equation of the axis of symmetry. | Given that $f(x)-f(2-x)=0$ holds for any $x \in \mathbf{R}$, we have $f(1+x)=f[2-(1+x)]$, which means $f(1+x)=f(1-x)$. Therefore, the graph of the function $y=f(x)$ is symmetric about the line $x=1$.
Key Points for Solving the Problem: For the function $y=f(x)(x \in D)$, if $f(a+x)=f(a-x)$ holds for any $x$ in the dom... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.49 Divide a cube into non-overlapping tetrahedra. What is the minimum number of tetrahedra needed? | [Solution] Each face of a tetrahedron is a triangle, while the 6 faces of a cube are all squares, and each square can be divided into at least two triangles, resulting in a total of at least 12 triangles. Among the 4 faces of the divided tetrahedra, at most 3 faces can be on the surface of the cube, meaning each tetrah... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 34 Let $n$ be a positive integer, $a, b$ be positive real numbers, and satisfy $a+b=2$, then the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is $\qquad$ . | Since $a, b>0$, we have $a b \leqslant\left(\frac{a+b}{2}\right)^{2}=1, a^{n} b^{n} \leqslant 1$, thus $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=$ $\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+a^{n} b^{n}} \geqslant 1$, equality holds when $a=b=1$. Therefore, the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is 1. | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. In a tetrahedron $ABCD$ with volume 12, points $E, F, G$ are on edges $AB, BC, AD$ respectively, and $AE = 2EB, BF = FC, AG = 2GD$. A section $EFHG$ is made through points $E, F, G$. The distance from point $C$ to this section is 1. Then the area of this section is $\qquad$ . | 6. 7 As shown in Figure (1), it is easy to know that $G E / / D B$, so $D B / /$ plane $F E H G$, then $H F / / B D$, thus, $H$ is the midpoint of $D C$.
Let the distances from points $A, B, C, D$ to the section $E F H G$ be $h_{A}, h_{B}, h_{C}, h_{D}$, respectively, with $h_{B}=h_{C}=h_{D}=\frac{1}{2} h_{A}$. Given $... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 3 (Olympiad Training Problem from "Intermediate Mathematics" Issue 5, 2004) Let $a \in \mathbf{R}, A=\left\{x \mid 2^{1+x}+2^{1-x}=a\right\}, B=\{\sin \theta \mid$ $\theta \in \mathbf{R}\}$. If $A \cap B$ contains exactly one element, find the range of values for $a$. | Note that $B=[-1,1], 0 \in B$, obviously $a=2^{1+x}+2^{1-x} \geqslant 2 \sqrt{2^{1+x} \cdot 2^{1-x}}=4$, equality holds if and only if $2^{1+x}=2^{1-x}$, i.e., $x=0$. Therefore, $a=4$ satisfies the problem's requirements.
Next, we prove that $a>4$ do not satisfy the problem's requirements.
Assume $a>4$, and there is a ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. (16 points) Find the maximum possible value of the real number $k$ such that the inequality $\frac{2\left(a^{2}+k a b+b^{2}\right)}{(k+2)(a+b)} \geqslant \sqrt{a b}$ holds for any positive real numbers $a, b$.
Try to find the maximum possible value of the real number $k$ such that the inequality $\frac{2\left(a^{2}... | 9. The maximum possible value of $k$ is 6.
If $k=2$, the inequality is equivalent to $\frac{a+b}{2} \geqslant \sqrt{a b}$.
Now consider the case where $k>2$, then $k+2>0$, the original inequality is equivalent to
$$
\begin{array}{l}
2\left(a^{2}+k a b+b^{2}\right) \geqslant(k+2)(a+b) \sqrt{a b} \\
\Leftrightarrow 2\le... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. (6 points) Define the operation: $a \odot b=(a \times 2+b) \div 2$. Then $(4 \odot 6) \odot 8=$ | 3. (6 points) Define the operation: $a \odot b=(a \times 2+b) \div 2$. Then $(4 \odot 6) \odot 8=11$.
$$
\begin{array}{l}
=[(4 \times 2+6) \div 2] \odot 8, \\
=7 \odot 8 \\
=(7 \times 2+8) \div 2, \\
=22 \div 2, \\
=11
\end{array}
$$
Therefore, the answer is: 11. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
128 A cross-section of a parallelepiped is a pentagon, the ratio of any two sides of which is 1, 2, or 0.5, then this pentagon
A. has 4 angles of $120^{\circ}$, one angle of $60^{\circ}$
B. has 3 angles of $120^{\circ}$, two angles of $60^{\circ}$
C. all 5 angles are $108^{\circ}$
D. the size of the angles cannot be de... | 128 A. The ratio of the smallest side $x$ to any other side is 1 or 0.5, so the lengths of the sides of the pentagon can only be $x$ and $2x$. Since there are 3 pairs of parallel faces in a parallelepiped, there are two pairs of parallel sides in the pentagon. Let the vertices of the pentagon be $A, B, C, D, E$, and $B... | 1 | Geometry | MCQ | Yes | Yes | olympiads | false |
6. Positive numbers $a, b, c$ satisfy $a^{2}+b^{2}=100, a^{2}+c^{2}=81, b^{2}+c^{2}=m^{2}$, then the number of possible positive integer values for $m$ is $\qquad$. | $9$ | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. The factorial of a positive integer $m$ is the product of all positive integers up to $m$, denoted as $m!$ (for example, $2!=1 \times 2=2$, $(3!)!-(1 \times 2 \times 3)!=6!=1 \times 2 \times 3 \times 4 \times 5 \times 6=720$). If $((n!)!)!$ is a factor of $(2021!)!$, what is the maximum value of $n$? $\qquad$ . | 【Answer】6
【Analysis】For natural numbers $a, b$, $a$ ! is a factor of $b$ ! if and only if $a \leq b$. Since (( $n!)$ !)! is a factor of (2021!)!, we have $(n!)!\leq 2021!$, which means $n!\leq 2021$. $6!-1 \times 2 \times 3 \times 4 \times 5 \times 6=720,7!-7 \times 720>2021$, so the maximum value of $n$ is 6. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. On an island, there are 10 people, some of whom are honest people who always tell the truth, and others are liars who always lie. Each of them has thought of an integer. Then, the first person says: “My number is greater than 1.” The second person says: “My number is greater than 2.”...... The tenth person says: “My... | 2. There can be at most 8 honest people.
