Given a positive integer $n \geqslant 2$. Let integers $a_{0}, a_{1}, \cdots$, $a_{n}$ satisfy $0=a_{0}<a_{1}<\cdots<a_{n}=2 n-1$. Find the minimum possible number of elements in the set $\left\{a_{i}+a_{j} \mid 0 \leqslant i \leqslant j \leqslant n\right\}$.

Solution:
Hint: The minimum possible value is $3n$, start with an estimate.
First, prove: $a_{i}+a_{j}(0 \leqslant i \leqslant j \leqslant n-1)$ runs through a complete residue system modulo $2n-1$.

Since $a_{i}+a_{j}(0 \leqslant i \leqslant j \leqslant n)$ runs through a complete residue system modulo $2n-1$, thus, each of the other $n-1$ residue classes contains at least one number, resulting in $n-1$ numbers that are all distinct from the $2n$ numbers.

Additionally, $a_{n}+a_{n}=4n-2$ is the largest number and is necessarily distinct from the other numbers. Therefore, there are at least $2n+n-1+1=3n$ numbers. When taking $a_{i}=i(i=0,1, \cdots, n-1), a_{n}=2n-1$, the values of $a_{i}+a_{j}(0 \leqslant i \leqslant j)$ are
$$
0,1,2, \cdots, 2n-2,2n-1, \cdots, 3n-2,4n-2
$$
a total of $3n$ numbers.

In summary, the minimum possible value of the number of elements in the set $\left\{a_{i}+a_{j} \mid 0 \leqslant i \leqslant j \leqslant n\right\}$ is $3n$.

Answer: 3n


A reservoir in the form of a hemisphere with an internal radius of $R(m)$ is being filled with water at a rate of $Q(l)$ per second. Determine the rate of increase of the water level in the reservoir at the moment when it is equal to $0.5 R$.

Solution:
Solution. Let $h$ be the water level in $m$ and $v$ its volume in $\mu^{3}$. We will find the dependence between the variables $h$ and $v$ using the formula for the volume of a spherical cap:

$$
v=\pi h^{2}\left(R-\frac{h}{3}\right)
$$

Differentiating this equality with respect to time $t$, we will find the dependence between the rates of change of the variables $h$ and $v$:

$$
\frac{d v}{d t}=\frac{d v}{d h} \cdot \frac{d h}{d t}=\pi\left[2 h\left(R-\frac{h}{3}\right)-\frac{1}{3} h^{2}\right] \frac{d h}{d t}=\pi\left(2 R h-h^{2}\right) \frac{d h}{d t}
$$

Assuming, according to the condition, $\frac{d v}{d t}=0.001 Q\left(\frac{\mu^{3}}{\text { sec }}\right)$, we get $\frac{d h}{d t}=\frac{0.001 Q}{\pi h(2 R-h)}\left(\frac{M}{\text { sec }}\right)$.

When $h=\frac{R}{2}$, we get $\frac{d h}{d t}=\frac{0.004 Q}{3 \pi R^{2}}\left(\frac{M}{\text { sec }}\right)$.

Answer: \frac{0.004Q}{3\piR^{2}}


Answer: 1011


Given line segments $b, c$. Find the line segment $a$ such that
$$
\frac{1}{a}=\frac{1}{b}+\frac{1}{c} .
$$

Solution:
Analysis: Consider $\mathrm{a}$ as $\mathrm{f}$, $\mathrm{b}$ as $\mathrm{u}$, and $\mathrm{c}$ as $\mathrm{v}$. Using the above physical construction method, we can construct $\mathrm{a}$.
Construction:
(1) Draw a line 1, take a point 0 on 1, and on both sides of 0 on 1, mark $O D=b$ and $O E=c$;
(2) Draw perpendiculars DM, OT, and EN from D, O, and E to 1, and take an appropriate length for DM;
(3) Extend MO to intersect EN at N;
(4) Draw MT $\downarrow \mathrm{TO}$ at T;
(5) Connect $\mathbf{T N}$ to intersect 1 at $\mathrm{F}$.
Then OF is the desired length (Figure 2).
This construction method is more complex than the usual algebraic method, which is to construct $\mathrm{a}=\frac{\mathrm{bc}}{\mathrm{b}+\mathrm{c}}$. However, it is more intuitive and does not require algebraic manipulation. Is there a simpler construction method? Yes, but we need to prove a proposition first: If points $\mathrm{C}$ and $\mathrm{B}$ are collinear, then $\frac{1}{\mathrm{OC}}=\frac{1}{\mathrm{OA}}+\frac{1}{\mathrm{OB}}$.
Proof: Extend $\mathrm{BO}$ to $\mathrm{D}$ such that $\mathrm{OD}=\mathrm{OA}$, and connect $\mathrm{DA}$.
Since $\angle \mathrm{AOD}=180^{\circ}-\angle \mathrm{AOB}=60^{\circ}$ and $\mathrm{OD}=\mathrm{OA}$, $\triangle O A D$ is an equilateral triangle.
- Therefore, $\angle \mathrm{ADO}=60^{\circ}=\frac{1}{2} \angle \mathrm{AOB}=\angle \mathrm{CUB}$, so OC is parallel to AD.
Thus, $\triangle B C O \sim \triangle B B$,
$$
\begin{array}{l}
\frac{O B}{B D}=\frac{O C}{A D}, \quad \frac{O B}{O B+O A}=\frac{O C}{O A}, \\
O C=\frac{O A \cdot O B}{O A+O B} . \\
\end{array}
$$

Therefore, $\frac{1}{O C}=\frac{1}{O A}+\frac{1}{O B}$.
According to this proposition, the construction for Example 1 is easily solved.

Answer: a=\frac{bc}{b+c}


In $\triangle ABC, AB=8, BC=11$, and $AC=6$. The points $P$ and $Q$ are on $BC$ such that $\triangle PBA$ and $\triangle QAC$ are each similar to $\triangle ABC$. What is the length of $PQ$?

![](https://cdn.mathpix.com/cropped/2024_04_17_d13f9eaf1d614231018ag-3.jpg?height=360&width=534&top_left_y=300&top_left_x=855)

Solution:
From the similarity of $\triangle P B A$ and $\triangle A B C$, we get $\frac{P B}{A B}=\frac{A B}{C B}$.

Substituting, we get $\frac{P B}{8}=\frac{8}{11}$ which can be rearranged to get $P B=\frac{64}{11}$.

From the similarity of $\triangle Q A C$ and $\triangle A B C$, we get $\frac{Q C}{A C}=\frac{A C}{B C}$.

Substituting, we get $\frac{Q C}{6}=\frac{6}{11}$ which can be rearranged to get $Q C=\frac{36}{11}$.

We now know the lengths of $P B, Q C$, and $B C$. Since $P Q=B C-P B-Q C$, we get

$$
P Q=11-\frac{64}{11}-\frac{36}{11}=\frac{121-64-36}{11}=\frac{21}{11}
$$

ANSWER: $\frac{21}{11}$

Answer: \frac{21}{11}


There are $2022$ signs arranged in a straight line. Mark tasks Auto to color each sign with either red or blue with the following condition: for any given sequence of length $1011$ whose each term is either red or blue, Auto can always remove $1011$ signs from the line so that the remaining $1011$ signs match the given color sequence without changing the order. Determine the number of ways Auto can color the signs to satisfy Mark's condition.

Solution:
1. **Understanding the Problem:**
   We need to color 2022 signs in such a way that for any given sequence of length 1011 (each term being either red or blue), we can always remove 1011 signs from the line so that the remaining 1011 signs match the given sequence without changing the order.

2. **Initial Observations:**
   - The total number of signs is 2022.
   - We need to ensure that any sequence of 1011 signs can be matched by removing 1011 signs from the original 2022 signs.

3. **Key Insight:**
   - To satisfy the condition, each possible sequence of 1011 signs must be able to be formed from the remaining signs after removing 1011 signs.
   - This implies that the original sequence must be such that any subsequence of length 1011 can be formed.

4. **Pairing Strategy:**
   - Consider pairing the signs into chunks of 2, creating 1011 pairs of adjacent signs.
   - Call a chunk "colorful" if it contains both a red and a blue sign.

5. **Proof by Contradiction:**
   - Suppose there exists a chunk that is not colorful, i.e., it contains two signs of the same color (both red or both blue).
   - Without loss of generality, assume the chunk contains two red signs.
   - Consider the sequence given by Mark: 1011 blue signs followed by 1010 red signs.
   - Since the chunk with two red signs cannot contribute to the sequence of 1011 blue signs, we need to find 1011 blue signs in the remaining 1010 chunks.
   - This is impossible because there are only 1010 chunks left, and each chunk can contribute at most one blue sign.

6. **Conclusion from the Proof:**
   - Therefore, every chunk must be colorful to ensure that any sequence of 1011 signs can be formed.
   - If every chunk is colorful, Auto can create any sequence that Mark gives by choosing the appropriate color from each chunk.

7. **Counting the Colorings:**
   - Each chunk must contain one red and one blue sign.
   - There are 2 ways to color each chunk (red-blue or blue-red).
   - Since there are 1011 chunks, the total number of ways to color the signs is \(2^{1011}\).

The final answer is \(\boxed{2^{1011}}\)

Answer: 2^{1011}


The number of ordered pairs $(a, b)$ that satisfy $(a+b \mathrm{i})^{6}=a-b \mathrm{i}$ (where $\left.a, b \in \mathbf{R}, \mathrm{i}^{2}=-1\right)$ is

Solution:
Given $a+b \mathrm{i}=r(\cos \theta+\mathrm{i} \sin \theta)$, then $r^{6}(\cos 6 \theta+\mathrm{i} \sin 6 \theta)=r(\cos \theta-\mathrm{i} \sin \theta)$
$$
\Rightarrow\left\{\begin{array}{l}
r=1, \\
\cos 6 \theta=\cos \theta, \Rightarrow 6 \theta=2 k \pi-\theta \Rightarrow \theta=\frac{2 k \pi}{7}(k=0,1,2,3,4,5,6) . \\
\sin 6 \theta=-\sin \theta
\end{array}\right.
$$

Notice that 0 also satisfies the equation, so there are 8 pairs of $(a, b)$ that meet the requirements.

Answer: 8


Solution:
2. If $x>0$ applying the inequality of means we have

$4^{x} \cdot 9^{\frac{1}{x}}+9^{x} \cdot 4^{\frac{1}{x}}+6^{x+\frac{1}{x}} \geq 3 \sqrt[3]{4^{x} \cdot 9^{\frac{1}{x}} \cdot 9^{x} \cdot 4^{\frac{1}{x}} \cdot 6^{x+\frac{1}{x}}}=3 \cdot 6^{x+\frac{1}{x}} \geq 3 \cdot 6^{2}=108$

and equality holds if and only if $x=1$

$.5 p$

Answer: 1


The real root situation of the equation $\frac{1}{x}-2=x^{2}-2 x$ is ( ).
(A) It has three real roots
(B) It has only two real roots
(C) It has only one real root
(D) It has no real roots

Solution:
4.C.

The original equation is transformed into $\frac{1}{x}=(x-1)^{2}+1$.
Let $y_{1}=\frac{1}{x}, y_{2}=(x-1)^{2}+1$.
In the same coordinate system, draw the graphs of the inverse proportion function $y_{1}=\frac{1}{x}$ and the parabola $y_{2}=(x-1)^{2}+1$. At this point, the solution to the equation is converted to the x-coordinates of the intersection points of the two graphs. Clearly, the two graphs have only one intersection point in the first quadrant.

Answer: C


Find all pairs of real numbers $x$ and $y$ which satisfy the following equations:
\begin{align*}
  x^2 + y^2 - 48x - 29y + 714 & = 0 \\
  2xy - 29x - 48y + 756 & = 0
\end{align*}

Solution:
1. **Combine the two equations:**
   \[
   \begin{cases}
   x^2 + y^2 - 48x - 29y + 714 = 0 \\
   2xy - 29x - 48y + 756 = 0
   \end{cases}
   \]
   Add the two equations together:
   \[
   x^2 + y^2 - 48x - 29y + 714 + 2xy - 29x - 48y + 756 = 0
   \]
   Simplify the combined equation:
   \[
   x^2 + y^2 + 2xy - 77x - 77y + 1470 = 0
   \]
   Notice that the left-hand side can be written as a perfect square:
   \[
   (x + y - \frac{77}{2})^2 = \frac{49}{4}
   \]
   Therefore:
   \[
   x + y - \frac{77}{2} = \pm \frac{7}{2}
   \]
   This gives us two cases:
   \[
   x + y = \frac{77}{2} + \frac{7}{2} = 42 \quad \text{or} \quad x + y = \frac{77}{2} - \frac{7}{2} = 35
   \]

2. **Subtract the second equation from the first:**
   \[
   x^2 + y^2 - 48x - 29y + 714 - (2xy - 29x - 48y + 756) = 0
   \]
   Simplify the resulting equation:
   \[
   x^2 + y^2 - 2xy - 19x + 19y - 42 = 0
   \]
   Notice that the left-hand side can be written as a perfect square:
   \[
   (x - y - \frac{19}{2})^2 = \frac{529}{4}
   \]
   Therefore:
   \[
   x - y - \frac{19}{2} = \pm \frac{23}{2}
   \]
   This gives us two cases:
   \[
   x - y = \frac{19}{2} + \frac{23}{2} = 21 \quad \text{or} \quad x - y = \frac{19}{2} - \frac{23}{2} = -2
   \]

3. **Solve the system of linear equations:**
   We now have four systems of linear equations to solve:
   \[
   \begin{cases}
   x + y = 42 \\
   x - y = 21
   \end{cases}
   \]
   Adding these equations:
   \[
   2x = 63 \implies x = 31.5
   \]
   Substituting back:
   \[
   31.5 + y = 42 \implies y = 10.5
   \]
   So, one solution is:
   \[
   (x, y) = (31.5, 10.5)
   \]

   Next system:
   \[
   \begin{cases}
   x + y = 42 \\
   x - y = -2
   \end{cases}
   \]
   Adding these equations:
   \[
   2x = 40 \implies x = 20
   \]
   Substituting back:
   \[
   20 + y = 42 \implies y = 22
   \]
   So, another solution is:
   \[
   (x, y) = (20, 22)
   \]

   Next system:
   \[
   \begin{cases}
   x + y = 35 \\
   x - y = 21
   \end{cases}
   \]
   Adding these equations:
   \[
   2x = 56 \implies x = 28
   \]
   Substituting back:
   \[
   28 + y = 35 \implies y = 7
   \]
   So, another solution is:
   \[
   (x, y) = (28, 7)
   \]

   Last system:
   \[
   \begin{cases}
   x + y = 35 \\
   x - y = -2
   \end{cases}
   \]
   Adding these equations:
   \[
   2x = 33 \implies x = 16.5
   \]
   Substituting back:
   \[
   16.5 + y = 35 \implies y = 18.5
   \]
   So, the final solution is:
   \[
   (x, y) = (16.5, 18.5)
   \]

The final answer is \(\boxed{(31.5, 10.5), (20, 22), (28, 7), (16.5, 18.5)}\)

Answer: (31.5, 10.5), (20, 22), (28, 7), (16.5, 18.5)


For a regular octagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7} A_{8}$ with side length 1, if any two points $A_{i} A_{j}$ are chosen, then the maximum value of $\overrightarrow{A_{i} A_{j}} \cdot \overrightarrow{A_{1} A_{2}}$ is .

