Tourist Nikolai Petrovich was late by $\Delta t=5$ minutes for the departure of his riverboat, which had set off downstream. Fortunately, the owner of a fast motorboat agreed to help Nikolai Petrovich. Catching up with the riverboat and disembarking the unlucky tourist, the motorboat immediately set off on its return journey. How much time passed from the departure of the motorboat until its return? Assume that the speed of the riverboat relative to the water is $k=3$ times the speed of the river current, and the speed of the motorboat is $n=5$ times the speed of the river current.

Solution: 1. $s=\left(3 v_{\mathrm{T}}+v_{\mathrm{T}}\right) \cdot\left(t_{1}+\Delta t\right)=\left(5 v_{\mathrm{T}}+v_{\mathrm{T}}\right) \cdot t_{1} \rightarrow 4 \Delta t=2 t_{1} \rightarrow t_{1}=2 \Delta t=10$ min $t_{2}=\frac{s}{5 v_{\mathrm{T}}-v_{\mathrm{T}}}=\frac{6 v_{\mathrm{T}} \cdot t_{1}}{4 v_{\mathrm{T}}}=\frac{3}{2} t_{1}=3 \Delta t=15$ min $\rightarrow$ $T=t_{1}+t_{2}=5 \Delta t=25$ min

Answer: 25


In $\triangle A B C$, $A B=1, B C=2$, then the range of $\angle C$ is
A. $\left(0, \frac{\pi}{6}\right]$
B. $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
C. $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$
D. $\left(0, \frac{\pi}{2}\right)$

Solution:
Answer
As shown in the figure, point $A$ moves on a circle centered at $B$ with a radius of 1. When $AC$ is tangent to circle $B$, $\angle C$ reaches its maximum value. At this time, $\sin C=\frac{1}{2}$, so $C \in\left(0, \frac{\pi}{6}\right]$, hence the answer is $A$.

Answer: A


How many digits are needed to write down the numbers from 1 to 6786?

Solution:
Among the numbers ranging from 1 to 6786, there are 9 single-digit, 90 double-digit, 900 triple-digit, and 5687 four-digit numbers. Therefore, to write these down,

$$
9 \cdot 1 + 90 \cdot 2 + 900 \cdot 3 + 5687 \cdot 4 = 26037
$$

digits are required.

(József Löwy, Losoncz.)

Number of solutions: 36.

Answer: 26037


Find all natural numbers $n$ such that
$$
\min _{k \in \mathbb{N}}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991,
$$

where $\left[\frac{n}{k^{2}}\right]$ denotes the greatest integer not exceeding $\frac{n}{k^{2}}$, and $\mathbf{N}$ is the set of natural numbers.

Solution:
The condition $\min _{k \in \mathrm{N}}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991$ is equivalent to the following two conditions:
(1) For any $k \in \mathbf{N}$, we have $k^{2}+\frac{n}{k^{2}} \geqslant 1991$;
(2) There exists $k_{0} \in \mathbf{N}$, such that $k_{0}^{2}+\frac{n}{k_{0}^{2}}<1992$.

Conditions (1) and (2) are also respectively equivalent to:
(1) For any $k \in \mathbf{N}$, we have $k^{4}-1991 k^{2}+n \geqslant 0$;
(2) There exists $k_{0} \in \mathbf{N}$, such that $k_{0}^{4}-1992 k_{0}^{2}+n<0$.
We first solve the inequality in (1), which gives
$$
\left(k^{2}-\frac{1991}{2}\right)^{2}+n-\frac{1991^{2}}{4} \geqslant 0 \text {, }
$$

Since the closest perfect square to $\frac{1991}{2}=995.5$ is $32^{2}=1024$, from (1) we have
$$
\left(1024-\frac{1991}{2}\right)^{2}+n-\frac{1991^{2}}{4} \geqslant 0 \text {. }
$$

Solving this, we get $n \geqslant 1024 \times 967=990208$.
Condition (2) is equivalent to
$$
\min _{k \in \mathrm{N}}\left\{\left(k^{2}-996\right)^{2}+n-996^{2}\right\}<0,
$$

Since the minimum value of $\left(k^{2}-996\right)^{2}$ is $28^{2}$, from (2) we get
$$
n<996^{2}-28^{2}=1024 \times 968=991232 \text {. }
$$

In summary, the range of $n$ that satisfies the problem's requirements is $990208 \leqslant n \leqslant 991231$.