First, prove: No honest person will say "My number is greater than 9" or "My number is greater than 10."
In fact, if an honest person said either of these statements, it would mean that the number they are thinking of is not less than 10. Therefore, they would not be able to s... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
(3) Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}$ is a positive integer,
$$
a_{n+1}=\left\{\begin{array}{ll}
\frac{a_{n}}{2}, & a_{n} \text { is even, } \\
3 a_{n}+1, & a_{n} \text { is odd, }
\end{array}\right.
$$
If $a_{1}+a_{2}+a_{3}=29$, then $a_{1}=$ $\qquad$ | (3) 5 | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 1 Given $x^{2}+x+1=0$, try to find the value of the rational expression $x^{14}+\frac{1}{x^{14}}$. | Solution: Clearly, $x \neq 1$. From $x^{2}+x+1=\frac{x^{3}-1}{x-1}=0$, we get $x^{3}-1=0$, which means $x$ is a cube root of unity not equal to 1. Therefore, $x^{14}+\frac{1}{x^{14}}=x^{3 \times 4} \times x^{2}+\frac{x}{x^{3 \times 5}}=x^{2}+x=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
【Question 6】Using 2 unit squares (unit squares) can form a 2-connected square, which is commonly known as a domino. Obviously, dominoes that can coincide after translation, rotation, or symmetry transformation should be considered as the same one, so there is only one domino.
Similarly, the different 3-connected square... | Exam Point: Shape Cutting and Pasting
Analysis: There are 12 different 5-ominoes (polyominoes made of 5 unit squares) that can be intuitively remembered using the English letters F, I, L, P, T, U, V, W, X, Y, Z, which represent 11 of these shapes. | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (3 points) In the sprint stage of the 400-meter race, A is 30 meters ahead of B, C is 60 meters behind D, and B is 20 meters ahead of C. At this moment, the two leading students are ( ) meters apart.
A. 10
B. 20
C. 50
D. 60 | 【Analysis】According to the problem, draw the line segment diagram as follows: The two students running in front are Ding and Jia, with a difference of $60-20-30=$ 10 meters.
【Solution】Solution: From the analysis, we get: The two students running in front are Ding and Jia, with a difference of $60-20-30=10$ (meters). A... | 10 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
15 boys and 15 girls form a circle. Three adjacent children are called a group, making a total of 30 different groups. Among them, 10 groups are all boys, and 5 groups have 2 boys and 1 girl. Therefore, there are $\qquad$ groups that are all girls. | $10$ | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. If the functions $f(x)$ and $g(x)$ are defined on $\mathbf{R}$, and $f(x-y)=f(x) g(y)-g(x) f(y)$, if $f(-2)=f(1) \neq 0$, then $g(1)+g(-1)=$ (answer with a number).
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 4. $\begin{array}{l} \because f(x-y)=f(x) g(y)-\dot{g}(x) f(y), \\ \therefore f(y-x)=f(y) g(x)-g(y) f(x)=-[f(x) g(y)-g(x) f(y)] . \\ \therefore f(x-y)=-f(y-x)=-f[-(x-y)], \\ \therefore f(-x)=-f(x), \text{ i.e., } f(x) \text{ is an odd function. } \\ \therefore f(1)=f(-2)=f(-1-1)=f(-1) g(1)-g(-1) f(1) \\ =-f(1)[g(1)+g(-... | -1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.95 For the number $123456789101112 \cdots 9989991000$, perform the following operation: strike out two adjacent digits $a$ and $b$ ($a$ is before $b$), and then insert the number $a+2b$ at their original position (the first digit of the number can be taken as $b$, and the digit before the number can be taken as $a$).... | ［Proof］First note:
(1) Adding 0 before a 1-digit number and then performing the operation, the resulting number is larger than the original number, and this is not the case for other operations;
(2) The only number group that remains unchanged under this operation is 19.
Given these two points, we perform the operation... | 1 | Number Theory | proof | Yes | Yes | olympiads | false |
East and West towns are 45 kilometers apart. Person A and Person B start from the two towns at the same time and walk towards each other. A walks 1 kilometer more per hour than B. After 5 hours, the two meet, so A walks $\qquad$ kilometers per hour.
| 【Analysis and Solution】
Journey, Meeting Problem.
$v_{\text {甲 }}+v_{\text {乙 }}=45 \div 5=9$ km/h
$v_{\text {甲 }}-v_{\text {乙 }}=1$ km/h;
$v_{\text {甲 }}=(9+1) \div 2=5$ km/h. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Question: $p$ is a prime number, and the sum of all positive divisors of $p^{4}$ is exactly a perfect square. Then the number of prime numbers $p$ that satisfy the above condition is ( ).
(A) 3 ;
(B) 2 ;
(C) 1 ;
(D) 0 . | Select (C).
Since $p$ is a prime number, $p^{4}$ has 5 positive divisors: $1, p, p^{2}, p^{3}, p^{4}$.