Solution: 1. $\sqrt{2}+1$ Detailed explanation: According to the geometric meaning of the dot product of vectors, it is only necessary to look at the projection of the vector in the direction of $\overrightarrow{A_{1} A_{2}}$, the maximum value is $\sqrt{2}+1$

Answer: \sqrt{2}+1


For any integer n > 1, let p(n) be the smallest prime which does not divide n and let q(n) = the product of all primes less than p(n), or 1 if p(n) = 2. Define the sequence a 0 , a 1 , a 2 , ... by a 0 = 1 and a n+1 = a n p(a n )/q(a n ). Find all n such that a n = 1995.

Solution: : 142. We calculate the first few terms: 1, 2, 3, 2.3, 5, 2.5, 3.5, 2.3.5, 7, 2.7, 3.7, ... . On that basis we might guess that if p m is the (m+1)th prime and n = b r ... b 0 in binary, then a n = product of primes p i for which b i = 1. q(n) is a product of primes which divide n, so n/q(n) is an integer. Hence a n are all integral. Moreover it is easy to show by induction that a n is square-free. It is true for n = 1. Suppose it is true for n. Since a n+1 is obtained from a n by dropping some prime factors and adding a prime factor not already present, it is true for n+1. Hence for all n. So we can certainly represent a n as a binary number b r ... b 0 , where b m = 1 iff p m divides a m . Now p(a n ) = m, where m is the smallest subscript such that b m = 0 (with the obvious meaning m if a n is represented by m 1s: 1 ... 1). Hence a n+1 is represented by the binary number obtained by adding 1 to that for a n . But a 1 is represented by 1, so a n is represented by the binary expression for n. 1995 = 3.5.7.19, so it is represented by the binary number 10001110 (19 1 17 0 13 0 11 0 7 1 5 1 3 1 2 0 ). Hence the only possible n with a n = 1995 is 10001110 = 128 + 8 + 4 + 2 = 142. 36th IMO shortlist 1995 © John Scholes jscholes@kalva.demon.co.uk 8 Oct 2002

Answer: 142


Given $\sin \alpha-\cos \alpha=\frac{1}{2}$. Then the value of $\sin ^{3} \alpha-\cos ^{3} \alpha$ is $\qquad$.

Solution:
$$
\begin{array}{l}
\text { 12. } \frac{11}{16} \\
\sin ^{3} \alpha-\cos ^{3} \alpha \\
=(\sin \alpha-\cos \alpha)\left(\sin ^{2} \alpha+\sin \alpha \cdot \cos \alpha+\cos ^{2} \alpha\right) \\
=(\sin \alpha-\cos \alpha)(1+\sin \alpha \cdot \cos \alpha)
\end{array}
$$

Given $\sin \alpha-\cos \alpha=\frac{1}{2}$, and
$$
(\sin \alpha-\cos \alpha)^{2}=\sin ^{2} \alpha-2 \sin \alpha \cdot \cos \alpha+\cos ^{2} \alpha=\frac{1}{4} \text {, }
$$

thus $\sin \alpha \cdot \cos \alpha=\frac{3}{8}$.
Therefore, $\sin ^{3} \alpha-\cos ^{3} \alpha=\frac{11}{16}$.

Answer: \frac{11}{16}


Given that $[x]$ represents the greatest integer not exceeding $x$, the number of integer solutions to the equation $3^{2 x}-\left[10 \cdot 3^{x+1}\right]+ \sqrt{3^{2 x}-\left[10 \cdot 3^{x+1}\right]+82}=-80$ is $\qquad$

Solution:
4. 2 .
$$
3^{2 x}-\left[10 \cdot 3^{x+1}\right]+82+\sqrt{3^{2 x}-\left[10 \cdot 3^{x+1}\right]+82}-2=0 \text {, i.e., }\left(\sqrt{3^{2 x}-\left[10 \cdot 3^{x+1}\right]+82}+\right.
$$
2) $\cdot\left(\sqrt{3^{2 x}-\left[10 \cdot 3^{x+1}\right]+82}-1\right)=0$, so $3^{2 x}-\left[10 \cdot 3^{x+1}\right]+81=0$, then $3^{2 x}-10 \cdot 3^{x+1}+$ $81 \leqslant 0$, which gives $3 \leqslant 3^{x} \leqslant 27$, hence $1 \leqslant x \leqslant 3$. Upon verification: when $x=1,3$, the original equation holds, and when $x=2$, the original equation has no solution.

Answer: 2


Let $D$ be the midpoint of the hypotenuse $BC$ of the right triangle $ABC$. On the leg $AC$, a point $M$ is chosen such that $\angle AMB = \angle CMD$. Find the ratio $\frac{AM}{MC}$.

Solution:
Answer: $1: 2$.

Solution. Let point $E$ be the intersection of rays $B A$ and $D M$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9c0cded10981f14efe49g-02.jpg?height=610&width=588&top_left_y=278&top_left_x=820)

Then angles $A M E$ and $C M D$ are equal as vertical angles. Therefore, angles $A M B$ and $A M E$ are also equal. Hence, triangles $A M B$ and $A M E$ are congruent (by the common side $A M$ and the adjacent angles). This means $B A = E A$. Consider triangle $E B C$. In this triangle, point $M$ is the intersection of medians $C A$ and $E D$. By the property of the centroid, $A M: M C = 1: 2$.

Evaluation. Full solution: 14 points.

Answer: 1:2


Six weights of 2, 3, 5, 7, 9, 10 hectograms are available. Five of them are placed on the two plates of a balance in such a way that it is in equilibrium. Which weight is excluded?
(A) 2
(B) 3
(C) 5
(D) 10
(E) cannot be determined.

Solution: 20) The answer is (A). The sum of all weights is 36 hectograms. If the balance is to be in equilibrium, the sum of the 5 chosen weights must be even, so only an even weight can be removed. Excluding the 10 hectogram weight, 13 hectograms should be placed on each plate, and this cannot be achieved with the 5 remaining weights. Excluding instead the 2 hectogram weight, 17 hectograms should be placed on each plate, which can be achieved by placing 3, 5, and 9 hectograms on one plate and 7 and 10 hectograms on the other.

Answer: A


Solution:
4. (8 points) Given that the average of four numbers $a, b, c, d$ is $12.345, a>b>c>d$, then $b(\quad)$
A. Greater than 12.345
B. Less than 12.345
C. Equal to 12.345
D. Cannot be determined
【Solution】Since the average of the four numbers $a, b, c, d$ is $12.345, a>b>c>d$,
Therefore, $a$ must be greater than $12.345, d$ must be less than 12.345,
However, the value of $b$ cannot be determined, $b$ could be greater than 12.345, or less than 12.345, or equal to 12.345. Hence, the answer is: $D$.

Answer: D


Let the complex number $z$ satisfy $|z|=1$, such that the equation $z x^{2}+2 \bar{z} x+2=0$ has real roots, then the sum of such complex numbers $z$ is $\qquad$ .

Solution:
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R})$, then $(a+b \mathrm{i}) x^{2}+2(a-b \mathrm{i}) x+2=0$
$$
\Rightarrow\left\{\begin{array}{ll}
a x^{2}+2 a x+2=0, & \text { (1) } \\
b x^{2}-2 b x=0 & \text { (2) }
\end{array} \Rightarrow b x(x-2)=0\right. \text {. }
$$

When $b=0$, from $a^{2}+b^{2}=1 \Rightarrow a= \pm 1$. Substituting into (1) gives $a=1$ with no solution for $x$, so $a=-1$ $\Rightarrow z=a+b \mathrm{i}=-1$; When $x=0$, substituting into (1) gives $2=0$ with no solution;
When $x=2$, substituting into (1) gives $a=-\frac{1}{4} \Rightarrow b= \pm \frac{\sqrt{15}}{4} \Rightarrow z=a+b \mathrm{i}=-\frac{1}{4} \pm \frac{\sqrt{15}}{4} \mathrm{i}$. Therefore, the sum of the complex numbers $z$ that satisfy the condition is $-\frac{3}{2}$.

Answer: -\frac{3}{2}


For a positive integer $n$, let $\sigma (n)$ be the sum of the divisors of $n$ (for example $\sigma (10) = 1 + 2 + 5 + 10 = 18$). For how many $n \in \{1, 2,..., 100\}$, do we have $\sigma (n) < n + \sqrt{n}$?

Solution:
1. **Understanding the problem**: We need to find how many integers \( n \) in the set \(\{1, 2, \ldots, 100\}\) satisfy the condition \(\sigma(n) < n + \sqrt{n}\), where \(\sigma(n)\) is the sum of the divisors of \(n\).

2. **Analyzing the condition for composite numbers**: Assume \( n > 1 \) and \( n \) is not a prime number. Then \( n \) has a divisor \( d \) such that \( \sqrt{n} \le d < n \). We can choose \( d = \frac{n}{p} \), where \( p \) is the smallest prime factor of \( n \).

3. **Lower bound for \(\sigma(n)\)**: Since \( d \) is a divisor of \( n \), we have:
   \[
   \sigma(n) \ge d + n
   \]
   Substituting \( d = \frac{n}{p} \), we get:
   \[
   \sigma(n) \ge \frac{n}{p} + n
   \]
   Since \( p \) is the smallest prime factor, \( p \ge 2 \). Therefore:
   \[
   \sigma(n) \ge \frac{n}{2} + n = \frac{3n}{2}
   \]

4. **Comparing \(\sigma(n)\) with \( n + \sqrt{n} \)**: We need to check if:
   \[
   \frac{3n}{2} \ge n + \sqrt{n}
   \]
   Simplifying, we get:
   \[
   \frac{3n}{2} - n \ge \sqrt{n}
   \]
   \[
   \frac{n}{2} \ge \sqrt{n}
   \]
   Squaring both sides:
   \[
   \left(\frac{n}{2}\right)^2 \ge n
   \]
   \[
   \frac{n^2}{4} \ge n
   \]
   \[
   n^2 \ge 4n
   \]
   \[
   n(n - 4) \ge 0
   \]
   Since \( n > 0 \), this inequality holds for \( n \ge 4 \).

5. **Conclusion for composite numbers**: For \( n \ge 4 \), if \( n \) is composite, \(\sigma(n) \ge n + \sqrt{n}\). Therefore, composite numbers \( n \ge 4 \) do not satisfy the condition \(\sigma(n) < n + \sqrt{n}\).

6. **Checking prime numbers and small values**:
   - For \( n = 1 \):
     \[
     \sigma(1) = 1 < 1 + \sqrt{1} = 2
     \]
     So, \( n = 1 \) satisfies the condition.
   - For prime numbers \( n \):
     \[
     \sigma(n) = 1 + n
     \]
     We need:
     \[
     1 + n < n + \sqrt{n}
     \]
     Simplifying:
     \[
     1 < \sqrt{n}
     \]
     Squaring both sides:
     \[
     1 < n
     \]
     So, all prime numbers \( n > 1 \) satisfy the condition.

7. **Counting the primes in \([2, 100]\)**: We need to count the prime numbers between 2 and 100. The prime numbers in this range are:
   \[
   2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
   \]
   There are 25 prime numbers in this range.

8. **Adding the count for \( n = 1 \)**: Including \( n = 1 \), we have:
   \[
   25 + 1 = 26
   \]

The final answer is \(\boxed{26}\)

Answer: 26


The relationship between the sets $M=\{u \mid u=12 m+8 n+4 l$, $m, n, l \in \mathbf{Z}\}$ and $N=\{u \mid u=20 p+16 q+12 r$, $p, q, r \in \mathbf{Z}\}$ is ( ).
(A) $M=N$
(B) $M \nsubseteq N, N \nsubseteq M$
(C) $M \subset N$
(D) $M \supset N$

Solution:
Explanation: Starting from the expressions of elements in two sets.
Since $12 m+8 n+4 l=4(3 m+2 n+l)$,
$$
20 p+16 q+12 r=4(5 p+4 q+3 r) \text {, }
$$

and $(3,2,1)=1,(5,4,3)=1$, by Bézout's theorem, $3 m+2 n+l$ and $5 p+4 q+3 r$ can both represent all integers. Therefore, $M=N=\{k \mid k=4 l, l \in \mathbf{Z}\}$.
Hence, the answer is (A).

Answer: A


Let $ABCD$ be a rectangle with an area of 2, and let $P$ be a point on side $CD$. Let $Q$ be the point where the incircle of $\triangle PAB$ touches side $AB$. The product $PA \cdot PB$ varies with the changes in rectangle $ABCD$ and point $P$. When $PA \cdot PB$ is minimized,
(1) Prove: $AB \geqslant 2BC$;
(2) Find the value of $AQ \cdot BQ$.

Solution:
(1) $S_{\triangle A P B}=\frac{1}{2} S_{\text {trapezoid } A B C D}=1 \Rightarrow \frac{1}{2} P A \cdot P B \sin \angle A P B=1$, which means $P A \cdot P B=\frac{2}{\sin \angle A P B} \geqslant 2$, equality holds if and only if $\angle A P B=90^{\circ}$. This indicates that point $P$ lies on the circle with $A B$ as its diameter, which should intersect with $C D$. Therefore, when $P A \cdot P B$ reaches its minimum value, it should have $B C \leqslant \frac{A B}{2}$;

(2) Let the inradius of $\triangle A P B$ be $r$, then $P A \cdot P B=(r+A Q)(r+B Q)=r(r+A Q+B Q)+A Q \cdot B Q$, and $P A \cdot P B=2 S_{\triangle A P B}, r(r+A Q+B Q)=S_{\triangle A P B}$, thus $A Q \cdot B Q=S_{\triangle A B P}=1$.

Answer: 1


The venusian prophet Zabruberson sent to his pupils a $10000$-letter word, each letter being $A$ or $E$: the [i]Zabrubic word[/i]. Their pupils consider then that for $1 \leq k \leq 10000$, each word comprised of $k$ consecutive letters of the Zabrubic word is a [i]prophetic word[/i] of length $k$. It is known that there are at most $7$ prophetic words of length $3$. Find the maximum number of prophetic words of length $10$.

Solution:
1. **Understanding the Problem:**
   We are given a 10000-letter word composed of the letters 'A' and 'E'. We need to find the maximum number of distinct prophetic words of length 10, given that there are at most 7 distinct prophetic words of length 3.

2. **Analyzing the Constraints:**
   Since there are at most 7 distinct prophetic words of length 3, one of the possible 8 combinations of length 3 words (AAA, AAE, AEA, AEE, EAA, EAE, EEA, EEE) must be missing. This missing word will influence the structure of the longer prophetic words.

3. **Symmetry Considerations:**
   By symmetry, we can interchange 'A' and 'E' without changing the problem. This reduces the number of cases we need to consider. Specifically, we need to consider the cases where AAA, AAE, or AEA is missing (since the other cases are equivalent under reversal or swapping of 'A' and 'E').

4. **Recursive Enumeration:**
   We need to enumerate the number of 10-letter words that do not contain the missing 3-letter word. We will do this for each of the three cases (AAA, AAE, AEA).