Answer: 990208\leqslantn\leqslant991231


Calculate the definite integral:

$$
\int_{\pi / 4}^{\operatorname{arctg} 3} \frac{1+\operatorname{ctg} x}{(\sin x+2 \cos x)^{2}} d x
$$

Solution:
## Solution

$$
\begin{aligned}
& \int_{\pi / 4}^{\operatorname{arctg} 3} \frac{1+\operatorname{ctg} x}{(\sin x+2 \cos x)^{2}} d x=\int_{\pi / 4}^{\operatorname{arctg} 3} \frac{1+\operatorname{ctg} x}{\sin ^{2} x+4 \sin x \cos x+4 \cos ^{2} x} d x= \\
& =\int_{\pi / 4}^{1+2 \sin 2 x+3 \cos ^{2} x} \frac{1+\operatorname{ctg} x}{1 x} d x
\end{aligned}
$$

We will use the substitution:

$$
t=\operatorname{tg} x
$$

From which:

$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, \cos ^{2} x=\frac{1}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1 \\
& x=\operatorname{arctg} 3 \Rightarrow t=\operatorname{tg}(\operatorname{arctg} 3)=3
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& =\int_{1}^{3} \frac{1+\frac{1}{t}}{1+\frac{4 t}{1+t^{2}}+\frac{3}{1+t^{2}}} \cdot \frac{d t}{1+t^{2}}=\int_{1}^{3} \frac{\frac{t+1}{t}}{1+t^{2}+4 t+3} d t= \\
& =\int_{1}^{3} \frac{1+t}{t\left(t^{2}+4 t+4\right)} d t=\int_{1}^{3} \frac{1+t}{t(t+2)^{2}} d t=
\end{aligned}
$$

Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

$$
\frac{1+t}{t(t+2)^{2}}=\frac{A}{t}+\frac{B_{1}}{t+2}+\frac{B_{2}}{(t+2)^{2}}=\frac{A(t+2)^{2}+B_{1} t(t+2)+B_{2} t}{t(t+2)^{2}}=
$$

$$
\begin{aligned}
& =\frac{A\left(t^{2}+4 t+4\right)+B_{1}\left(t^{2}+2 t\right)+B_{2} t}{t(t+2)^{2}}=\frac{\left(A+B_{1}\right) t^{2}+\left(4 A+2 B_{1}+B_{2}\right) t+4 A}{t(t+2)^{2}} \\
& \left\{\begin{array}{l}
A+B_{1}=0 \\
4 A+2 B_{1}+B_{2}=1 \\
4 A=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=-\frac{1}{4} \\
2 B_{1}+B_{2}=0 \\
A=\frac{1}{4}
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=-\frac{1}{4} \\
B_{2}=\frac{1}{2} \\
A=\frac{1}{4}
\end{array}\right. \\
& \frac{1+t}{t(t+2)^{2}}=\frac{1}{4 t}-\frac{1}{4(t+2)}+\frac{1}{2(t+2)^{2}} \\
& =\int_{1}^{3}\left(\frac{1}{4 t}-\frac{1}{4(t+2)}+\frac{1}{2(t+2)^{2}}\right) d t=\left.\left(\frac{1}{4} \ln |t|-\frac{1}{4} \ln |t+2|-\frac{1}{2(t+2)}\right)\right|_{1} ^{3}= \\
& =\left(\frac{1}{4} \ln |3|-\frac{1}{4} \ln |3+2|-\frac{1}{2(3+2)}\right)-\left(\frac{1}{4} \ln |1|-\frac{1}{4} \ln |1+2|-\frac{1}{2(1+2)}\right)= \\
& =\frac{\ln 3}{4}-\frac{\ln 5}{4}-\frac{1}{10}+\frac{\ln 3}{4}+\frac{1}{6}=\frac{1}{4} \ln \frac{9}{5}+\frac{1}{15}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-10 »