According to the problem,
then
$$
1+p+p^{2}+p^{3}+p^{4}=n^{2} \quad(n \in \mathbf{N}) .
$$
then
$$
\begin{aligned}
(2 n)^{2} & >4 p^{4}+4 p^{3}+p^{2}=\left(2 p^{2}+p\right)^{2} . \\
(2 n)^{2} & >4 p^{4}+p^{2}+4+4 p^... | 1 | Number Theory | MCQ | Yes | Yes | olympiads | false |
9. (3 points) Four children, A, B, C, and D, are having a chess competition. Each pair of them plays one game. The winner gets 3 points, a draw gets 1 point, and the loser gets 0 points. In the end, D gets 6 points, B gets 4 points, and C gets 2 points. So, how many points does A get? $\qquad$ points. | 【Answer】Solution: Each two people play one game, which means a total of 6 games are played, and each person plays three games;
Ding getting six points indicates: winning two games and losing one game $(3+3+0=6)$;
Yi getting four points indicates: winning one game, drawing one game, and losing one game $(3+1+0=4)$;
Bing... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9. In the sequence $\left\{a_{n}\right\}$, $a_{4}=1, a_{11}=9$, and the sum of any three consecutive terms is 15. Then $a_{2016}=$ $\qquad$ | For any $n \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
a_{n}+a_{n+1}+a_{n+2}=a_{n+1}+a_{n+2}+a_{n+3}=15 \\
\Rightarrow a_{n+3}=a_{n} .
\end{array}
$$
Then $a_{1}=a_{4}=1, a_{2}=a_{11}=9$,
$$
a_{3}=15-a_{1}-a_{2}=5 \text {. }
$$
Therefore, $a_{2016}=a_{3 \times 672}=a_{3}=5$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 12 The sports meet lasted for $n$ days $(n>1)$, and a total of $m$ medals were awarded. On the first day, 1 medal was awarded, and then $\frac{1}{7}$ of the remaining $m-1$ medals were awarded. On the second day, 2 medals were awarded, and then $\frac{1}{7}$ of the remaining medals were awarded, and so on, unti... | Let $a_{k}$ be the number of medals awarded on the $k$-th day, then
$$
a_{1}=1+\frac{1}{7}(m-1)=\frac{1}{7}(m+6), a_{k}=k+\frac{1}{7}\left(m-a_{1}-a_{2}-\cdots-a_{k-1}-k\right) \text {, }
$$
Thus, $a_{k}-a_{k-1}=1+\frac{1}{7}\left(-a_{k-1}-1\right)$, which means $a_{k}=\frac{6}{7} a_{k-1}+\frac{6}{7}$.
Since the fixed... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
19.2.3 * For a positive integer $k$, there exist positive integers $n, m$, such that $\frac{1}{n^{2}}+\frac{1}{m^{2}}=\frac{k}{n^{2}+m^{2}}$, find all such $k$. | From the original equation, we get $\left(\frac{n}{m}+\frac{m}{n}\right)^{2}=k$. Without loss of generality, assume $m$ and $n$ are coprime (otherwise, factor out the common divisor). Since $\frac{n}{m}+\frac{m}{n}$ is an integer, $n+\frac{m^{2}}{n}$ is also an integer, thus $n=1$. Similarly, $m=1$. Therefore, $k=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Define the length of intervals $(c, d),[c, d],(c, d],[c, d)$ to be $d-c$, where $d>c$. Given real numbers $a > b$, the length sum of intervals formed by $x$ that satisfy $\frac{1}{x-a}+\frac{1}{x-b} \geqslant 1$ is $\qquad$ . | 8. 2 Detailed Explanation: From $\frac{1}{x-a}+\frac{1}{x-b} \geqslant 1$, we get $\frac{x-b+x-a-[(x-a)(x-b)]}{(x-a)(x-b)} \geqslant 0$. Simplifying, we obtain $-\frac{x^{2}-(a+b+2) x+a b+a+b}{(x-a)(x-b)} \geqslant 0$. Let $g(x)=x^{2}-(a+b+2) x+a b+a+b$, then $g(a)=b-a<0$, and $g(b)=a-b<0$. Since $a>b$, we have $b<x_{1... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let $F_{1}, F_{2}$ be the left and right foci of the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, respectively. A line passing through point $F_{1}$ intersects the ellipse $E$ at points $A$ and $B$, with $\left|A F_{1}\right|=3\left|B F_{1}\right|$ and $\cos \angle A F_{2} B=\frac{3}{5}$. As shown ... | 3. -1
Analysis: Let $\left|F_{1} B\right|=k$, then $k>0$ and $\left|A F_{1}\right|=3 k,|A B|=4 k$. By the definition of an ellipse, we have $\left|A F_{2}\right|=2 a-3 k,\left|B F_{2}\right|=2 a-k$. In $\triangle A B F_{2}$, by the cosine rule, we get $|A B|^{2}=\left|A F_{2}\right|^{2}+\left|B F_{2}\right|^{2}-2\left... | -1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$8 \cdot 7$ If $x, y$ are integers, the number of solutions to the equation $(x-8)(x-10)=2^{y}$ is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) More than 3.
(13th American High School Mathematics Examination, 1962) | ［Solution］ The given equation can be transformed into
$$
x^{2}-18 x+80-2^{y}=0 \text {. }
$$
Considering it as an equation in $x$, we have
$$
x=9 \pm \sqrt{1+2^{y}} \text {, }
$$
where $x$ is an integer if and only if $1+2^{y}$ is a perfect square.
That is, $1+2^{y}=n^{2}$ or $n^{2}-1=2^{y}$.
Since $n^{2}-1=(n-1)(n+1... | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
15. Given that $F$ is the right focus of the ellipse $\frac{x^{2}}{1+a^{2}}+y^{2}=1(a>0)$, points $M(m, 0), N(0, n)$ are points on the $x$-axis and $y$-axis respectively, and satisfy $\overrightarrow{M N} \cdot \overrightarrow{N F}=0$. If point $P$ satisfies $\overrightarrow{O M}=2 \overrightarrow{O N}+\overrightarrow{... | (1) $F(a, 0)$, from $\overrightarrow{M N} \cdot \overrightarrow{N F}=-m a+n(-n)=0$
$\Rightarrow m=-\frac{n^{2}}{a}$, thus $P(-m, 2 n)=\left(\frac{n^{2}}{a}, 2 n\right)$.