5. **Case Analysis:**
   - **Case 1: Missing AAA**
     - We need to count the number of 10-letter words that do not contain the substring 'AAA'.
     - This can be done using a recursive approach or dynamic programming to count valid sequences.
     - Let \( f(n) \) be the number of valid sequences of length \( n \) that do not contain 'AAA'.
     - We can build \( f(n) \) using the recurrence relation:
       \[
       f(n) = f(n-1) + f(n-2) + f(n-3)
       \]
       where:
       - \( f(n-1) \) corresponds to appending 'E' to a valid sequence of length \( n-1 \),
       - \( f(n-2) \) corresponds to appending 'AE' to a valid sequence of length \( n-2 \),
       - \( f(n-3) \) corresponds to appending 'AAE' to a valid sequence of length \( n-3 \).
     - Initial conditions:
       \[
       f(0) = 1, \quad f(1) = 2, \quad f(2) = 4
       \]
     - Using this recurrence, we can compute \( f(10) \).

6. **Computing \( f(10) \):**
   \[
   \begin{align*}
   f(3) &= f(2) + f(1) + f(0) = 4 + 2 + 1 = 7 \\
   f(4) &= f(3) + f(2) + f(1) = 7 + 4 + 2 = 13 \\
   f(5) &= f(4) + f(3) + f(2) = 13 + 7 + 4 = 24 \\
   f(6) &= f(5) + f(4) + f(3) = 24 + 13 + 7 = 44 \\
   f(7) &= f(6) + f(5) + f(4) = 44 + 24 + 13 = 81 \\
   f(8) &= f(7) + f(6) + f(5) = 81 + 44 + 24 = 149 \\
   f(9) &= f(8) + f(7) + f(6) = 149 + 81 + 44 = 274 \\
   f(10) &= f(9) + f(8) + f(7) = 274 + 149 + 81 = 504
   \end{align*}
   \]

7. **Conclusion:**
   The maximum number of distinct prophetic words of length 10, given that there are at most 7 distinct prophetic words of length 3, is 504.

The final answer is \(\boxed{504}\).

Answer: 504


Solve the following equation over the set of integer pairs:

$$
(x+2)^{4}-x^{4}=y^{3} \text {. }
$$

Solution:
Solution. After performing the exponentiation and subtracting $x^{4}$:

$$
8 x^{3}+24 x^{2}+32 x+16=y^{3}
$$

Since 8 is a cube, let $8 b^{3}=y^{3}$, thus

$$
\begin{aligned}
8 x^{3}+24 x^{2}+32 x+16 & =8 b^{3} \\
x^{3}+3 x^{2}+4 x+2 & =b^{3}
\end{aligned}
$$

Transform the obtained equation:

$$
(x+1)^{3}+(x+1)=b^{3}
$$

Now let $a=x+1$, thus

$$
\begin{aligned}
a^{3}+a & =b^{3} \\
a\left(a^{2}+1\right) & =b^{3}
\end{aligned}
$$

Since $a$ and $a^{2}+1$ are relatively prime, $a\left(a^{2}+1\right)$ can only be a cube if both $a$ and $a^{2}+1$ are cubes. This is only possible if $a^{2}=0$ and $a^{2}+1=1$. Therefore, $x+1=0$, so $x=-1, y=0$.

Answer: -1,0


Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$ be $n$ real numbers satisfying the following condition: for any integer $k>0$, we have $a_{1}^{k}+a_{2}^{k}+\cdots+a_{n}^{k} \geqslant 0$. Then, $p=\max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\cdots,\left|a_{n}\right|\right\}=$ $\qquad$

Solution:
2. $a_{1}$.

Obviously, $p$ is either equal to $a_{1}$, or equal to $\left|a_{n}\right|$.
Assume $p \neq a_{1}$, then $p=\left|a_{n}\right|$, and $p>a_{1}, a_{n}a_{n-k+1}=$ $a_{n-k+2}=\cdots=a_{n}$, then $\left|\frac{a_{1}}{a_{n}}\right|0 .
\end{array}
$$

Thus, $a_{1}^{2 l+1}+\cdots+a_{n}^{2 l+1}$
$$
\begin{array}{l}
=a_{n}^{2 l+1} \cdot\left[\left(\frac{a_{1}}{a_{n}}\right)^{2 l+1}+\cdots+\left(\frac{a_{n}}{a_{n}}\right)^{2 l+1}\right] \\
<0 .
\end{array}
$$

This contradicts the given information.
Therefore, it must be that $p=a_{1}$.

Answer: a_{1}


Find the distance between the points of intersection of the three lines: \(y = 3x\), \(y = 3x - 6\), and \(y = 1975\).

Solution: 14. 2. Instruction. Show that the desired distance is equal to the distance between points \(A(0 ; 0)\) and \(B(2 ; 0)\).

Answer: 2


Students $A$ and $B$ each have 3 cards. They play a game by tossing a fair coin. When the coin lands heads up, $A$ wins a card from $B$, otherwise, $B$ wins a card from $A$. If one person wins all the cards, the game ends. What is the probability that the game ends exactly after 5 coin tosses? ( )
(A) $\frac{1}{16}$
(B) $\frac{1}{32}$
(C) $\frac{1}{8}$
(D) $\frac{3}{16}$

Solution:
6.D.

Toss a coin 5 times. If $A$ wins, then there is exactly 1 tail in the first 3 tosses, and the remaining 4 tosses are all heads; if $B$ wins, then there is exactly 1 head in the first 3 tosses, and the remaining 4 tosses are all tails.
Thus, the required probability is $P=\frac{3+3}{2^{5}}=\frac{3}{16}$.

Answer: D


The base of the pyramid is an equilateral triangle with a side length of 6. One of the lateral edges is perpendicular to the base plane and equals 4. Find the radius of the sphere circumscribed around the pyramid.

Solution:
Let $Q$ be the center of the base $ABC$ of the triangular pyramid $ABCD$, where the lateral edge $AD$ is perpendicular to the plane of the base, and $AB = BC = AC = a = 6$, $AD = h = 4$. The center $O$ of the sphere with radius $R$, circumscribed around the pyramid $ABCD$, lies on the line perpendicular to the plane of the base $ABC$ passing through the point $Q$, and also in the plane perpendicular to the line $AD$ and passing through the midpoint $P$ of the segment $AD$. The lines $AP$ and $OQ$ are parallel (since they are perpendicular to the same plane), and the quadrilateral $APOQ$ is a rectangle. In the right triangle $AOQ$, it is known that

$$
AQ = \frac{a \sqrt{3}}{3}, OQ = AP = \frac{h}{2}
$$

Therefore,

$$
R = OA = \sqrt{AQ^2 + OQ^2} = \sqrt{\left(\frac{a \sqrt{3}}{3}\right)^2 + \left(\frac{h}{2}\right)^2} = \sqrt{12 + 4} = 4
$$

## Answer

4.00

Answer: 4


For what value of the real number $\mathrm{x}$, does $\sqrt{4 \mathrm{x}^{2}-4 \mathrm{x}+1} - \sqrt{x^{2}-2 x+1}(0 \leqslant x \leqslant 2)$ attain its maximum and minimum values? And find the maximum and minimum values.

Solution:
Let $y=\sqrt{4 x^{2}-4 x+1}-\sqrt{x^{2}-2 x+1}$
$$
\begin{aligned}
x & \sqrt{(2 x-1)^{2}}=\sqrt{(x-1)^{2}} \\
\text { Then } y & =|2 x-1|-|x-1|, \quad(0 \leqslant x \leqslant 2)
\end{aligned}
$$
$$
\therefore y=\left\{\begin{array}{ll}
-x, & \text { (when } 0 \leqslant x \leqslant \frac{1}{2} \text { ) } \\
3 x-2, & \text { (when } \frac{1}{2}<x<1 \text { ) } \\
x . & \text { (when } 1 \leqslant x \leqslant 2 \text { ) }
\end{array}\right.
$$

Figure 1
From Figure 1, $v^{\prime} 4 x^{2}-4 x+1 \sqrt{x^{2}-2 x+1}$ $(0 \leqslant x \leqslant 2)$

When $z=\frac{1}{2}$, it achieves the minimum value $-\frac{1}{2}$; when $x=2$, it achieves the maximum value 2.

Answer: 2


Given that $[x]$ represents the greatest integer not exceeding $x$, such as $[1.5]=1,[2]=2$, then the last two digits of $\left[\frac{10^{2040}}{10^{60}-3}\right]$ are . $\qquad$

Answer: 23


A deck of playing cards, excluding the joker, has 4 suits totaling 52 cards, with each suit having 13 cards, numbered from 1 to 13. Feifei draws 2 hearts, 3 spades, 4 diamonds, and 5 clubs. If the sum of the face values of these 14 cards Feifei drew is exactly 35, then how many of them are 1?

$\qquad$ cards are 1.

Solution:
【Solution】Solution: According to the analysis, the minimum sum of the two hearts is $1+2=3$, the minimum sum of the three spades is $1+2+3=6$,

the minimum sum of the four diamonds is $1+2+3+4=10$, and the sum of the five clubs is: $1+2+3+4+5=15$, so the minimum sum of the 14 cards is: $3+6+10+15=34$,
(1) If only $1 \sim 2$ cards are 1, the sum of the card faces is clearly greater than 35;
(2) If three are 1, then the minimum sum of the card faces is: $2+3+1+2+3+1+2+3+4+1+2+3+4+5=36>35$, which contradicts the problem statement;
(3) If four are 1, then the minimum sum of the card faces is: $3+6+10+15=34<35$, which meets the requirement.

Therefore: There are four 1s.
The answer is: 4.

Answer: 4


The infinite series
$$\sum_{k=4}^{\infty}\frac{1}{(T_{k-1} - 1)(T_k - 1)(T_{k+1} - 1)}$$
has the value $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution:
1. First, we need to simplify the expression for \( T_k - 1 \). Recall that \( T_k = \frac{k(k+1)}{2} \), so:
   \[
   T_k - 1 = \frac{k(k+1)}{2} - 1 = \frac{k^2 + k - 2}{2} = \frac{(k-1)(k+2)}{2}
   \]
   Therefore, for \( k-1 \), \( k \), and \( k+1 \), we have:
   \[
   T_{k-1} - 1 = \frac{(k-2)(k+1)}{2}, \quad T_k - 1 = \frac{(k-1)(k+2)}{2}, \quad T_{k+1} - 1 = \frac{k(k+3)}{2}
   \]

2. Substituting these into the summation, we get:
   \[
   \sum_{k=4}^{\infty} \frac{1}{(T_{k-1} - 1)(T_k - 1)(T_{k+1} - 1)} = \sum_{k=4}^{\infty} \frac{1}{\left(\frac{(k-2)(k+1)}{2}\right) \left(\frac{(k-1)(k+2)}{2}\right) \left(\frac{k(k+3)}{2}\right)}
   \]
   Simplifying the denominator:
   \[
   = \sum_{k=4}^{\infty} \frac{8}{(k-2)(k+1)(k-1)(k+2)k(k+3)}
   \]

3. Next, we perform partial fraction decomposition on:
   \[
   \frac{8}{(k-2)(k+1)(k-1)(k+2)k(k+3)}
   \]
   We assume:
   \[
   \frac{8}{(k-2)(k+1)(k-1)(k+2)k(k+3)} = \frac{A}{k-2} + \frac{B}{k-1} + \frac{C}{k} + \frac{D}{k+1} + \frac{E}{k+2} + \frac{F}{k+3}
   \]

4. To find the coefficients \( A, B, C, D, E, F \), we multiply both sides by the common denominator and solve for each \( k \):
   \[
   8 = A(k-1)k(k+1)(k+2)(k+3) + B(k-2)k(k+1)(k+2)(k+3) + C(k-2)(k-1)(k+1)(k+2)(k+3) + D(k-2)(k-1)k(k+2)(k+3) + E(k-2)(k-1)k(k+1)(k+3) + F(k-2)(k-1)k(k+1)(k+2)
   \]

5. By substituting specific values for \( k \) (e.g., \( k = 2, 1, 0, -1, -2, -3 \)), we solve for each coefficient:
   \[
   A = \frac{1}{15}, \quad B = -\frac{1}{3}, \quad C = \frac{2}{3}, \quad D = -\frac{2}{3}, \quad E = \frac{1}{3}, \quad F = -\frac{1}{15}
   \]

6. Substituting these coefficients back into the sum, we get:
   \[
   \sum_{k=4}^{\infty} \left( \frac{1}{15} \left( \frac{1}{k-2} - \frac{1}{k+3} \right) - \frac{1}{3} \left( \frac{1}{k-1} - \frac{1}{k+2} \right) + \frac{2}{3} \left( \frac{1}{k} - \frac{1}{k+1} \right) \right)
   \]

7. Each of these sums is a telescoping series:
   \[
   \sum_{k=4}^{\infty} \frac{1}{15} \left( \frac{1}{k-2} - \frac{1}{k+3} \right) - \frac{1}{3} \left( \frac{1}{k-1} - \frac{1}{k+2} \right) + \frac{2}{3} \left( \frac{1}{k} - \frac{1}{k+1} \right)
   \]

8. Evaluating the telescoping series, we get:
   \[
   \frac{1}{15} \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \right) - \frac{1}{3} \left( \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \right) + \frac{2}{3} \left( \frac{1}{4} \right)
   \]

9. Simplifying the fractions:
   \[
   \frac{1}{15} \cdot \frac{29}{20} - \frac{1}{3} \cdot \frac{47}{60} + \frac{2}{3} \cdot \frac{1}{4} = \frac{87 - 235 + 150}{900} = \frac{2}{900} = \frac{1}{450}
   \]

10. Therefore, the final answer is:
   \[
   m = 1, \quad n = 450 \quad \Rightarrow \quad m + n = 451
   \]

The final answer is \(\boxed{451}\).

Answer: 451


Find a number $\mathrm{N}$ with five digits, all different and none zero, which equals the sum of all distinct three digit numbers whose digits are all different and are all digits of $\mathrm{N}$.

Solution:
## Solution

Answer: 35964

There are $4.3=12$ numbers with a given digit of $n$ in the units place. Similarly, there are 12 with it in the tens place and 12 with it in the hundreds place. So the sum of the 3 digit numbers is $12.111(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e})$, where $\mathrm{n}=\mathrm{abcde}$. So $8668 \mathrm{a}=332 \mathrm{~b}+1232 \mathrm{c}+1322 \mathrm{~d}+$ 1331e. We can easily see that $\mathrm{a}=1$ is too small and $\mathrm{a}=4$ is too big, so $\mathrm{a}=2$ or 3 . Obviously e must be even. 0 is too small, so $\mathrm{e}=2,4,6$ or 8 . Working mod 11 , we see that $0=2 \mathrm{~b}+2 \mathrm{~d}$, so $\mathrm{b}+\mathrm{d}=11$. Working $\bmod 7$, we see that $2 \mathrm{a}=3 \mathrm{~b}+6 \mathrm{~d}+\mathrm{e}$. Using the $\bmod 11$ result, $\mathrm{b}=2, \mathrm{~d}=$ 9 or $b=3, d=8$ or $b=4, d=7$ or $b=5, d=6$ or $b=6, d=5$ or $b=7, d=4$ or $b=8, d=3$ or $\mathrm{b}=9, \mathrm{~d}=2$. Putting each of these into the $\bmod 7$ result gives $2 \mathrm{a}-\mathrm{e}=4,1,5,2,6,3,0,4$ mod 7. So putting $\mathrm{a}=2$ and remembering that $\mathrm{e}$ must be $2,4,6,8$ and that all digits must be different gives a, b, d, e = 2,4, 7, 6 or 2, 7, 4, 8 or 2, 8, 3, 4 as the only possibilities. It is then straightforward but tiresome to check that none of these give a solution for $\mathrm{c}$. Similarly putting $\mathrm{a}=4$, gives $\mathrm{a}, \mathrm{b}, \mathrm{d}, \mathrm{e}=3,4,7,8$ or $3,5,6,4$ as the only possibilities. Checking, we find the solution above and no others.