Categories: Kuznetsov's Problem Book Integrals Problem 9| Integrals

- Last edited: 20:45, 11 May 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 9-11

## Material from PlusPi

Answer: \frac{1}{4}\ln\frac{9}{5}+\frac{1}{15}


Let $ABCDEF$ be a regular hexagon, and let $G$, $H$, $I$, $J$, $K$, and $L$ be the midpoints of sides $AB$, $BC$, $CD$, $DE$, $EF$, and $FA$, respectively. The intersection of lines $\overline{AH}$, $\overline{BI}$, $\overline{CJ}$, $\overline{DK}$, $\overline{EL}$, and $\overline{FG}$ bound a smaller regular hexagon. Find the ratio of the area of the smaller hexagon to the area of $ABCDEF$.

Solution:
1. **Define the Problem and Setup:**
   Let \( ABCDEF \) be a regular hexagon with side length \( s \). Let \( G, H, I, J, K, \) and \( L \) be the midpoints of sides \( AB, BC, CD, DE, EF, \) and \( FA \), respectively. The lines \( \overline{AH}, \overline{BI}, \overline{CJ}, \overline{DK}, \overline{EL}, \) and \( \overline{FG} \) intersect to form a smaller regular hexagon. We need to find the ratio of the area of the smaller hexagon to the area of \( ABCDEF \).

2. **Assign Side Length:**
   Without loss of generality, let the side length of the hexagon be \( s = 2 \).

3. **Identify Key Points and Angles:**
   Note that \( \angle ABH = \angle BMH = 120^\circ \). Since \( \triangle ABH \) and \( \triangle BCI \) are rotational images of one another, we get that \( \angle MBH = \angle HAB \). Thus, \( \triangle ABH \sim \triangle BMH \sim \triangle BCI \).

4. **Use Similarity and Law of Cosines:**
   Since \( \triangle ABH \sim \triangle BMH \sim \triangle BCI \), we can use the Law of Cosines to find \( BI \):
   \[
   BI = \sqrt{2^2 + 1^2 - 2 \cdot 2 \cdot 1 \cdot \cos(120^\circ)} = \sqrt{4 + 1 + 2} = \sqrt{7}
   \]

5. **Calculate Lengths:**
   Since \( \triangle ABH \sim \triangle BMH \sim \triangle BCI \), we have:
   \[
   \frac{BC + CI}{BI} = \frac{3}{\sqrt{7}}
   \]
   Therefore,
   \[
   BM + MH = \frac{3BH}{\sqrt{7}} = \frac{3}{\sqrt{7}}
   \]

6. **Find \( MN \):**
   \[
   MN = BI - (BM + MH) = \sqrt{7} - \frac{3}{\sqrt{7}} = \frac{4}{\sqrt{7}}
   \]

7. **Calculate the Ratio of Areas:**
   The ratio of the area of the smaller hexagon to the area of the larger hexagon is given by the square of the ratio of their corresponding side lengths:
   \[
   \left( \frac{MN}{BC} \right)^2 = \left( \frac{\frac{4}{\sqrt{7}}}{2} \right)^2 = \left( \frac{2}{\sqrt{7}} \right)^2 = \frac{4}{7}
   \]

The final answer is \(\boxed{\frac{4}{7}}\).

Answer: \frac{4}{7}


As shown in Figure 3, the circumference of a circle with center $O$ is divided into 12 equal arcs, and the division points are labeled as $A, B, \cdots, L$. Then $\alpha+\beta=$ ( ).
(A) $75^{\circ}$
(B) $80^{\circ}$
(C) $90^{\circ}$
(D) $120^{\circ}$
(E) $150^{\circ}$

Solution:
15. C.

Divide the circumference of $\odot O$ into 12 equal parts, with each arc having a degree measure of $\frac{360^{\circ}}{12}=30^{\circ}$.
Then $\alpha=\frac{1}{2} \angle G O E=30^{\circ}$,
$$
\beta=\frac{1}{2} \angle A O I=60^{\circ} \text {. }
$$

Thus, $\alpha+\beta=90^{\circ}$.