Therefore, the equation of the trajectory $C$ of point $P$ is $y^{2}=4 a x$.
(2) Let $A B: x=k y+a$, combining $\left\{\begin{array}{l}x=k y+a, \\ y^{... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. In Rt $\triangle A B C$, $\angle A C B=\frac{\pi}{2}, A C=B C$ $=2, P$ is a point on the hypotenuse $A B$, and $B P=2 P A$. Then $\overrightarrow{C P} \cdot \overrightarrow{C A}+\overrightarrow{C P} \cdot \overrightarrow{C B}=$ $\qquad$ | 9.4 .
Notice,
$$
\begin{array}{l}
\overrightarrow{C P}=\overrightarrow{C A}+\overrightarrow{A P}=\overrightarrow{C A}+\frac{1}{3} \overrightarrow{A B} \\
=\overrightarrow{C A}+\frac{1}{3}(\overrightarrow{A C}+\overrightarrow{C B})=\frac{2}{3} \overrightarrow{C A}+\frac{1}{3} \overrightarrow{C B} \\
\text { Therefore, ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. The last digit of $\sum_{k=0}^{201}(10 k+7)^{k+1}$ is $\qquad$ | 7. 6 .
$$
\text{It is easy to see that } \sum_{k=0}^{201}(10 k+7)^{k+1} \equiv \sum_{k=0}^{201} 7^{k+1}(\bmod 10) \text{.}
$$
Notice that, for any natural number $n$, the last digits of $7^{4 n+1}$, $7^{4 n+2}$, $7^{4 n+3}$, and $7^{4 n+4}$ are $7$, $9$, $3$, and $1$ respectively, and the last digit of their sum is 0.... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
15 Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}, a \neq 0)$ satisfy the following conditions:
（1）For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and $f(x) \geqslant x$;
（2）For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.
Find the l... | Analyze: First, derive the analytical expression of $f(x)$ from the given conditions, then discuss $m$ and $t$, and finally determine the maximum value of $m$.
Solution: Since $f(x-4) = f(2-x)$ for $x \in \mathbf{R}$, it is known that the axis of symmetry of the quadratic function $f(x)$ is $x = -1$. From (3), we know... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 2 The equation about $x$ and $y$
$$
x^{2}+x y+2 y^{2}=29
$$
has ( ) groups of integer solutions $(x, y)$.
(A) 2
(B) 3
(C) 4
(D) infinitely many
(2009, National Junior High School Mathematics Competition) | Solve: Regarding the original equation as a quadratic equation in $x$, it can be transformed into
$$
x^{2}+y x+\left(2 y^{2}-29\right)=0 \text {. }
$$
Since the equation has integer roots, the discriminant $\Delta$ must be a perfect square.
$$
\begin{array}{l}
\text { By } \Delta=y^{2}-4\left(2 y^{2}-29\right) \\
=-7 ... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
(4) If real numbers $a$, $b$, $c$ satisfy: $a+b+c=a^{2}+b^{2}+c^{2}$, then the maximum value of $a+b+c$ is $\qquad$ . | 4 3 Hint: According to the Cauchy-Schwarz inequality,
$$
3(a+b+c)=\left(1^{2}+1^{2}+1^{2}\right)\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+b+c)^{2},
$$
thus $a+b+c \leqslant 3$, with equality when $a=b=c=1$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$17 \cdot 78$ Two rays starting from $O$ form an angle of $30^{\circ}$, point $A$ is on one ray, point $B$ is on the other, and $A B=1$. Then the maximum length of $O B$ is
(A) 1 .
(B) $\frac{1+\sqrt{3}}{\sqrt{2}}$
(C) $\sqrt{3}$.
(D) 2 .
(E) $\frac{4}{\sqrt{3}}$.
(46th American High School Mathematics Examination, 199... | [Solution] Let $O A=x, O B=y$. In $\triangle A B O$, by the cosine rule we have
$$
x^{2}+y^{2}-\sqrt{3} x y=1 \text {, }
$$
which simplifies to $x^{2}-\sqrt{3} y x+\left(y^{2}-1\right)=0$.
Considering the above equation as a quadratic in $x$ with real roots, we have
$$
\Delta=3 y^{2}-4\left(y^{2}-1\right) \geqslant 0 ... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
2. Let $\triangle A B C$ have internal angles $A, B, C$ with opposite sides $a, b, c$ respectively, and satisfy $a \cos B - b \cos A = \frac{3}{5} c$. Then the value of $\frac{\tan A}{\tan B}$ is . $\qquad$ | $$
\begin{array}{l}
a \cos B - b \cos A = \frac{3}{5} c \Rightarrow \sin A \cos B - \sin B \cos A = \frac{3}{5}(\sin A \cos B + \sin B \cos A) \\
\Rightarrow 2 \sin A \cos B = 8 \sin B \cos A \Rightarrow \tan A = 4 \tan B \Rightarrow \frac{\tan A}{\tan B} = 4
\end{array}
$$ | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
17・193 In $\triangle A B C$ (as shown in the figure) $\angle A B C=$ $60^{\circ}, \angle A C B=45^{\circ}$, and $A D, C F$ are both altitudes of the triangle, intersecting at $P$, while the angle bisector $B E$ of $\angle A B C$ intersects $A D, C F$ at $Q, S$. The number of isosceles triangles in the figure is
(A) 2.
... | [Solution] It is easy to know that $\triangle Q A B, \triangle S B C, \triangle B A E, \triangle P Q S$ and $\triangle D A C$ are all isosceles triangles, a total of 5.
Therefore, the answer is $(D)$ | 5 | Geometry | MCQ | Yes | Yes | olympiads | false |
1. (10 points) A four-digit number, with all digits being distinct, the sum of all digits equals 6, and the number is a multiple of 11. How many such four-digit numbers are there?