Answer: 35964


Before the football match between teams "North" and "South," five predictions were made:

a) there will be no draw;

b) goals will be scored in "South's" goal;

c) "North" will win;

d) "North" will not lose;

e) exactly 3 goals will be scored in the match.

After the match, it turned out that exactly three of the predictions were correct. What was the final score of the match?

Solution:
Suppose "North" won. Then 4 predictions ("a", "b", "c", and "d") would be correct, which contradicts the condition.

Suppose the match ended in a draw. Then the predictions "a", "c", and "d" (in a draw, the total number of goals is even) are definitely incorrect, which also contradicts the condition.

Therefore, "North" lost the match. Then predictions "c" and "d" are incorrect, and the remaining 3 predictions are correct. Specifically: there was no draw, goals were scored against "South", and exactly 3 goals were scored in the match. Therefore, the match ended with a score of $1: 2$.

## Answer

$1: 2$.

Answer: 1:2


Given that $a$ is a root of the equation $x^{2}-5 x+1=0$. Then the unit digit of $a^{4}+a^{-4}$ is $\qquad$ .

Solution:
ニ、1.7.
Obviously, $a^{-1}$ is another root of the equation. Then $a+a^{-1}=5$.
Thus, $a^{2}+a^{-2}=\left(a+a^{-1}\right)^{2}-2=23$, $a^{4}+a^{-4}=\left(a^{2}+a^{-2}\right)^{2}-2=527$.

Answer: 7


Given: Triangular pyramid $S-ABC$, the base is an equilateral triangle with side length $4\sqrt{2}$, the edge $SC$ has a length of 2 and is perpendicular to the base, $E$ and $D$ are the midpoints of $BC$ and $AB$ respectively. Find: The distance between $CD$ and $SE$.

Solution:
[3D Geometry Solution] Since $CD$ and $SE'$ are two skew lines, it is not obvious where their common perpendicular intersects these two lines. Therefore, the problem needs to be transformed into whether a plane can be found that passes through $SE$ and is parallel to $CD$. $E$ is the midpoint of $BC$. Find the midpoint $F$ on $BD$, and connect $EF$. $EF$ is the midline, so $EF \parallel CD$, hence $CD \parallel$ plane $SEF$. Then,
the distance from $CD$ to plane $SEF$ is the distance between the two skew lines (by Theorem 1). The distance between a line and a plane can also be converted to the distance from a point $C$ on the line to the plane $SEF$. Let $CG$ be the distance from point $C$ to plane $SEF$.

Connect $CF$, using the fact that the volumes of tetrahedra $S-CFE$ and $C-EFS$ are the same, we can find the length of $CG$.

Given $BC = 4\sqrt{2}$, $E$ and $F$ are the midpoints of $BC$ and $BD$, respectively, we can get $CD = 2\sqrt{6}$, $EF = \sqrt{6}$, $DF = \sqrt{2}$. Also, $SC = 2$, so
\[
V_{C-EFS} = \frac{1}{3} \left( \frac{1}{2} \sqrt{6} \cdot \sqrt{2} \right) \cdot 2 = \frac{2\sqrt{3}}{3}.
\]
On the other hand,
\[
\text{Let } a = SE - EFS = \frac{1}{3} S_{\triangle BFS} \cdot CG.
\]
\[
b = \sqrt{4 + 8} = \sqrt{2} \cdot \sqrt{6},
\]
\[
c = EF = \sqrt{4 + 26} = \sqrt{2} \cdot \sqrt{15},
\]
\[
s = \frac{1}{2}(a + b + c) = \frac{\sqrt{2}}{2}(\sqrt{3}).
\]

Using Heron's formula to find the area of $\triangle EFS$:
\[
S_{\triangle EFS}^2 = s(s - a)(s - b)(s - c)
\]
\[
= \frac{1}{4} \left[ (\sqrt{3} + \sqrt{15})^2 - 3 \right] \cdot \left[ 3 - (\sqrt{6} - \sqrt{15})^2 \right]
\]
\[
= \frac{1}{4} (6 + 15 + 2\sqrt{90} - 3) \cdot (3 - 6 - 15 + 2\sqrt{90})
\]
\[
= \frac{1}{4} (6\sqrt{10} + 18)(6\sqrt{10} - 18)
\]
\[
= y.
\]
\[
V_{C-EFS} = \frac{1}{3} \cdot 3 \cdot CG = CG.
\]
\[
\because V_{C-EFS} = V_{S-GPE},
\]
\[
\frac{2\sqrt{3}}{3}.
\]

Therefore, $\triangle EFS = 3$.
\[
\therefore CG = \frac{2\sqrt{3}}{3}, \text{ i.e., the distance between } CD \text{ and } SE \text{ is } \frac{2\sqrt{3}}{3}.
\]

[Analytical Method]
Establish a spatial rectangular coordinate system, with $C$ as the origin, the line containing $CD$ as the $x$-axis, and the line containing $SC$ as the $z$-axis (as shown in the figure). Use the midpoints $E$ and $F$ of $BC$ and $BD$.

To find the distance from $C$ to the plane $SEF$, we can use the analytical method and the point-to-plane distance formula. From the given conditions, we determine: $C(0,0,0)$, $S(0,0,2)$, $F(2\sqrt{6}, \sqrt{2}, 0)$, $E'(\sqrt{6}, \sqrt{2}, 0)$. The plane equation determined by points $S$, $E$, and $F$ is:
\[
\left| \begin{array}{ccc}
x - 0 & y - 0 & z - 2 \\
\sqrt{6} & \sqrt{2} & -2 \\
2\sqrt{6} & \sqrt{2} & 2
\end{array} \right| = 0,
\]
which simplifies to
\[
2y + \sqrt{2}z - 2\sqrt{2} = 0.
\]
Let the distance from $C$ to the plane $SEF$ be $d$, then
```
[Algebraic Method] Let the common perpendicular segment of $SE$ and $CD$ be $FG$ (as shown in the figure). Draw a perpendicular from $F$ to $GH$, then
$GH \parallel SO$. Connect $IH$.
    Since $OSC \perp$ plane $ABC$,
    and $HF \subset$ plane $ABC$,
    we know $\angle BCD = 30^\circ$. Let $FH = x$, then $CH = 2x$. Therefore,
    \[
    FG^2 = FH^2 + GH^2 = x^2 + (2 - \sqrt{2}x)^2 = 3x^2.
    \]
    \[
    SF^2 = SC^2 + CF^2 = 4 + (\sqrt{3}x)^2 = 3x^2.
    \]
    \[
    SE = \sqrt{4 + (2\sqrt{2})^2} = \sqrt{12}.
    \]
    \[
    SG = \sqrt{12} \cdot \frac{2x}{2\sqrt{2}} = \sqrt{6}x.
    \]
    From (1) and (2),
    \[
    3x^2 = 3x^2.
    \]
    Solving for $x$,
    \[
    x = \frac{2\sqrt{2}}{3}.
    \]
```

Answer: \frac{2\sqrt{3}}{3}


Given the function $f(x)=|x+1|+|a x+1|$ has a minimum value of $\frac{3}{2}$. Then the value of the real number $a$ is $\qquad$ .

Solution:
1. $-\frac{1}{2}$ or -2.
When $|a| \leqslant 1$, $f(-1)=\frac{3}{2}$, we get
$$
a=-\frac{1}{2} ;
$$

When $|a|>1$, $f\left(-\frac{1}{a}\right)=\frac{3}{2}$, we get $a=-2$.
Therefore, $a=-\frac{1}{2}$ or -2.

Answer: -\frac{1}{2}or-2


Bus tickets have numbers from 000001 to 999999. A number is considered lucky if the first three digits are odd and different, the second three digits are even, and 7 and 8 do not stand next to each other. How many different lucky numbers exist?

Solution:
Solution. The number of numbers satisfying the first two conditions of the problem (but perhaps not satisfying the third condition) can be calculated as follows: the first digit of such a number can be chosen in five ways, the second in four ways, the third in three ways, and each subsequent digit in five ways, so the total number of such tickets is 7500 $(5 \cdot 4 \cdot 3 \cdot 5 \cdot 5 \cdot 5=7500)$.

Among the considered numbers, those that do not satisfy the third condition of the problem are the numbers where the third digit is 7 and the fourth digit is 8. The number of such numbers is calculated in the same way as above: it is equal to $300(4 \cdot 3 \cdot 5 \cdot 5=300)$.

Therefore, the number of different lucky numbers is $7200(7500-300=7200)$.

Answer: 7200.

Answer: 7200


From the vertex $B$ of an equilateral triangle $A B C$, a perpendicular $B K$ is erected to the plane $A B C$, such that $|B K|=|A B|$. We will find the tangent of the acute angle between the lines $A K$ and $B C$.

Solution:
Solution. Let's construct the image of the given figure and find its parametric number (Fig. 4).

Considering an arbitrary triangle $ABC$ as the image of an equilateral triangle,

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-014.jpg?height=605&width=671&top_left_y=246&top_left_x=1024)

Fig. 4, we spend two parameters. Considering the segment $BK$ as the image of the perpendicular to the plane $ABC$, we also spend two parameters, and finally, considering that $|BK|:|AB|=1:1$, we spend one more parameter; thus, $p=5$. Therefore, the given image is metrically defined, and any further metric constructions must not be arbitrary.

Let's perform the following additional constructions: in the plane $ABC$, draw a line $l \parallel (BC)$ through point $A$. Then the angle between the lines $AK$ and $BC$ is equal to the angle between the lines $AK$ and $l$. To find the angle between the lines $AK$ and $l$, it is obviously advisable to include it in some right triangle. The line $DK \perp l$ can be constructed as an oblique line, the projection of which is perpendicular to the line $l$. To construct this projection - the line $BD$ - we can use the fact that the triangle $ABC$ is equilateral and, therefore, its median $AM$ is also a perpendicular to the segment $BC$. Thus, by constructing the median $AM$ of the triangle $ABC$, and then $(BD) \parallel (AM)$ and $[DK]$, we will obtain the right triangle $ADK$. The ratio $|DK|:|AD|=\operatorname{tg} \widehat{DAK}$ will be the desired one. Setting $|AB|=a$, we find:

$$
|BK|=a,|BD|=\frac{a \sqrt{3}}{2},|DK|=\frac{a \sqrt{7}}{2},|AD|=|BM|=\frac{a}{2} \quad \text { and, }
$$

thus, $\operatorname{tg} \widehat{DAK}=\sqrt{7}$, and therefore, the tangent of the angle between the lines $AK$ and $BC$ is $\sqrt{7}$.

Answer: \sqrt{7}


If for the lengths $a, b, c$ of the sides of a triangle it holds that $(a+b+c)(a+b-c)=3ab$, determine the angle opposite side $c$.

Solution:
## Solution.

The given expression can be written in the form $(a+b)^{2}-c^{2}=3 a b$, or $a^{2}+b^{2}-c^{2}=a b$.

According to the cosine rule, $\cos \gamma=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{a b}{2 a b}=\frac{1}{2}$, so $\gamma=60^{\circ}$.

For 4 points: correct up to the result $\cos \gamma=\frac{1}{2}$.

For 8 points: correct result and procedure.

Answer: 60


Let $x$ and $y$ be positive real numbers such that $x^{2}+y^{2}=1$ and $\left(3 x-4 x^{3}\right)\left(3 y-4 y^{3}\right)=-\frac{1}{2}$. Compute $x+y$.

Solution:
Answer: $\sqrt{\frac{\sqrt{6}}{2}}$ Solution 1: Let $x=\cos (\theta)$ and $y=\sin (\theta)$. Then, by the triple angle formulae, we have that $3 x-4 x^{3}=-\cos (3 \theta)$ and $3 y-4 y^{3}=\sin (3 \theta)$, so $-\sin (3 \theta) \cos (3 \theta)=-\frac{1}{2}$. We can write this as $2 \sin (3 \theta) \cos (3 \theta)=\sin (6 \theta)=1$, so $\theta=\frac{1}{6} \sin ^{-1}(1)=\frac{\pi}{12}$. Thus, $x+y=\cos \left(\frac{\pi}{12}\right)+\sin \left(\frac{\pi}{12}\right)=$ $\frac{\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{\sqrt{6}}{2}$.

Solution 2: Expanding gives $9 x y+16 x^{3} y^{3}-12 x y^{3}-12 x^{3} y=9(x y)+16(x y)^{3}-12(x y)\left(x^{2}+y^{2}\right)=-\frac{1}{2}$, and since $x^{2}+y^{2}=1$, this is $-3(x y)+16(x y)^{3}=-\frac{1}{2}$, giving $x y=-\frac{1}{2}, \frac{1}{4}$. However, since $x$ and $y$ are positive reals, we must have $x y=\frac{1}{4}$. Then, $x+y=\sqrt{x^{2}+y^{2}+2 x y}=\sqrt{1+2 \cdot \frac{1}{4}}=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2}$.

Answer: \frac{\sqrt{6}}{2}


Given that $a, b, c, d$ are all positive integers, and $\log _{a} b=\frac{3}{2}, \log _{c} d=\frac{5}{4}$. If $a-c=9$, then $b-d=$ $\qquad$

Solution: $\log _{a} b=\frac{3}{2} \Rightarrow b=a^{\frac{3}{2}}, \log _{c} d=\frac{5}{4} \Rightarrow d=c^{\frac{5}{4}}$, also $a, b, c, d$ are all positive integers, set $a=m^{2}, c=n^{4} \Rightarrow m^{2}-n^{4}=\left(m+n^{2}\right)\left(m-n^{2}\right)=9 \Rightarrow\left\{\begin{array}{l}m-n^{2}=1, \\ m+n^{2}=9\end{array}\right.$ $\Rightarrow\left\{\begin{array}{l}m=5, \\ n=2\end{array} \Rightarrow\left\{\begin{array}{l}a=25, \\ c=16\end{array} \Rightarrow\left\{\begin{array}{l}b=125, \\ d=32\end{array} \Rightarrow b-d=93\right.\right.\right.$.

Answer: 93


Xiao Ming divided a cube with an edge length of 4 into 29 smaller cubes with integer edge lengths. Then the number of smaller cubes with an edge length of 1 is ( ).
(A) 22
(B) 23
(C) 24
(D) 25

Solution:
6. C.

If a cube with an edge length of 3 is separated out, then there can only be 1 cube with an edge length of 3, and the remaining ones are all cubes with an edge length of 1, totaling 37, which does not meet the requirement.

Let the number of cubes with an edge length of 2 be $x$, and the number of cubes with an edge length of 1 be $y$. Then
$$
\left\{\begin{array} { l }
{ x + y = 2 9 , } \\
{ 8 x + y = 6 4 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x=5, \\
y=24 .
\end{array}\right.\right.
$$

Answer: C


A teacher at a summer math camp took with him for the whole summer several shirts, several pairs of pants, several pairs of shoes, and two jackets. At each lesson, he wore pants, a shirt, and shoes, and he wore a jacket on some lessons. On any two lessons, at least one of the items of his clothing or shoes was different. It is known that if he had taken one more shirt, he could have conducted 18 more lessons; if he had taken one more pair of pants, he could have conducted 63 more lessons; if he had taken one more pair of shoes, he could have conducted 42 more lessons. What is the maximum number of lessons he could conduct under these conditions?

Solution:
Answer: 126.

Solution: Let the teacher bring $x$ shirts, $y$ pairs of trousers, $z$ pairs of shoes, and 2 jackets. Then he can conduct $3 x y z$ lessons (the number 3 means: 2 lessons in each of the jackets and 1 lesson without a jacket). If he has one more shirt, the number of lessons will increase by $3 y z$. If he has one more pair of trousers, the number of lessons will increase by $3 x z$. If he has one more pair of shoes, the number of lessons will increase by $3 y z$. Thus, we get a system of three equations: $3 y z=18, 3 x z=63, 3 x y=42$. Therefore, $y z=6, x z=21, x y=14$, and thus $(x y z)^{2}=6 \cdot 21 \cdot 14, x y z=42$ (although it is not necessary, but we can calculate that $x=7, y=2, z=3$). The desired value: $3 x y z=126$.

Remark. There is a possible interpretation of the problem condition in which it is required to find the maximum number of lessons given that the teacher will take with him one more shirt, one more pair of trousers, and one more pair of shoes. The corresponding answer $3(x+1)(y+1)(z+1)=288$ is also considered correct.

Answer: 126


Assuming the Earth rotates once around the axis connecting the North Pole and the South Pole in 23 hours 56 minutes 4 seconds, and the Earth's equatorial radius is $6378.1 \mathrm{~km}$. Then, when you stand on the equator and rotate with the Earth, the linear velocity around the axis is $\qquad$ meters/second (rounded to meters, $\pi$ taken as 3.1416$).

Solution: 1. 465

Answer: 465


A bucket that is $\frac{2}{3}$ full contains $9 \mathrm{~L}$ of maple syrup. What is the capacity of the bucket, in litres?

Solution:
Since the bucket is $\frac{2}{3}$ full and contains $9 \mathrm{~L}$ of maple syrup, then if it were $\frac{1}{3}$ full, it would contain $\frac{1}{2}(9 \mathrm{~L})=4.5 \mathrm{~L}$.

Therefore, the capacity of the full bucket is $3 \cdot(4.5 \mathrm{~L})=13.5 \mathrm{~L}$.

ANSWER: $13.5 \mathrm{~L}$

Answer: 13.5\mathrm{~L}


Integers, the decimal representation of which reads the same from left to right and from right to left, we will call symmetric. For example, the number 513315 is symmetric, while 513325 is not. How many six-digit symmetric numbers exist such that adding 110 to them leaves them symmetric?

Solution: Answer: 81 numbers of the form $\overline{a b 99 b a}$, where $a=1,2, \ldots, 9, b=0,1,2, \ldots, 8$.

Answer: 81


JenÃk wrote several different prime numbers (at least three) on the board. When he added any two of them and then decreased this sum by 7, the resulting number was among those written. Which numbers could have been on the board?

Solution:
2. Let's denote the prime numbers written on the board as $p_{1}<p_{2}<\ldots<p_{k}$. From these numbers, due to the assumption $k \geqq 3$, it is possible to form $k$ different numbers

$$
p_{1}+p_{2}-7<p_{1}+p_{3}-7<\ldots<p_{1}+p_{k}-7<p_{k-1}+p_{k}-7,
$$

all of which must be among the numbers written on the board. Therefore, they sequentially equal the prime numbers $p_{1}<p_{2}<\ldots<p_{k}$. Precisely, the smallest of them equals $p_{1}$, i.e., $p_{1}+p_{2}-7=p_{1}$, and thus $p_{2}=7$. For the second smallest number in the inequalities (1), we have $p_{1}+p_{3}-7=p_{2}=7$, and thus $p_{1}+p_{3}=14$. For the prime number $p_{1}<p_{2}=7$, we have only three possibilities $p_{1} \in\{2,3,5\}$, which correspond to the values $p_{3}=14-p_{1} \in\{12,11,9\}$, of which only $p_{3}=11$ is a prime number, so $p_{1}=3$. Suppose there is another prime number $p_{4}$ written on the board. Then the third smallest number in the inequalities (1) must be $11=p_{3}=p_{1}+p_{4}-7$, from which, given $p_{1}=3$, it follows that $p_{4}=15$, which is not a prime number.

Answer: Only three prime numbers 3, 7, and 11 could be written on the board.

Let's add that the conclusion $k=3$ can be justified differently by noting that the same $k$-tuple of prime numbers in the group (1) must be obtained even if the last of the numbers is replaced by $p_{2}+p_{k}-7$; thus, it must hold that $p_{k-1}=p_{2}$ or $k=3$.

Another solution. Let's denote the prime numbers written on the board as $p_{1}<p_{2}<\ldots<p_{k}$. If Jeník selects the two smallest and the two largest prime numbers, their sum reduced by 7 is on the board, and thus $p_{1}+p_{2}-7 \geqq p_{1}$ and $p_{k}+p_{k-1}-7 \leqq p_{k}$, from which it follows that $7 \leqq p_{2}<$ $<p_{k-1} \leqq 7$. To avoid a contradiction, we must have $k \leqq 3$. According to the problem, there are at least three prime numbers on the board, so there are exactly three prime numbers $p_{1}<p_{2}<p_{3}$. Since $p_{1}+p_{2}<p_{1}+p_{3}<p_{2}+p_{3}$, it must hold that

$$
p_{1}+p_{2}-7=p_{1}, \quad p_{1}+p_{3}-7=p_{2}, \quad p_{2}+p_{3}-7=p_{3}
$$

From the first (and last) equation, we get $p_{2}=7$ and from the middle equation $p_{1}+p_{3}=14$, with $p_{1}<7=p_{2}$. By testing all possibilities $p_{1} \in\{2,3,5\}$, we get the only solution $p_{1}=3$ and $p_{3}=11$.

For a complete solution, award 6 points, for solving the situation with three prime numbers, award 3 points. Award another 3 points if the student proves that there could be at most 3 prime numbers on the board. For guessing the solution $\{3,7,11\}$, award 2 points.

Answer: 3,7,11


At a rectangular table, eight chairs are placed, four on one side and four opposite them on the other side. In how many ways can eight friends be seated at this table so that Ana and Bane sit opposite each other, and Vesna and Goran sit next to each other? (It is known that all friends have distinct names.)

Solution: 4. We can choose the country for Vesna and Goran in 2 ways, their position as a pair in 3 ways, and their mutual arrangement in 2 ways, i.e., we can arrange them in 12 ways. Now let's arrange Ana and Banet. On the side where Vesna and Goran are, there are two free seats, so we can choose the position of Ana and Banet in 2 ways, and then who sits on which side in 2 ways. Thus, we can arrange them in 4 ways. For the remaining 4 guests, there are 4 free seats, so we can arrange them in 4! ways. Therefore, the total number of arrangements is $12 \cdot 4 \cdot 4!=1152$.

Answer: 1152


Person A departs from point A to point B, while at the same time, person B departs from point B to point A. After 12 minutes, the two have passed their meeting point and are 100 meters apart. It is known that person A takes 20 minutes to complete the journey, and person B walks at a speed of 65 meters per minute. Therefore, the distance between points A and B is ______ meters.

Solution: Reference answer: 1700

Answer: 1700


The first term of the sequence is 934. Each subsequent term is equal to the sum of the digits of the previous term, multiplied by 13.

Find the 2013th term of the sequence.

Solution:
Let's compute the first few terms of the sequence: $a_{2}=16 \cdot 13=208, a_{3}=10 \cdot 13=130, a_{4}=4 \cdot 13=52, a_{5}$ $=7 \cdot 13=91, a_{6}=10 \cdot 13=130=a_{3}$. Since each subsequent number is calculated using only the previous number, the terms of the sequence will repeat with a period of 3. The number 2013 is divisible by 3, so $a_{2013}=a_{3}=130$.

## Answer

130.

Answer: 130


One day, the boss gives letters to the secretary to type at different times, each time placing the letter on top of the secretary's pile. The secretary, when available, takes the letter from the top of the pile to type. Suppose there are 5 letters, and the boss gives them in the order $1,2,3,4,5$. Among the following sequences, which one cannot be the order in which the secretary types the letters?
(A) 12345.
(B) 24351.
(C) 32415.
(D) 45231.
(E) 54321.

Solution:
［Solution］(A) is possible: after typing 1, type 2; after typing 2, type 3; $\cdots$, after typing 4, type 5, and finally type 5;
(B) is possible: when typing 2 (1 is at the bottom), 3 and 4 come; after finishing 4, when typing 3, 5 comes, then continue to type 5 and 1;
(C) is possible: start typing in the order of 3, 2, 1; when typing 2, 4 comes; after finishing 4 and 1, 5 comes;
(E) is possible: after all five letters arrive, type them in sequence;
(D) is impossible: after typing 4 and 5, the next should be 3.
Therefore, the answer is $(D)$.

Answer: D


Let $m$ be an integer greater than 1, and the sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{l}
a_{0}=m, a_{1}=\varphi(m), \\
a_{2}=\varphi^{(2)}(m)=\varphi(\varphi(m)), \cdots, \\
a_{n}=\varphi^{(n)}(m)=\varphi\left(\varphi^{(n-1)}(m)\right),
\end{array}
$$

where $\varphi(m)$ is the Euler's totient function.
If for any non-negative integer $k$, we have $a_{k+1} \mid a_{k}$, find the largest positive integer $m$ not exceeding 2016.

Solution:
3. It is known that if $a \mid b$, then $\varphi(a) \mid \varphi(b)$.

Moreover, when $m>2$, $\varphi(m)$ is even.
Therefore, it is only necessary to find integers $m$ such that $a_{1} \mid a_{0}$, which implies $a_{k+1} \mid a_{k}$.
If $m$ is an odd number greater than 2, it is impossible for $a_{1} \mid m$.
Let $m=2^{\alpha} \prod_{k=1}^{s} p_{k}^{\alpha_{k}}$, where $p_{1}, p_{2}, \cdots, p_{s}$ are all odd primes. Then
$$
\varphi(m)=m\left(1-\frac{1}{2}\right) \prod_{k=1}^{s}\left(1-\frac{1}{p_{k}}\right) .
$$

Thus, only when $s=1$ and $p_{1}=3$, we have $a_{1} \mid a_{0}$.
Therefore, $m$ must be of the form $2^{\alpha} \times 3^{\beta}$.
Hence, the largest positive integer value not exceeding 2016 is
$$
m=2^{3} \times 3^{5}=1944 \text {. }
$$

Answer: 1944


Among the four-digit numbers composed of the digits $2,0$ and $1,2$, those greater than 1000 are ( ) in number.
(A) 6
(B) 7
(C) 8
(D) 9
(E) 12

Solution:
1. D.

The four-digit numbers that meet the criteria are:
$$
\begin{array}{l}
2012, 2021, 2102, 2120, 2201, 2202, \\
2210, 1022, 1202, 1220 .
\end{array}
$$

Answer: D


If the system of equations concerning $x$ and $y$
$$
\left\{\begin{array}{l}
\sin x=m \sin ^{3} y, \\
\cos x=m \cos ^{3} y
\end{array}\right.
$$

has real solutions, then the range of positive real number $m$ is
$\qquad$.

Solution:
14. $[1,2]$.

Squaring both equations and eliminating $x$, we get
$$
\begin{array}{l}
\sin ^{2} 2 y=\frac{4}{3}\left(1-m^{2}\right) \in[0,1] \\
\Rightarrow 1 \leqslant m^{2} \leqslant 2 .
\end{array}
$$

Conversely, when $1 \leqslant m^{2} \leqslant 2$, there exists $\left(x_{0}, y_{0}\right)$ that satisfies the equations.
Therefore, the positive real number $m \in[1,2]$.

Answer: [1,2]


a) Determine the set $M=\{(x, y) \in \mathbb{Z} \times \mathbb{Z} \mid 5 x+7 y=1\}$.

b) Determine the largest natural number $n$ that cannot be written in the form $n=5 x+7 y$, where $x, y \in \mathbb{N}$. Justify your answer.

Solution:
Solution. It's almost not worth mentioning... How can such a problem be signed? I assume the authors of the two problems presented here are related; perhaps more care should be taken to avoid such things.

a) It is a classic result that if $p, q \in \mathbb{Z}^{*} \times \mathbb{Z}^{*}$ with greatest common divisor (gcd) $(p, q)=1$, then there exists $(x, y) \in \mathbb{Z} \times \mathbb{Z}$ such that $p x + q y = 1$, and specifically, if $\left(x_{0}, y_{0}\right)$ is a solution, then there exists $m \in \mathbb{Z}$ for which $x = x_{0} - m q$ and $y = y_{0} + m p$. Since $\left(x_{0}, y_{0}\right) = (3, -2)$ is a solution for $(p, q) = (5, 7)$, it means that

$$
M = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} \mid 5 x + 7 y = 1\} = \{(3 - 7 m, -2 + 5 m) \mid m \in \mathbb{Z}\}
$$

b) It is the same classic result, known as SYLVESTER, Frobenius, or the "Chicken McNugget" theorem, which states that under the above conditions, the largest natural number $n$ that cannot be written in the form $n = p x + q y$, with $x, y \in \mathbb{N}$, is $n = p q - p - q$ (and thus all numbers $n \geq (p-1)(q-1)$ can be represented in this way). Therefore, in our case, $n = 23$.

Remark. The results are as expected under these conditions

| Points | $7-6$ | $5-4$ | $3-2$ | $1-0$ |
| :---: | :---: | :---: | :---: | :---: |
| \# Scores |  | 3 | 4 | 100 |

Such theoretical results, or are known to competitors (which does not seem to be the case here), or have a minimal chance of being "discovered" in the heat of battle, during a competition.

Answer: 23


The area of square $ABCD$ is $196 \mathrm{~cm}^{2}$. Point $E$ is inside the square, at the same distances from points $D$ and $C$, and such that $\angle DEC=150^{\circ}$. What is the perimeter of $\triangle ABE$ equal to? Prove your answer is correct.

Solution:
D/2 Solution 1: Since the area of square $A B C D$ is $196=14^{2}\left(\mathrm{~cm}^{2}\right)$, then the side of square $A B C D$ is $14 \mathrm{~cm}$.
We claim that $\triangle A B E$ is equilateral. To prove this, we make a reverse construction, starting from an equilateral $\triangle A B E^{\prime}$, building up to square $A B C D$, and eventually showing that points $E$ and $E^{\prime}$ coincide. Thus,

Step 1. Let $E^{\prime}$ be a point inside square $A B C D$ such that $\triangle A B E^{\prime}$ is equilateral. Hence, in particular, $\angle A B E^{\prime}=60^{\circ}$.

Step 2. Then $\angle E^{\prime} B C=90^{\circ}-60^{\circ}=30^{\circ}$.
Step 3. Since $B E^{\prime}=A B=B C$ ( $\triangle A B E^{\prime}$ is equilateral and $A B C D$ is a square), we conclude that $\triangle C E^{\prime} B$ is isosceles with $B E^{\prime}=B C$ and $\angle E^{\prime} B C=30^{\circ}$.