A. 6
B. 7
C. 8
D. 9 | 【Analysis】Given this four-digit number, all digits are different, and the sum of all digits equals 6. Therefore, the four digits that make up the four-digit number are $0, 1, 2, 3$. This number is a multiple of 11, so the sum of the digits in the odd positions equals the sum of the digits in the even positions, which i... | 6 | Number Theory | MCQ | Yes | Yes | olympiads | false |
4. For real numbers $a, b, c$, satisfying:
$$
a+b, b+c, c+a \in [0,1] \text{.}
$$
Find the maximum value of $a^{2}+b^{2}+c^{2}$. | 4. By symmetry, without loss of generality, let $a \geqslant b \geqslant c$.
Since $b-c$ and $b+c$ are both non-negative real numbers, we have
$$
b^{2} \geqslant c^{2} \text {. }
$$
Similarly, $1-b-a$ and $1-b+a$ are both non-negative real numbers.
$$
\begin{array}{l}
\text { Then }(1-b)^{2} \geqslant a^{2} \\
\Right... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
94. Fill in 9 different natural numbers in a 3x3 grid, satisfying: in each row, the sum of the two left numbers equals the rightmost number; in each column, the sum of the two upper numbers equals the bottom number. What is the smallest number that can be in the bottom-right corner? $\qquad$ | answer: 12 | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. In the sequence $1,3,2, \cdots$, each term after the first two is equal to the difference of the two preceding terms (the preceding term minus the term before it). The sum of the first 100 terms of this sequence is
A. 5
B. 4
C. 2
D. 1 | 2. According to the problem, the first eight terms of this sequence are $1,3,2,-1,-3,-2,1,3$. Since the seventh and eighth terms are the same as the first and second terms, the ninth term will be the same as the third term, and so on; that is, this sequence repeats every 6 terms, i.e.,
$a_{k}=a_{k+6}(k=1,2,3, \cdots) .... | 5 | Number Theory | MCQ | Yes | Yes | olympiads | false |
$33 \cdot 13$ If $N=\frac{1987^{4(\sqrt{3}+2)^{1987}+1988}}{1987^{(\sqrt{3}+2)^{1988}}+(\sqrt{3}+2)^{1988}}$, then the last digit of $N$ is
(A) 1 .
(B) 3 .
(C) 7 .
(D) 9 .
(China Sichuan Province Junior High School Mathematics Competition, 1988) | ［Solution］By the properties of exponential operations, we also have:
$$
\begin{array}{l}
=1987^{(\sqrt{3}+2)^{1986}}\left[4(\sqrt{3}+2)-(\sqrt{3}+2)^{2}-1\right]+1988 \\
=1987^{1988}=1987^{497 \cdot 4} \text {, } \\
\end{array}
$$
Thus, the last digit of $N$ is 1 (the last digit of $7^{k}$ cycles through $7,9,3,1$). T... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
Let $a, b, c, d$ be four distinct real numbers, satisfying: $a+b+c+d=3, a^{2}+$ $b^{2}+c^{2}+d^{2}=45$. Find $\frac{a^{5}}{(a-b)(a-c)(a-d)}+\frac{b^{5}}{(b-a)(b-c)(b-d)}+$ $\frac{c^{5}}{(c-a)(c-b)(c-d)}+\frac{d^{5}}{(d-a)(d-b)(d-c)}$.
| Let $S_{n}=\frac{a^{n}}{(a-b)(a-c)(a-d)}+\frac{b^{n}}{(b-a)(b-c)(b-d)}+$ $\frac{c^{n}}{(c-a)(c-b)(c-d)}+\frac{d^{n}}{(d-a)(d-b)(d-c)}$. By the Lagrange interpolation formula, we know
$$
\begin{array}{c}
\frac{(x-b)(x-c)(x-d)}{(a-b)(a-c)(a-d)} a^{3}+\frac{(x-a)(x-c)(x-d)}{(b-a)(b-c)(b-d)} b^{3}+ \\
\frac{(x-a)(x-b)(x-d)... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$23 . \star \star \triangle A B C$ is inscribed in a unit circle, the angle bisectors of the three interior angles $A, B, C$ are extended to intersect this circle at $A_{1}, B_{1}, C_{1}$. Then the value of $\frac{A A_{1} \cdot \cos \frac{A}{2}+B B_{1} \cdot \cos \frac{B}{2}+C C_{1} \cdot \cos \frac{C}{2}}{\sin A+\sin ... | Parse as shown in the figure, connect $B A_{1}$, then $A A_{1}=2 \sin \left(B+\frac{A}{2}\right)=$ $2 \sin \left(\frac{A+B+C}{2}+\frac{B}{2}-\frac{C}{2}\right)=2 \cos \left(\frac{B}{2}-\frac{C}{2}\right)$. Therefore,
$$
\begin{aligned}
A A_{1} \cdot \cos \frac{A}{2} & =2 \cos \left(\frac{B}{2}-\frac{C}{2}\right) \cos \... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
4. In a spinning wheel game, every spin can earn the player the score corresponding to the area the pointer lands on, and the scores are accumulated. A player wins if the total score is exactly 100. If Guangtou Qiang wins with the fewest number of spins, then the number of times he could get 17 points is $\qquad$ possi... | $4$ | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
13. (3 points) A cold drink shop launched a "Buy one get the second at half price" promotion. Xiaogang bought 2 cups of drinks and spent a total of 13 yuan 5 jiao. What is the original price of one cup of drink? $\qquad$ yuan. | 【Analysis】Regarding the original price of the first drink as unit "1", the price of the second drink is $\frac{1}{2}$ of the first drink. According to the problem: the price of the first drink $\left(1+\frac{1}{2}\right)$ is 13.5 yuan. Based on the known fraction of a number, to find the number, use division.