Step 4. This in turn implies that both base angles of $\triangle C E^{\prime} B$ are $\frac{1}{2}\left(180^{\circ}-30^{\circ}\right)=75^{\circ}$. In particular, $\angle B C E^{\prime}=75^{\circ}$. (Note: Since $\angle B C E^{\prime}0) .
$$

Now back to our problem. Drop a perpendicular from point $E$ to side $B C$ of the square and mark by $H$ the foot of this perpendicular. Since $E$ is at the same distances from $C$ and $D$, it easily follows that $E$ is half-way between parallel lines $A D$ and $B C$; i.e., $E H=\frac{1}{2} A B=7 \mathrm{~cm}$.
Just as above, from isosceles $\triangle D E C$ we know that $\angle E C D=15^{\circ}$. This means that $\angle C E H=15^{\circ}$ (EH\|DC,CE is a transversal, and $\angle E C D$ and $\angle C E H$ are alternating interior angles). Thus, from right $\triangle E H C$ we can calculate $C H=7 \tan 15^{\circ}$. This implies that $B H=B C-C H=14-7 \tan 15^{\circ}$.

We are ready to calculate $\cot \angle E B C$ from right $\triangle E H B$ :
$$
\cot \angle E B C=\cot \angle E B H=\frac{B H}{E H}=\frac{14-7 \tan 15^{\circ}}{7}=2-\tan 15^{\circ}=2-(2-\sqrt{3})=\sqrt{3} .
$$

But $\cot 30^{\circ}=\sqrt{3}$, so the acute $\angle E B C$ must be $30^{\circ}$. We proved what we aimed for!

To finish off the problem, note that $\angle A B E=90^{\circ}-\angle E B C=90^{\circ}-30^{\circ}=60^{\circ}$. By symmetry, $\angle A B E=60^{\circ}$ and $\triangle A B E$ is equilateral. Thus, its perimeter is $3 A B=42 \mathrm{~cm}$.

Solution 3: It is not hard to compute the ratios of a 15-75-90 right triangle without any trig, provided one knows the ratios of the 30-60-90 triangle. Just divide the 75-degree angle into 15 degree and 60-degree angles, dissecting the original triangle into a 15-15-150 and 30-60-90. Quite elementary from there.

Answer: 42\mathrm{~}


Solution:
## Solution.

Let $3 x^{2}-2 x=y$. With respect to $y$, the equation takes the form $\sqrt{y+15}+\sqrt{y+8}=7$. Squaring both sides of the equation, we get $y+15+2 \sqrt{(y+15)(y+8)}+y+8=49 \Leftrightarrow \sqrt{(y+15)(y+8)}=13-y$, where $13-y \geq 0, y \leq 13$. From this, $(y+15)(y+8)=(13-y)^{2}, y=1$.

Then $3 x^{2}-2 x=1,3 x^{2}-2 x-1=0$, from which $x_{1}=-\frac{1}{3}, x_{2}=1$.

Answer: $x_{1}=-\frac{1}{3}, x_{2}=1$.

Answer: x_{1}=-\frac{1}{3},x_{2}=1


Let $x, y, z$ be 3 non-zero real numbers. Find the maximum value of $\frac{x y+2 y z}{x^{2}+y^{2}+z^{2}}$.

Solution:
Introducing the parameter $\beta$, by the AM-GM inequality (here $\beta \in \mathbf{R}^{+}$, and $\beta^{2}<1$)
$$
x^{2}+\beta^{2} y^{2} \geqslant 2 \beta x y,\left(1-\beta^{2}\right) y^{2}+z^{2} \geqslant 2 \sqrt{1-\beta^{2}} y z
$$

Thus, $x^{2}+y^{2}+z^{2} \geqslant 2\left(\beta x y+\sqrt{1-\beta^{2}} y z\right)$. Let $\frac{\beta}{\sqrt{1-\beta^{2}}}=\frac{1}{2}$, solving gives $\beta=\frac{\sqrt{5}}{5}$. Therefore, $x^{2}+y^{2}+z^{2} \geqslant \frac{2}{5} \sqrt{5}(x y+2 y z)$, which means $\frac{x y+2 y z}{x^{2}+y^{2}+z^{2}} \leqslant \frac{\sqrt{5}}{2}$. When $x=k, y=\sqrt{5} k, z=2 k$ (here $k \neq 0$), the equality holds, and $\frac{x y+2 y z}{x^{2}+y^{2}+z^{2}}$ achieves $\frac{\sqrt{5}}{2}$, thus its maximum value is $\frac{\sqrt{5}}{2}$.

Answer: \frac{\sqrt{5}}{2}


Solution:
## Solution.

$\frac{\operatorname{ctg}\left(270^{\circ}-\alpha\right)}{1-\operatorname{tg}^{2}\left(\alpha-180^{\circ}\right)} \cdot \frac{\operatorname{ctg}^{2}\left(360^{\circ}-\alpha\right)-1}{\operatorname{ctg}\left(180^{\circ}+\alpha\right)}=\frac{\operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha} \cdot \frac{\operatorname{ctg}^{2} \alpha-1}{\operatorname{ctg} \alpha}=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha} \times$ $\times \frac{\operatorname{ctg}^{2} \alpha-1}{2 \operatorname{ctg} \alpha}=\operatorname{tg} 2 \alpha \operatorname{ctg} 2 \alpha=1$.

Answer: 1.

Answer: 1


Let $p_1,p_2,p_3,p_4$ be four distinct primes, and let $1=d_1<d_2<\ldots<d_{16}=n$ be the divisors of $n=p_1p_2p_3p_4$. Determine all $n<2001$ with the property that
$d_9-d_8=22$.

Solution:
1. We are given four distinct primes \( p_1, p_2, p_3, p_4 \) such that \( p_1 < p_2 < p_3 < p_4 \) and \( n = p_1 p_2 p_3 p_4 \). We need to find all \( n < 2001 \) such that \( d_9 - d_8 = 22 \), where \( 1 = d_1 < d_2 < \cdots < d_{16} = n \) are the divisors of \( n \).

2. Since \( d_i \cdot d_{17-i} = n \) for all \( 1 \leq i \leq 16 \), choosing \( i = 8 \) gives us \( d_8 \cdot d_9 = n \).

3. Assume \( q \in \{2, 11\} \) and \( q \mid n \). Then \( q \mid d_8 d_9 \) implies \( q \mid d_8 \) and \( q \mid d_9 \), leading to \( q^2 \mid n \). However, since \( n \) is square-free, this is impossible. Therefore, \( p_1 \neq 2 \) and \( p_i \neq 11 \) for \( i \in \{2, 3, 4\} \).

4. Assume \( p_1 > 3 \). Then \( n = p_1 p_2 p_3 p_4 \leq 5 \cdot 7 \cdot 13 \cdot 17 = 7735 > 2000 \), which is a contradiction. Hence, \( p_1 \leq 3 \). Since \( p_2 \neq 2 \), we have \( p_1 = 3 \).

5. Assume \( p_2 > 5 \). Then \( p_4 = \frac{n}{p_1 p_2 p_3} \leq \frac{2000}{3 \cdot 7 \cdot 13} = \frac{2000}{273} < 8 \), contradicting \( p_4 > 11 \). Hence, \( p_2 \leq 5 \), which means \( p_2 = 5 \).

6. Given \( p_1 = 3 \) and \( p_2 = 5 \), we have \( p_3^2 < p_3 p_4 = \frac{n}{3 \cdot 5} = \frac{n}{15} \leq \frac{2000}{15} < 144 = 12^2 \), yielding \( p_3 < 12 \). Therefore, \( p_3 = 7 \).

7. Thus, \( n = 3 \cdot 5 \cdot 7 \cdot p_4 = 105 p_4 \). This implies \( 105 p_4 \leq 2000 \), so \( p_4 \leq \frac{2000}{105} < 20 \). Since \( p_3 = 7 < p_4 \leq 19 \) and \( p_4 \neq 11 \), we have \( p_4 \in \{13, 17, 19\} \).

8. Since \( d_8 d_9 = n \) and \( n \) is odd, \( d_8 \) must be odd. Given \( d_9 - d_8 = 22 \), we have \( d_8 d_9 - d_8^2 = 22 d_8 \), leading to \( d_8^2 + 22 d_8 = n \). Thus, \( (d_8 + 11)^2 = 105 p_4 + 121 \).

9. Since \( d_8 \) is odd, \( 4 \mid 105 p_4 + 121 \), implying \( 4 \mid p_4 + 1 \). Given \( p_4 \in \{13, 17, 19\} \), we find \( p_4 = 19 \).

10. Therefore, \( n = 105 \cdot 19 = 1995 \). This means \( d_8 = 5 \cdot 7 = 35 \) and \( d_9 = 3 \cdot 19 = 57 \), yielding \( d_9 - d_8 = 57 - 35 = 22 \).

The final answer is \( \boxed{ n = 1995 } \).

Answer: n = 1995


Given a triangle $ABC$ with $\measuredangle CBA=20^{\circ}, \measuredangle ACB=40^{\circ}$ and $\overline{AD}=2 \text{~cm}$, where $D$ is the intersection point of the angle bisector of the angle at vertex $A$ and side $BC$. Determine the difference $\overline{BC}-\overline{AB}$.

Solution:
Solution. On side $BC$, we choose a point $E$ such that $\measuredangle AEB=80^{\circ}$. Then $\measuredangle BAE=80^{\circ}$, so triangle $ABE$ is isosceles, i.e., $\overline{AB}=\overline{AE}$. Triangle $AEC$ is isosceles because $\measuredangle EAC=40^{\circ}$, so $\overline{AE}=\overline{CE}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_1cb7beb611b796adc950g-08.jpg?height=285&width=927&top_left_y=653&top_left_x=457)

Triangle $ADE$ is isosceles because $\measuredangle ADE=80^{\circ}$, so $\overline{AE}=\overline{AD}=$ $2 \text{ cm}$. Therefore, $\overline{BC}-\overline{AB}=\overline{BC}-\overline{BE}=\overline{CE}=\overline{AE}=\overline{AD}=2 \text{ cm}$

## VII Section

Answer: 2


Find all positive integers $x, y$ such that $x^{2}+3 y, y^{2}+3 x$ are both perfect squares.

Solution:
5. Let $\left\{\begin{array}{l}x^{2}+3 y=u^{2} \\ y^{2}+3 x=v^{2}\end{array}\right.$ Since $x, y$ are positive integers, we have $u>x, v>y$. We set $u=x+a, v=y+b$, where $a, b$ are positive integers. From
$$\left\{\begin{array}{l}
x^{2}+3 y=(x+a)^{2} \\
y^{2}+3 x=(y+b)^{2}
\end{array}\right.$$

we can derive
$$\left\{\begin{array}{l}
3 y=2 a x+a^{2} \\
3 x=2 b y+b^{2}
\end{array}\right.$$

Solving this system of linear equations in $x, y$ yields $\left\{\begin{array}{l}x=\frac{2 a^{2} b+3 b^{2}}{9-4 a b}, \\ y=\frac{2 b^{2} a+3 a^{2}}{9-4 a b} .\end{array}\right.$ Since $x, y$ are positive integers, we have $9-4 a b>0$. Since $a, b$ are positive integers, we have $a b=1$ or 2, i.e., $(a, b)=(1,1),(1,2)$, $(2,1)$. Correspondingly, we obtain $(x, y)=(1,1),(16,11),(11,16)$.

Answer: (1,1),(16,11),(11,16)


We cut an $8 \times 8$ chessboard into $p$ rectangular pieces such that no single square is cut in half. Each such cutting must satisfy the following conditions:

(a) Each rectangle must contain an equal number of white and black squares.

(b) If $a_{i}$ denotes the number of white squares in the $i$-th rectangle, then the sequence $a_{1}<a_{2}<\ldots<a_{p}$ must hold.

Find the largest value of $p$ for which such a cutting exists.

Furthermore, provide all sequences of $a_{1}, a_{2}, \ldots, a_{p}$ corresponding to this value of $p$.

Solution:
Since there are 32 white squares on a chessboard,

$$
a_{1}+a_{2}+\ldots+a_{p}=32
$$

It follows that $p$ can be at most 7, because according to condition $b$, if $p \geq 8$,

$$
a_{1}+a_{2}+\ldots+a_{p} \geq 1+2+\ldots+8=36>32
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_d2753a75c03c296983f7g-1.jpg?height=233&width=246&top_left_y=539&top_left_x=522)
a)

![](https://cdn.mathpix.com/cropped/2024_05_02_d2753a75c03c296983f7g-1.jpg?height=233&width=241&top_left_y=539&top_left_x=801)
b)

![](https://cdn.mathpix.com/cropped/2024_05_02_d2753a75c03c296983f7g-1.jpg?height=243&width=225&top_left_y=529&top_left_x=1080)
c)

![](https://cdn.mathpix.com/cropped/2024_05_02_d2753a75c03c296983f7g-1.jpg?height=247&width=241&top_left_y=527&top_left_x=1338)
d)

We need to examine whether $p=7$ is possible. After some trial and error, we can find a partition that meets the conditions of the problem, as shown in part a) of the figure. This answers the first part of the problem: the maximum value of $p$ is 7.

To generate all possible sequences for $p=7$, we need to determine which numbers can appear among $a_{1}, \ldots, a_{7}$. Any number from 1 to 10 can appear. 11 cannot, because a rectangle of 22 squares cannot be cut from an $8 \times 8$ square. Any number larger than 11 cannot appear either, because then the sum of the other six different natural numbers could be at most $32-12=20$, but $1+2+\ldots+6=21$.

At least one of the numbers 8, 9, 10 must appear, because the sum of the first 7 natural numbers is only 28. The numbers 8 and 10, or 9 and 10, cannot appear together, because the sum of the other five different terms would have to be 14 or 13, which is not possible because $1+2+3+4+5=15$. Depending on which of the numbers 8, 9, 10 appear, we have four possible cases:

a) Only the 8 appears: $1+2+\ldots+8=36$, and we need to remove one of these to get a sum of 32. This can only be achieved by removing the 4. The resulting sequence is: 1, 2, 3, 5, 6, $7,8$.

b) Only the 9 appears: Similarly, by removing the 5, we get the sequence 1, 2, 3, 4, 6, 7, 9.

c) Only the 10 appears: By removing the 6, we get the sequence $1,2,3,4,5,7,10$.

d) Both 8 and 9 appear. We need to remove two numbers from $1,2, \ldots, 8,9$ to reduce the sum by $45-32=13$, but we must keep the 8 and 9. This can only be achieved by removing the 6 and 7, resulting in the sequence: $1,2,3,4,5,8,9$.

Thus, at most the four sequences listed above can correspond to the value $p=7$. It remains to show that these sequences can indeed be obtained by partitioning. A partition for each sequence is shown in the figure.

Neumer Attila (Budapest, Fazekas M. Gyak. Gimn., I. o. t.)

Answer: 7


The sides of the bases of a regular truncated quadrilateral pyramid are 2 and 1 cm, and the height is $3 \mathrm{~cm}$. A plane is drawn through the point of intersection of the diagonals of the pyramid, parallel to the bases of the pyramid, dividing the pyramid into two parts. Find the volume of each of the resulting parts.

Solution:
Solution.

$O$ and $O_{1}$ are the centers of the lower and upper bases of the right truncated pyramid $A B C D A_{1} B_{1} C_{1} D_{1}, O_{2}$ is the point of intersection of its diagonals $A C_{1}$ and $A_{1} C$ (Fig. 11.19), $O O_{1}=3 \text{ cm}, A D=2 \text{ cm}$, $A_{1} D_{1}=1$ cm. The square $A_{2} B_{2} C_{2} D_{2}$ is the section of the pyramid by the plane mentioned in the problem condition.

The diagonal section of the truncated pyramid is an isosceles trapezoid $A A_{1} C_{1} C, \Delta A_{1} O_{2} C_{1} \sim \Delta A O_{2} C$. Then $\frac{O_{1} O_{2}}{O_{2} O}=\frac{A_{1} C_{1}}{A C}=\frac{A_{1} D_{1}}{A D}=\frac{1}{2}$.

Since $O O_{1}=3 \text{ cm}$, then $O_{1} O_{2}=1 \text{ cm}, O O_{2}=2 \text{ cm} . \Delta A_{1} A C_{1} \sim \triangle A_{2} A O_{2}$ and $\frac{A_{2} O_{2}}{A_{1} C_{1}}=\frac{O O_{2}}{O O_{1}}=\frac{2}{3}$. Then $A_{2} O_{2}=\frac{2}{3} A_{1} C_{1}=\frac{2 \sqrt{2}}{3}$ cm and $A_{2} C_{2}=\frac{4 \sqrt{2}}{3} \text{ cm}$, $A_{2} D_{2}=\frac{4}{3}$ cm. The areas of the squares $A B C D, A_{1} B_{1} C_{1} D_{1}, A_{2} B_{2} C_{2} D_{2}$ are respectively $S=4 \text{ cm}^{2}, S_{1}=1 \text{ cm}^{2}, S_{2}=\frac{16}{9} \text{ cm}^{2}$. The volume of the truncated pyramid with height $O_{1} O_{2}: V_{1}=\frac{O_{1} O_{2}}{3}\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right)=\frac{1}{3}\left(1+\frac{16}{9}+\frac{4}{3}\right)=\frac{37}{27}$ cm $^{3}$.

The volume of the truncated pyramid with height $\mathrm{O}_{2} \mathrm{O}$ :

$V_{2}=\frac{O_{2} O}{3}\left(S+S_{2}+\sqrt{S S_{2}}\right)=\frac{2}{3}\left(4+\frac{16}{9}+\frac{8}{3}\right)=\frac{152}{27} \text{ cm}^{3}$.

Answer: $\frac{37}{27} \text{ cm}^{3} ; \frac{152}{27} \text{ cm}^{3}$.

Answer: \frac{37}{27}


A finite sequence satisfies: the sum of any 3 consecutive terms is negative, and the sum of any 4 consecutive terms is positive. The maximum number of terms in this sequence is $\qquad$ .

Solution:
Solution: 5
On the one hand, a sequence of 5 terms can be constructed: $2,2,-5,2,2$ which meets the requirements;
On the other hand, prove that the sequence satisfying the conditions does not exceed 5 terms. Otherwise, take the first 6 terms, and arrange them as follows:
$$
\begin{array}{l}
a_{1}, a_{2}, a_{3} \\
a_{2}, a_{3}, a_{4} \\
a_{3}, a_{4}, a_{5} \\
a_{4}, a_{5}, a_{6}
\end{array}
$$

From the fact that the sum of each row is negative, we know that the sum of these 12 numbers is negative; from the fact that the sum of each column is positive, we know that the sum of these 12 numbers is positive. This is a contradiction.

Answer: 5


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 $n-2$ people is equal, and it is $3^{k}$ times, where $k$ is a positive integer. Find all possible values of $n$.

Assume the $n$ people are $A_{1}, A_{2}, \cdots, A_{n}$, and the number of calls made by $A_{i}$ is $m_{i}$. The number of calls between $A_{i}$ and $A_{j}$ is $\lambda_{i j}(1 \leqslant i, j \leqslant n)$, where $\lambda_{i j}=0$ or 1. Thus, the quantitative characteristic is
$$
m_{i}+m_{j}-\lambda_{i j}=\frac{1}{2} \sum_{s=1}^{n} m_{s}-3^{k}=c,
$$

where $c$ is a constant. Consequently, we conjecture that $m_{i}(i=1,2, \cdots, n)$ is a constant.

Solution:
Let $n$ people be denoted as $A_{1}, A_{2}, \cdots, A_{n}$. Let the number of calls made by $A_{i}$ be $m_{i}$, and the number of calls between $A_{i}$ and $A_{j}$ be $\lambda_{i j}(1 \leqslant i, j \leqslant n)$, where $\lambda_{i j}=0$ or 1.
Clearly, $n \geqslant 5$. Therefore,
$$
\begin{array}{l}
\left|m_{i}-m_{j}\right|=\left|\left(m_{i}+m_{s}\right)-\left(m_{j}+m_{s}\right)\right| \\
=\left|\lambda_{i s}-\lambda_{j}\right| \leqslant 1,1 \leqslant i, j, s \leqslant n .
\end{array}
$$

Let $m_{i}=\max \left\{m_{s}, 1 \leqslant s \leqslant n\right\}$,
$$
m_{j}=\min \left\{m_{s}, 1 \leqslant s \leqslant n\right\} .
$$

Thus, $m_{i}-m_{j} \leqslant 1$.
If $m_{i}-m_{j}=1$, then for any $s \neq i, j, 1 \leqslant s \leqslant n$, we have
$$
\begin{array}{l}
\left(m_{i}+m_{s}-\lambda_{i s}\right)-\left(m_{j}+m_{s}-\lambda_{j s}\right) \\
=1-\left(\lambda_{i s}-\lambda_{j s}\right)=0,
\end{array}
$$

which means $\lambda_{i s}-\lambda_{j s} \equiv 1,1 \leqslant i, j, s \leqslant n$.
Therefore, $\lambda_{i s} \equiv 1, \lambda_{j i} \equiv 0, s \neq i, j$, and $1 \leqslant s \leqslant n$.
Thus, $m_{i} \geqslant n-2, m_{j} \leqslant 1$.
Hence, $m_{i}-m_{j} \geqslant n-3 \geqslant 2$, which is a contradiction.
Therefore, $m_{i}-m_{j}=0$.
Then $\lambda_{i j} \equiv 0$ or $\lambda_{i j} \equiv 1$.
If $\lambda_{i j} \equiv 0$, then $m_{s} \equiv 0,1 \leqslant s \leqslant n$, which is a contradiction.
If $\lambda_{i j} \equiv 1$, then $m_{s}=n-1,1 \leqslant s \leqslant n$.
Thus, $(n-2)(n-3)=3^{k} \times 2$.
Let $n-2=2 \times 3^{k_{1}}, n-3=3^{k_{2}}, k_{1} \geqslant k_{2}$.

From $2 \times 3^{k_{1}}-3^{k_{2}}=1$, we get
$$
3^{k_{2}}=1 \text{. }
$$

Therefore, $k_{1}=k_{2}=0$. This contradicts $k \geqslant 1$.
Let $n-2=3^{k_{1}}, n-3=2 \times 3^{k_{2}}, k_{1} \geqslant k_{2}+1$.
From $3^{k_{1}}-2 \times 3^{k_{2}}=1$, we get
$$
k_{2}=0, k_{1}=1 \text{. }
$$

Therefore, $n=5$ is the solution.

Answer: 5


Determine all natural numbers $n$ for which among the numbers $n, 4^{n}+1$, and $n^{2}+2$ at least two are prime numbers.

Solution:
## Solution.

First, note that the number $4^{n}+1$ can only be prime if $n=1$ or if $n$ is even.

Indeed, for the values $n=1,2,3,4, \ldots$, the expression $4^{n}$ gives remainders $4,1,4,1, \ldots$ when divided by 5. Thus, when $n$ is odd, the number $4^{n}+1$ is divisible by 5, and thus is a composite number (except for $n=1$ when $4^{n}+1=5$).

For the last part, notice that if the number $n$ is not divisible by 3, then it is of the form $n=3k+1$ or $3k-1$. Then we have

$$
n^{2}+2=(3k \pm 1)^{2}+2=9k^{2} \pm 6k+3
$$

which means $n^{2}+2$ is a number divisible by 3.

1 point

Thus, $n^{2}+2$ can only be prime if $n$ is divisible by 3 (or if $n=1$, when this expression equals 3).

1 point

Now let's consider all cases:

1) The numbers $4^{n}+1$ and $n^{2}+2$ are both prime. For $n=1$, we get a solution (among the three numbers in the problem, two are prime: 5 and 3).

1 point

For $n>1$, from the above conclusions, we get that the number cannot be either even or odd, which means there are no more solutions in this case.

2) The numbers $4^{n}+1$ and $n$ are both prime. From the above conclusions, we know that the number cannot be an odd number. Therefore, $n$ must be even and prime, for which the only possibility is $n=2$. This is indeed a solution (among the three numbers, two are prime: 2 and 17).

2 points

3) The numbers $n^{2}+2$ and $n$ are both prime. From the above conclusions, we know that the number must be divisible by 3. However, $n$ is prime and divisible by 3 only in the case $n=3$. This is indeed a solution (among the three numbers, two are prime: 3 and 11).

Therefore, the desired numbers are $n=1,2,3$.

Note: The solution can be grouped into cases in many ways, but all solutions must be scored as above. Specifically:

- The statement that $4^{n}+1$ cannot be prime if $n$ is an odd number greater than 1 (total 2 points, where the statement and proof are worth 1 point each)
- The statement that $n^{2}+2$ cannot be prime if $n$ is even (1 point)
- The statement that $n^{2}+2$ cannot be prime if $n$ is not divisible by 3, or equal to 1 (total 2 points, where the statement and proof are worth 1 point each)
- The conclusion that $n=1$ is a solution (1 point), and for $n \neq 1$, the numbers $4^{n}+1$ and $n^{2}+2$ are not both prime (0 points)
- The conclusion that $n=2$ is a solution (1 point), and for $n \neq 2$, the numbers $4^{n}+1$ and $n$ are not both prime (1 point)
- The conclusion that $n=3$ is a solution (1 point), and for $n \neq 3$, the numbers $n^{2}+2$ and $n$ are not both prime (1 point).

The proof of the compositeness of $4^{n}+1$ can also be derived by factorization: if $n$ is an odd number greater than 1, then

$$
\left(4^{n}+1\right)=(4+1)\left(4^{n-1}-4^{n-2}+\cdots+1\right)
$$

is clearly a composite number.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

2nd grade - high school - A variant

## February 17, 2021.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO SCORE AND EVALUATE THAT APPROACH APPROPRIATELY.

Answer: 1,2,3


For each integer $n \geq 1$, define $a_{n}=\left[\frac{n}{[\sqrt{n}]}\right]$, where $[x]$ denotes the largest integer not exceeding $x$, for any real number $x$. Find the number of all $n$ in the set $\{1,2,3, \ldots, 2010\}$ for which $a_{n}>a_{n+1}$.

Solution:
Solution: Let us examine the first few natural numbers: $1,2,3,4,5,6,7,8,9$. Here we see that $a_{n}=1,2,3,2,2,3,3,4,3$. We observe that $a_{n} \leq a_{n+1}$ for all $n$ except when $n+1$ is a square in which case $a_{n}>a_{n+1}$. We prove that this observation is valid in general. Consider the range

$$
m^{2}, m^{2}+1, m^{2}+2, \ldots, m^{2}+m, m^{2}+m+1, \ldots, m^{2}+2 m
$$

Let $n$ take values in this range so that $n=m^{2}+r$, where $0 \leq r \leq 2 m$. Then we see that $[\sqrt{n}]=m$ and hence

$$
\left[\frac{n}{[\sqrt{n}]}\right]=\left[\frac{m^{2}+r}{m}\right]=m+\left[\frac{r}{m}\right]
$$

Thus $a_{n}$ takes the values $\underbrace{m, m, m, \ldots, m}_{m \text { times }}, \underbrace{m+1, m+1, m+1, \ldots, m+1}_{m \text { times }}, m+2$, in this range. But when $n=(m+1)^{2}$, we see that $a_{n}=m+1$. This shows that $a_{n-1}>a_{n}$ whenever $n=(m+1)^{2}$. When we take $n$ in the set $\{1,2,3, \ldots, 2010\}$, we see that the only squares are $1^{2}, 2^{2}, \ldots, 44^{2}$ (since $44^{2}=1936$ and $\left.45^{2}=2025\right)$ and $n=(m+1)^{2}$ is possible for only 43 values of $m$. Thus $a_{n}>a_{n+1}$ for 43 values of $n$. (These are $2^{2}-1$, $3^{2}-1, \ldots, 44^{2}-1$.)

Answer: 43


Suppose you have $27$ identical unit cubes colored such that $3$ faces adjacent to a vertex are red and the other $3$ are colored blue. Suppose further that you assemble these $27$ cubes randomly into a larger cube with $3$ cubes to an edge (in particular, the orientation of each cube is random). The probability that the entire cube is one solid color can be written as $\frac{1}{2^n}$ for some positive integer $n$. Find $n$.

Solution:
1. **Identify the types of unit cubes in the larger cube:**
   - There are 27 unit cubes in total.
   - These can be categorized into:
     - 8 corner cubes
     - 12 edge cubes
     - 6 face-center cubes
     - 1 center cube

2. **Determine the probability for each type of cube to be oriented correctly:**
   - **Corner cubes:** Each corner cube has 3 red faces adjacent to a vertex. For it to be oriented correctly, the vertex with the red faces must be at the corner of the larger cube. There are 8 possible orientations for a corner cube, but only 1 correct orientation. Thus, the probability is:
     \[
     P(\text{corner}) = \frac{1}{8}
     \]
   - **Edge cubes:** Each edge cube has 3 red faces adjacent to a vertex. For it to be oriented correctly, the vertex with the red faces must be on the edge of the larger cube. There are 8 possible orientations for an edge cube, but only 2 correct orientations. Thus, the probability is:
     \[
     P(\text{edge}) = \frac{2}{8} = \frac{1}{4}
     \]
   - **Face-center cubes:** Each face-center cube has 3 red faces adjacent to a vertex. For it to be oriented correctly, the vertex with the red faces must be on the face of the larger cube. There are 8 possible orientations for a face-center cube, but only 4 correct orientations. Thus, the probability is:
     \[
     P(\text{face-center}) = \frac{4}{8} = \frac{1}{2}
     \]
   - **Center cube:** The center cube is always oriented correctly because it is never seen from the outside. Thus, the probability is:
     \[
     P(\text{center}) = 1
     \]

3. **Calculate the probability that the entire larger cube is solid red:**
   - The probability that all 8 corner cubes are oriented correctly is:
     \[
     \left(\frac{1}{8}\right)^8
     \]
   - The probability that all 12 edge cubes are oriented correctly is:
     \[
     \left(\frac{1}{4}\right)^{12}
     \]
   - The probability that all 6 face-center cubes are oriented correctly is:
     \[
     \left(\frac{1}{2}\right)^6
     \]
   - The probability that the center cube is oriented correctly is:
     \[
     1
     \]

4. **Combine these probabilities:**
   \[
   P(\text{solid red}) = \left(\frac{1}{8}\right)^8 \cdot \left(\frac{1}{4}\right)^{12} \cdot \left(\frac{1}{2}\right)^6 \cdot 1
   \]

5. **Simplify the expression:**
   \[
   P(\text{solid red}) = \left(\frac{1}{8}\right)^8 \cdot \left(\frac{1}{2^2}\right)^{12} \cdot \left(\frac{1}{2}\right)^6
   \]
   \[
   = \left(\frac{1}{2^3}\right)^8 \cdot \left(\frac{1}{2^2}\right)^{12} \cdot \left(\frac{1}{2}\right)^6
   \]
   \[
   = \frac{1}{2^{24}} \cdot \frac{1}{2^{24}} \cdot \frac{1}{2^6}
   \]
   \[
   = \frac{1}{2^{24+24+6}} = \frac{1}{2^{54}}
   \]

6. **Account for the possibility of the entire cube being solid blue:**
   - The probability of the entire cube being solid blue is the same as the probability of it being solid red.
   - Therefore, the total probability of the entire cube being one solid color (either red or blue) is:
     \[
     2 \cdot \frac{1}{2^{54}} = \frac{1}{2^{53}}
     \]

The final answer is \( \boxed{53} \)

Answer: 53


Find the maximum possible area of a quadrilateral in which the product of any two adjacent sides is 1.