【Solutio... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Let the complex number $z=\cos \theta+\mathrm{i} \sin \theta\left(0^{\circ} \leqslant \theta \leqslant 180^{\circ}\right)$, and the complex numbers $z 、(1+\mathrm{i}) z 、 2 \bar{z}$ correspond to three points $P 、 Q 、 R$ on the complex plane. When $P 、 Q 、 R$ are not collinear, the fourth vertex of the parallelogram... | $$
\begin{array}{l}
(1+\mathrm{i}) z=(\cos \theta-\sin \theta)+(\cos \theta+\sin \theta) \mathrm{i}, 2 \bar{z}=2 \cos \theta-2 \mathrm{i} \sin \theta, \\
\Rightarrow P(\cos \theta, \sin \theta), Q(\cos \theta-\sin \theta, \cos \theta+\sin \theta), R(2 \cos \theta,-2 \sin \theta) . \\
\text { Thus } \overrightarrow{P S}... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$36 \cdot 8$ Let $A$ be an $n$-digit positive integer, $n \geqslant 2$, and $B$ be a $k$-digit positive integer, $k \geqslant 1$. Then there are $n-1$ ways to insert $B$ entirely between two adjacent digits of $A$, resulting in an $n+k$-digit positive integer $C$. For example, if $A=1991$ and $B=35$, there are three in... | ［Solution］By checking each number, we can easily find:
(1) 1 is a harmonious number;
(2) If $A$ is divisible by 2, then the last digit of $A$ is even. Inserting 2 arbitrarily into $A$ results in the last digit of $C$ still being even, so $C$ is divisible by 2. Therefore, 2 is a harmonious number;
(3) Note that: A natur... | 11 | Number Theory | MCQ | Yes | Yes | olympiads | false |
1. The minimum value of the function $f(x)=\frac{5-4 x+x^{2}}{2-x}$ on $(-\infty, 2)$ is
A. 0
B. 1
C. 2
D. 3 | $f(x)=2-x+\frac{1}{2-x} \geqslant 2$, equality holds when $x=1$. So the answer is $C$. | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
14. (12 points) From a large grid paper, cut out 5 connected squares (two squares that only share one vertex are not considered connected), so that the cut-out shape can be folded into an open-top cube. The total number of different shapes that can be cut out (shapes that are the same after rotation or flipping are con... | 【Answer】Solution: According to the problem, the cut-out shapes are as shown in the figure: there are 8 types in total.
Therefore, the correct choice is: A. | 8 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
$\begin{array}{l}\text { 1. Given } x+\lg x=10, y+10^{y}=10 \text {. Then } x+y \\ \text {. }\end{array}$ | 1. 10 .
Solution: Let $f(t)=t+10^{t}$, then $f(t)$ is a strictly increasing function, so there exists a unique real number $y$ such that $y+10^{y}=10$. Also, $\lg x=10$ $-x$, so $10^{10-x}=x$, which means $(10-x)+10^{10-x}=10$. Therefore, $10-x=y$, which implies $x+y=10$. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. If real numbers $x, y$ satisfy $(x+5)^{2}+(y-12)^{2}=14^{2}$, then the minimum value of $x^{2}+y^{2}$ is
(A) 2
(B) 1
(C) $\sqrt{3}$
(D) $\sqrt{2}$ | (B)
2.【Analysis and Solution】 $(x+5)^{2}+(y-12)^{2}=14^{2}$ is a circle with center at point $C(-5,12)$ and radius 14. Let $P$ be any point on the circle, then $|O P| \geqslant|C P|-|O C|=$ $14-13=1$
When points $C, O, P$ are collinear, the equality holds, so the minimum value of $P$ to point $O$ is 1, hence the answe... | 1 | Geometry | MCQ | Yes | Yes | olympiads | false |
5. Given the set $A=\left\{x \mid x^{2}-2 x-3=0\right\}, B=\{x \mid a x=1\}$. If $B \subseteq A$, then the product of all possible values of the real number $a$ is ( ).
A. -.1
B. $-\frac{1}{3}$
C. 0
D. None of the above | 5. C. When $a=0$, $B=\varnothing \subseteq A$; when $a \neq 0$, $B=\left\{\frac{1}{a}\right\}$, then $\frac{1}{a}=-1$ or 3, i.e., $a=-1$ or $\frac{1}{3}$. | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
9. Xiao Ming is playing a ring toss game, scoring 9 points for each chicken, 5 points for each monkey, and 2 points for each dog. Xiao Ming tossed 10 times, hitting every time, and each toy was hit at least once. Xiao Ming scored a total of 61 points in 10 tosses. Question: What is the minimum number of times the chick... | 9.【Solution】Catching a chick, a monkey, and a dog once each, for a total of $16(=9+5+2)$ points, so there are still $7(=10-3)$ times, totaling $45(=61-16)$ points. In these 7 times, the maximum number of times a chick can be caught is 4 (since $9 \times 5$ is already 45 points, so it's impossible to catch a chick 5 tim... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (8 points) In another 12 days, it will be 2016, Hao Hao sighs: I have only experienced 2 leap years so far, and the year I was born is a multiple of 9. So how old will Hao Hao be in 2016? | 4. (8 points) In another 12 days, it will be 2016, HaoHao sighs: I have only experienced 2 leap years so far, and the year I was born is a multiple of 9, so in 2016, HaoHao is $\qquad$ years old.
【Solution】Solution: HaoHao has only experienced 2 leap years so far. Counting backwards from 2015, the two leap years are 2... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13. Peter, Emma, and Keller play chess. Peter won 4 games and lost 2 games; Emma won 3 games and lost 3 games. If Keller lost 3 games, then he won ( ) games.
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 | 13. B.
It is clear that in each game, one side must win and the other must lose.
Therefore, from the total number of games played being equal to the total number of losses, we have
$$
(2+3+3)-(4+3)=1 \text{, }
$$
which means Keller won one game. | 1 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
16. Given $A=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}$, then the integer part of $A$ is | Analysis: Fractions with denominators in multiple relationships can be added together first.