Solution:
1. Find the maximum possible area of a quadrilateral for which the product of any two adjacent sides is 1.

OTBET: 1.

SOLUTION.

Let the quadrilateral have sides $a, b, c, d$. Then $a b=b c=c d=d a=1$. From the equality $a b=b c$, it follows that $a=c$, and from the equality $b c=c d$, we get that $b=d$. Therefore, the given quadrilateral is a parallelogram. Let $\alpha$ be the angle between sides $a$ and $b$. Then $S=a \cdot b \sin \alpha$, and the area is maximized if $\sin \alpha=1$. Thus, $S=a \cdot b=1$.

Answer: 1


We use $A, B, C, D, E$ to represent the number of gold coins held by students Ã§ÂÂ², Ã¤Â¹Â, Ã¤Â¸Â, Ã¤Â¸Â, Ã¦ÂÂ, respectively. The total number of these gold coins is denoted by $n$, which means $n=A+B+C+D+E$. It is known that:
(1) $A, B, C, D, E$ are all greater than or equal to 1;
(2) $A<B<C<D<E$;
(3) Each of the five students knows the value of $n$ and the number of gold coins they hold;
For a certain $n$, if there exists a distribution of gold coins (once the total number of gold coins is fixed, the number of gold coins each person holds can be allocated according to the above requirements) such that no student can guess the number of gold coins everyone holds, then $n$ is called a "central number". Find: the smallest "central number".

Answer: 19


When $m=$ $\qquad$, the polynomial
$$
12 x^{2}-10 x y+2 y^{2}+11 x-5 y+m
$$
can be factored into the product of two linear factors.

Solution:
Solution: First, factorize the quadratic term, we have
$$
12 x^{2}-10 x y+2 y^{2}=(3 x-y)(4 x-2 y) \text {. }
$$

Therefore, we can set
$$
\begin{array}{l}
12 x^{2}-10 x y+2 y^{2}+11 x-5 y+m \\
=(3 x-y+a)(4 x-2 y+b) \\
=(3 x-y)(4 x-2 y)+(4 a+3 b) x- \\
(2 a+b) y+a b .
\end{array}
$$

Comparing the coefficients of the linear terms, we get
$$
\left\{\begin{array}{l}
4 a+3 b=11, \\
2 a+b=5 .
\end{array}\right.
$$

Solving these, we find $a=2, b=1$.
Comparing the constant term, we get $m=a b=2$.

Answer: 2


Given a line $l$ intersects the ellipse $C: \frac{x^{2}}{2}+y^{2}=1$ at points $A, B$, and $O$ is the origin.
(1) Find the maximum value of the area of $\triangle A O B$, and the equation of the line $l$ at this time;
(2) Let the set $L=\{l \mid$ line $l$ makes the area of $\triangle A O B$ take the maximum value $\}, l_{1}, l_{2}, l_{3}, l_{4} \in L$ and satisfy $l_{1} / / l_{2}, l_{3} / / l_{4}$, $k_{1}, k_{2}, k_{3}, k_{4}$ are the slopes of $l_{1}, l_{2}, l_{3}, l_{4}$ respectively and satisfy $k_{1}+k_{2}+k_{3}+k_{4}=0$, find the minimum value of the area of the quadrilateral formed by these four lines.

Solution:
10. (1) When the slope of $l$ does not exist, we can set $l: x=t$, then
$$
S_{\triangle A O B}=\frac{1}{2}|t| \cdot 2 \sqrt{1-\frac{t^{2}}{2}}=\frac{\sqrt{2}}{2} \sqrt{t^{2}\left(2-t^{2}\right)} \leqslant \frac{\sqrt{2}}{2} .
$$

Equality holds if and only if $t^{2}=2-t^{2}$, i.e., $t= \pm 1$.
When the slope of $l$ exists, we can set $l: y=k x+m$, from
$$
\left\{\begin{array}{l}
y=k x+m \\
x^{2}+2 y^{2}=2
\end{array}\right.
$$

we get
$$
\left(1+2 k^{2}\right) x^{2}+4 k m x+2\left(m^{2}-1\right)=0,
$$

so
$$
\Delta=16 k^{2} m^{2}-8\left(1+2 k^{2}\right)\left(m^{2}-1\right)=8\left(1+2 k^{2}-m^{2}\right),
$$

so
$$
\begin{aligned}
S_{\triangle A O B} & =\frac{1}{2}|A B| \cdot h=\frac{1}{2} \sqrt{1+k^{2}} \cdot \frac{\sqrt{8\left(1+2 k^{2}-m^{2}\right)}}{1+2 k^{2}} \cdot \frac{|m|}{\sqrt{1+k^{2}}} \\
& =\sqrt{2} \frac{\sqrt{\left(1+2 k^{2}-m^{2}\right) m^{2}}}{1+2 k^{2}} \\
& \leqslant \sqrt{2} \cdot \frac{\frac{1}{2}\left(1+2 k^{2}\right)}{1+2 k^{2}}=\frac{\sqrt{2}}{2} .
\end{aligned}
$$

In the above, $h$ represents the distance from $O$ to $A B$. Equality holds if and only if $m^{2}=1+2 k^{2}-m^{2}$, i.e., $m^{2}=k^{2}+\frac{1}{2}$.

In summary, the maximum value of $S_{\triangle A O B}$ is $\frac{\sqrt{2}}{2}$, at this time the equation of the line $l$ is $x= \pm 1$ or $y=k x \pm \sqrt{k^{2}+\frac{1}{2}}$ $(k \in \mathbf{R})$.
(2) By symmetry, the lines
$$
\begin{array}{l}
l_{1}: y=k x+\sqrt{k^{2}+\frac{1}{2}}, \\
l_{2}: y=k x-\sqrt{k^{2}+\frac{1}{2}}, \\
l_{3}: y=-k x+\sqrt{k^{2}+\frac{1}{2}}, \\
l_{4}: y=-k x-\sqrt{k^{2}+\frac{1}{2}} .
\end{array}
$$

Here, $k>0$. So the quadrilateral formed by these four lines is a rhombus, then
$$
S=4 \cdot \frac{1}{2} \sqrt{k^{2}+\frac{1}{2}} \cdot \frac{\sqrt{k^{2}+\frac{1}{2}}}{k}=2\left(k+\frac{1}{2 k}\right) \geqslant 2 \sqrt{2},
$$

Equality holds if and only if $k=\frac{1}{2 k}$, i.e., $k=\frac{\sqrt{2}}{2}$. Therefore, the minimum value of the area of the quadrilateral is $2 \sqrt{2}$.

Answer: 2\sqrt{2}


Find the smallest distance from the point with coordinates $(10, 5, 10)$ to a point whose coordinates are positive and satisfy the inequality $(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq 9 \sqrt{1-(2x+y)^{2}}$. In your answer, write the square of the found distance.

Solution:
Solution. We use the Cauchy inequality $a+b+c \geq 3 \sqrt[3]{a b c}$, which holds for all positive values of $a, b, c$. Then, $x+y+z \geq 3 \sqrt[3]{x y z}, \quad \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq 3 \sqrt[3]{\frac{1}{x y z}}$. Since all parts of the inequalities are positive, we have

$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq 9 \sqrt[3]{x y z \cdot \frac{1}{x y z}}=9 . \quad \text { The expression }
$$

$9 \sqrt{1-(2 x+y)^{2}} \leq 9$ for all possible values of $x$ and $y$. The original inequality is valid for all positive $z$ and all positive $x$ and $y$ for which the inequality 1 $(2 x+y)^{2} \geq 0$ holds. Thus, the minimum distance from the point with coordinates $(10 ; 5 ; 10)$ to the point, the coordinates of which are positive and satisfy the original inequality, is the distance from the point (10; 5) in the Oxy plane to the line $2 x+y=1$. This distance squared is 115.2.

Answer: 115.2.

Answer: 115.2


$\lim _{x \rightarrow 0}(1-\cos x)^{x}$.

Solution:
Solution.

$$
\lim _{x \rightarrow 0}(1-\cos x)^{x}=\left(0^{0}\right)=e^{\lim _{x \rightarrow 0} x \ln (1-\cos x)}
$$

## Find separately

$$
\begin{aligned}
& \lim _{x \rightarrow 0} x \cdot \ln (1-\cos x)=(0 \cdot(-\infty))=\lim _{x \rightarrow 0} \frac{(\ln (1-\cos x))^{\prime}}{(1 / x)^{\prime}}=\lim _{x \rightarrow 0} \frac{\frac{\sin x}{1-\cos x}}{-\frac{1}{x^{2}}}= \\
& =-\lim _{x \rightarrow 0} \frac{x^{2} \cdot 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^{2} \frac{x}{2}}=-\lim _{x \rightarrow 0} 2 x \cdot \lim _{x / 2 \rightarrow 0} \frac{x / 2}{\sin \frac{x}{2}} \cdot \lim _{x \rightarrow 0} \cos \frac{x}{2}=0
\end{aligned}
$$

Here, the first remarkable limit was used

$$
\lim _{\alpha \rightarrow 0} \frac{\alpha}{\sin \alpha}=1
$$

From here $e^{\lim _{x \rightarrow 0} x \ln (1-\cos x)}=e^{0}=1$.

## INDETERMINACY OF THE FORM « 1 *

Answer: 1


Determine all pairs $(a, n)$ of positive integers such that

$$
3^{n}=a^{2}-16
$$

Solution:
Let $(a, n)$ be a solution to the equation.

Then, $3^{n}=(a-4)(a+4)$.

We deduce that $a-4$ and $a+4$ are powers of 3.

There exist integers $k$ and $l$ such that $a-4=3^{k}$ and $a+4=3^{l}$ (with $l \geqslant k$).

Thus, $8=(a+4)-(a-4)=3^{k}\left(3^{l-k}-1\right)$.

Since 3 does not divide 8, we get $k=0$ and $8=3^{l}-1$ hence $l=2$. Therefore, $a-4=1$ so $a=5$ and $n=2$. Conversely, the pair $(5,2)$ is a solution to the required equation.

Answer: (5,2)


On her birthday, Grandma told Xiaoming: "I have only had 18 birthdays since I was born." Grandma should be   years old this year.

Solution: 【Answer】Solution: $18 \times 4=72$ (years), Answer: Grandma should be 72 years old this year. Therefore, the answer is: 72.

Answer: 72


$M$ is a point inside the convex quadrilateral $A B C D$, and the points symmetric to $M$ with respect to the midpoints of the sides are $P, Q, R, S$. If the area of quadrilateral $A B C D$ is 1, then the area of quadrilateral $P Q R S$ is equal to

Solution: 6. 2 (Hint: Connect the midpoints of each pair of adjacent sides of quadrilateral $A B C D$, to get a parallelogram, whose area is easily known to be $\frac{1}{2}$. On the other hand, these four sides are precisely the midlines of four triangles that share $M$ as a common vertex and have the four sides of quadrilateral $P Q R S$ as their bases.)

Answer: 2


Let the set $A=\{1,2, \cdots, 366\}$. If a two-element subset $B=\{a, b\}$ of $A$ satisfies $17 \mid (a+b)$, then $B$ is said to have property $P$.
(1) Find the number of two-element subsets of $A$ that have property $P$;
(2) Find the number of a set of two-element subsets of $A$, which are pairwise disjoint and have property $P$.

Solution:
4. (1) Divide $1,2, \cdots, 366$ into 17 classes based on the remainder when divided by 17: $[0],[1], \cdots,[16]$. Number of elements.
(i) When $a, b \in[0]$, the number of subsets with property $P$ is $C_{21}^{2}=210$;
(ii) When $a \in[k], b \in[17-k], k=1,2, \cdots, 7$, the number of subsets with property $P$ is $C_{22}^{1} \cdot C_{21}^{1}=462$;
(iii) When $a \in[8], b \in[9]$, the number of subsets with property $P$ is $C_{22}^{1} \cdot C_{22}^{1}=484$.
Therefore, the number of subsets of $A$ with property $P$ is $210+462 \times 7+484=3928$.
(2) To ensure that the binary subsets do not intersect, when $a, b \in[0]$, 10 subsets can be paired;
When $a \in[8], b \in[9]$, 22 subsets can be paired.
Thus, the number of pairwise non-intersecting subsets with property $P$ is $10+21 \times 7+22=179$.

Answer: 179


In Rt $\triangle A B C$, $\angle A=30^{\circ}, \angle C=90^{\circ}$, equilateral triangles $\triangle A B D$ and $\triangle A C E$ are constructed outward from $A B$ and $A C$ respectively, and $D E$ is connected, intersecting $A C$ and $A B$ at points $F$ and $G$. Then the value of $\frac{A F}{A G}$ is ( ).
(A) $\frac{2 \sqrt{3}}{7}$
(B) $\frac{3 \sqrt{3}}{7}$
(C) $\frac{4 \sqrt{3}}{7}$
(D) $\frac{5 \sqrt{3}}{7}$

Solution:
4.C.

As shown in Figure 3, draw $E M \perp A C$ at $M$, extend $E M$ to intersect $A B$ at $N$, and connect $D N$. Then $M$ is the midpoint of $A C$, and $M N \parallel C B$. Therefore, $N$ is the midpoint of $A B$. Thus, $D N \perp A B$.
Also, $\angle E A N = \angle D A M = 60^{\circ} + 30^{\circ} = 90^{\circ}$, which gives $A E \parallel D N, A D \parallel E N$.
Therefore, quadrilateral $A D N E$ is a parallelogram.
Hence, $A G = \frac{1}{2} A N = \frac{1}{4} A B$.
Let $B C = a$. Then
$A D = A B = 2 a, A E = A C = \sqrt{3} a, A G = \frac{1}{2} a$.
In the right triangle $\triangle A E M$,
$A M = \frac{1}{2} A C = \frac{\sqrt{3}}{2} a, E M = \frac{3}{2} a$.
Since $A D \parallel E M$, we have $\triangle A F D \sim \triangle M F E$.
Thus, $\frac{A F}{F M} = \frac{A D}{E M}$, which means $\frac{A F}{A M} \approx \frac{A D}{E M + A D}$.
Therefore, $A F = \frac{A D \cdot A M}{E M + A D} = \frac{2 \sqrt{3}}{7} a$.
Hence, $\frac{A F}{A G} = \frac{4 \sqrt{3}}{7}$.

Answer: C