That is: $A=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\right)+\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{5}+\frac{1}{7}\right)=\frac{11}{18}+\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{5}+\frac{1}{7}\right)$,
By the Squeeze Theorem, we g... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
137 Given that $x, y, z$ are all real numbers, and $x^{2}+y^{2}+z^{2}=x+y+z$, then the sum of the maximum and minimum values of $x+y+z$ is
A. 2
B. $2 \sqrt{2}$
C. 3
D. $2 \sqrt{3}$ | 137 C. Solution 1: It is clear that the minimum value of $x+y+z$ is 0. Also, $x+y+z+3=\left(x^{2}+1\right)+$ $\left(y^{2}+1\right)+\left(z^{2}+1\right) \geqslant 2 \cdot x+2 \cdot y+2 \cdot z$, from which we get $x+y+z \leqslant 3$. The equality holds when $x=$ $y=z=1$ and satisfies the given equation, so the maximum v... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. $\triangle A B C$ has an interior angle of $60^{\circ}, \sin 3 A+\sin 3 B+\sin 3 C=$ | 7. 0. $\sin 3 A+\sin 3 B+\sin 3 C=-4 \cos \frac{3 A}{2} \cos \frac{3 B}{2} \cos \frac{3 C}{2}=0$.
Translate the above text into English, keeping the original text's line breaks and format:
7. 0. $\sin 3 A+\sin 3 B+\sin 3 C=-4 \cos \frac{3 A}{2} \cos \frac{3 B}{2} \cos \frac{3 C}{2}=0$. | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.47 The number of real roots of the equation $x=\sin x+1993$ is .
(A) 0 .
(B) 1 .
(C) 2 .
(D) More than 2 .
(China Jiangsu Province Suzhou City High School Mathematics Competition, 1993) | [ Sol] Since the roots of the equation are the x-coordinates of the intersection points of the line $y=x$ and the curve $y=\sin x+1993$, the equation has at least one real root.
Assume the equation has two real roots $x_{1}, x_{2}$, and $x_{1} \neq x_{2}$, then
$$
\begin{array}{l}
x_{1}=\sin x_{1}+1993, \quad x_{2}=\si... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
6. (10 points) The sequence $1,1,2,3,5,8, \cdots$ is such that each term from the second onward is equal to the sum of the two preceding terms, forming the Fibonacci sequence. Each term in this sequence is called a Fibonacci number. It can be proven that "any positive integer $n$ can be expressed as the sum of several ... | 【Solution】Solution: First, list out the Fibonacci numbers less than 100. $1,1,2,3,5,8,13,21,34,55$, 89.
(1) $100=89+3+8$
(2) $=89+1+2+8$
(3) $=89+1+2+3+5$
(4) $=55+34+1+2+3+5$
(5) $=55+34+1+2+8$
(6) $=55+34+3+8$
(7) $=55+13+21+1+2+3+5$
(8) $=55+13+21+3+8$
(9) $=55+13+21+1+2+8$
Therefore, the answer is: 9 | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Given that the function $f(x+1)$ is an odd function, the function $f(x-1)$ is an even function, and $f(0)=2$, then $f(4)=$ | $$
f(4)=f(3+1)=-f(-3+1)=-f(-2)=-f(-1-1)=-f(1-1)=-f(0)=-2 .
$$ | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find the last non-zero digit of $50!=1 \times 2 \times \cdots \times 50$. | 4. Notice that, in the prime factorization of 50!, 5 appears 12 times, and 2 appears more than 12 times. Therefore, the last non-zero digit of 50! is the unit digit of $\frac{50!}{2^{12} \times 5^{12}}$, and it is an even number.
Since $2^{12} \equiv 1(\bmod 5)$, we have
$$
\frac{50!}{2^{12} \times 5^{12}} \equiv \frac... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (5 points) 2015 minus its $\frac{1}{2}$, then minus the remaining $\frac{1}{3}$, then minus the remaining $\frac{1}{4}, \cdots$, and finally minus the remaining $\frac{1}{2015}$, the final number obtained is $\qquad$ . | 6. (5 points) 2015 minus its $\frac{1}{2}$, then minus the remaining $\frac{1}{3}$, then minus the remaining $\frac{1}{4}, \cdots$, and finally minus the remaining $\frac{1}{2015}$, the final number obtained is $\qquad$ .
【Solution】Solution: $2015 \times\left(1-\frac{1}{2}\right) \times\left(1-\frac{1}{3}\right) \time... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$14 \cdot 10$ The numbers from 1 to 9 are written on nine pieces of paper placed in a hat. Jack randomly draws one and puts it back, then Jill also randomly draws one. What is the most likely digit in the units place of the sum of the numbers drawn by Jack and Jill?
(A) 0.
(B) 1.
(C) 8.
(D) 9.
(E) Each digit is equally... | [Solution] The unit digit of the sum of numbers $1 \sim 9$ and $1 \sim 9$ can be shown in the following table:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline$\langle m+n\rangle^{m}$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 0 \\
\hline 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 0 & 1 \\
\hli... | 0 | Number Theory | MCQ | Yes | Yes | olympiads | false |
9. (15 points) A store purchased a batch of pens. If it sells 20 pens at a retail price of 7 yuan each and 15 pens at a retail price of 8 yuan each, the profit earned is the same. What is the purchase price of each pen in yuan?
---
Please note that the translation retains the original format and structure, including ... | 9. (15 points) A store purchased a batch of pens. If it sells 20 pens at a retail price of 7 yuan each and 15 pens at a retail price of 8 yuan each, the profit earned is the same. What is the purchase price of each pen in yuan?
【Analysis】Using equations to solve this problem is easier to understand. Let the purchase pr... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. (20 points) Let $\lambda$ be a positive real number. For any pairwise distinct positive real numbers $a, b, c$, we have
$$
\frac{a^{3}}{(b-c)^{2}}+\frac{b^{3}}{(c-a)^{2}}+\frac{c^{3}}{(a-b)^{2}} \geqslant \lambda(a+b+c) \text {. }
$$
Find the maximum value of $\lambda$. | 10. Let $a=\frac{1}{2}, b=\frac{1}{2}-\varepsilon, c=\varepsilon\left(0<b<c$. We can set
$$
a=c+x, b=c+y(x>y>0) \text {. }
$$
Then the left side of equation (1)
$$
\begin{array}{l}
=\frac{(c+x)^{3}}{y^{2}}+\frac{(c+y)^{3}}{x^{2}}+\frac{c^{3}}{(x-y)^{2}} \\
=\frac{x^{3}+3 x^{2} c+3 x c^{2}+c^{3}}{y^{2}}+ \\
\frac{y^{3}... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. $A, B, C, D, E$ five people participate in a chess tournament together and find that their average age is exactly 28 years old. One year later, $A, B, C, D, F$ participate together and find that their average age is exactly 30 years old. How many years older is $F$ than $E$? | II. 1.5 years.
Let the age of $E$ in the first year of the competition be $x$, and the sum of the ages of the other four people be $a$. In the second year of the competition, the age of $F$ is $y$. Then, from the given information, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{a+x}{5}=28, \\
\frac{a+4+y}{5}... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. If $M=\left\{z \left\lvert\, z=\frac{t}{1+t}+i \frac{1+t}{t}\right., t \in \mathbf{R}, t \neq-1, t \neq 0\right\}, N=\{z \mid z=\sqrt{2}[\cos (\arcsin t)+i \cos$ $(\arccos t)], t \in \mathbf{R},|t| \leqslant 1\}$, then the number of elements in $M \cap N$ is
( A ) 0
(B) 1
(C) 2
(D) 4 | (A)
5 【Analysis and Solution】The points in $M$ lie on the curve
$M:\left\{\begin{array}{l}x=\frac{t}{1+t} \\ y=\frac{1+t}{t}\end{array} \quad(t \in \mathbf{R}, t \neq 0,-1)\right.$: The points in $N$ lie on the curve
$N:\left\{\begin{array}{l}x=\sqrt{2\left(1-t^{2}\right)} \\ y=\sqrt{2 t}\end{array} \quad(t \in \mathbf... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
5.69 11 theater troupes participate in a performance, with some of the troupes scheduled to perform each day, while the rest join the general audience. By the end of the performance, each troupe, apart from their own performance days, must have watched at least one performance by each of the other troupes. How many day... | [Solution] Let $N=\{1,2, \cdots, n\}$ be the set of performance dates.
Each theater group can be represented by a subset $A \subset N$: Theater group $A$ performs on day $x$ if and only if $x \in A$.
For two theater groups $A$ and $B$ to watch each other's performances at least once, there must be at least one day tha... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9 In the quadrilateral pyramid $P-ABCD$, $\overrightarrow{AB}=(4,-2,3)$, $\overrightarrow{AD}=(-4,1,0)$, $\overrightarrow{AP}=(-6,2,-8)$, then the height $h$ of this quadrilateral pyramid is $h=(\quad)$.
(A) 1
(B) 2
(C) 13
(D) 26 | (9) Let the height of the pyramid be $P H, \overrightarrow{P H}=(x, y, z)$, then
$$
\begin{array}{l}
\overrightarrow{P H} \cdot \overrightarrow{A B}=4 x-2 y+3 z=0, \\
\overrightarrow{P H} \cdot \overrightarrow{A D}=-4 x+y=0,
\end{array}
$$
Solving, we get $y=4 x, z=\frac{4}{3} x$. Therefore,
$$
\overrightarrow{P H}=\l... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
8. Let $a \in \mathbf{R}$, and the complex numbers
$$
z_{1}=a+\mathrm{i}, z_{2}=2 a+2 \mathrm{i}, z_{3}=3 a+4 \mathrm{i} \text {. }
$$
If $\left|z_{1}\right|, \left|z_{2}\right|, \left|z_{3}\right|$ form a geometric sequence, then the value of the real number $a$ is . $\qquad$ | 8. 0 .
From the condition, we know that $\left|z_{1}\right|\left|z_{3}\right|=\left|z_{2}\right|^{2}$.
Notice that, $\left|z_{2}\right|=2\left|z_{1}\right|$.
Then $\left|z_{3}\right|=4\left|z_{1}\right|$
$$
\begin{array}{l}
\Rightarrow \sqrt{9 a^{2}+16}=4 \sqrt{a^{2}+1} \\
\Rightarrow a=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. For any shape, the smallest positive integer $n$ that makes the total number of lattice points inside and on the boundary of a convex $n$-sided polygon $\geqslant n+1$ is ( )
A. 4
B. 5
C. 6
D. 7 | 1. B Consider the square $A B C D$ with vertices $A(0,0), B(0,1), C(1,1), D(1,0)$. There are no lattice points inside this square, so the required $n \geqslant 5$.
On the other hand, by classifying lattice points based on their parity, they can be divided into four categories: (odd, odd), (odd, even), (even, odd), (ev... | 5 | Geometry | MCQ | Yes | Yes | olympiads | false |
Example 32 (2004 Slovenia National Team Selection Test) Find all positive integers $n$, such that $n \cdot 2^{n-1} +$ 1 is a perfect square. | Let $n \cdot 2^{n-1} + 1 = m^2$, i.e.,
$$
n \cdot 2^{n-1} = (m+1)(m-1).
$$
It is evident that $n=1, n=2, n=3$ do not satisfy the condition.
Assume $n > 3$, then the value on the left side of equation (1) is even, hence $m$ must be odd.
Let $m = 2k + 1$, thus we have $n \cdot 2^{n-3} = k(k+1)$.
Since the consecutive in... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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