Answer: C


Solution:
【Solution】Solve: $3^{3}+4^{3}+5^{3}+6^{3}+7^{3}+8^{3}+9^{3}$
$$
\begin{array}{l}
=1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}+7^{3}+8^{3}+9^{3}-1^{3}-2^{3} \\
=(1+2+3+\cdots+9)^{2}-1-8 \\
=[(1+9) \times 9 \div 2]^{2}-9 \\
=45^{2}-9 \\
=2025-9 \\
=2016 ;
\end{array}
$$

Therefore, the answer is: 2016.

Answer: 2016


The sides of a triangle $ABC$ have lengths $AB = 26$ cm, $BC = 17$ cm, and $CA = 19$ cm. The bisectors of the angles $B$ and $C$ intersect at the point $I$. By $I$ a parallel to $BC$ is drawn that intersects the sides $AB$ and $BC$ at points $M$ and $N$ respectively. Calculate the perimeter of the triangle $AMN$.

Let $n$ be a positive integer. Determine the exact value of the following sum
$$\frac{1}{1+\sqrt3}+\frac{1}{\sqrt3+\sqrt5}+\frac{1}{\sqrt5+\sqrt7}+...+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}$$

Find the product of all positive integers that are less than $150$ and that have exactly $4$ positive divisors. Express your answer as a product of prime factors.

There is a board of $ n \times n$, where n is a positive integer and the numbers from $ 1$ to $n$ are written in each of its rows in some order, so that the arrangement obtained is symmetric with respect to the diagonal. It is desired that the diagonal of the board does not contain all the numbers from $ 1$ to $n$. Show that it is always possible when $n$ is even and that it is impossible when $n$ is odd.

There are six circles of radius $\frac12$, tangent to each other and to the sides of the rectangle, as shown in the figure. Also the centers $A$, $B$, $C$ and $D$ are collinear. Determine the lengths of the sides of the rectangle.

Solution:
1. Given the triangle \(ABC\) with sides \(AB = 26\) cm, \(BC = 17\) cm, and \(CA = 19\) cm, we need to find the perimeter of the triangle \(AMN\) where \(M\) and \(N\) are points on \(AB\) and \(AC\) respectively such that \(MN \parallel BC\).

2. Since \(MN \parallel BC\), triangles \(BMN\) and \(BIC\) are similar. The angle bisectors of \(B\) and \(C\) intersect at \(I\), the incenter of \(\triangle ABC\).

3. Let \(BM = x\) and \(NC = y\). Since \(MN \parallel BC\), we have:
   \[
   \frac{BM}{AB} = \frac{IN}{BC} \quad \text{and} \quad \frac{NC}{CA} = \frac{IM}{BC}
   \]
   Given that \(AB = 26\) cm, \(BC = 17\) cm, and \(CA = 19\) cm, we can write:
   \[
   \frac{x}{26} = \frac{y}{19}
   \]
   This implies:
   \[
   x = \frac{26y}{19}
   \]

4. Since \(MN \parallel BC\), the length of \(MN\) is proportional to the length of \(BC\). Let \(k\) be the proportionality constant:
   \[
   MN = k \cdot BC = k \cdot 17
   \]

5. The perimeter of \(\triangle AMN\) is given by:
   \[
   AM + MN + AN
   \]
   Since \(AM = AB - BM = 26 - x\) and \(AN = AC - NC = 19 - y\), we have:
   \[
   AM + AN = (26 - x) + (19 - y)
   \]

6. Substituting \(x = \frac{26y}{19}\) into the equation:
   \[
   AM + AN = 26 - \frac{26y}{19} + 19 - y = 45 - \left(\frac{26y}{19} + y\right)
   \]

7. To find \(MN\), we use the fact that \(MN \parallel BC\) and the similarity of triangles:
   \[
   MN = k \cdot BC = k \cdot 17
   \]

8. Since \(MN\) is parallel to \(BC\), the length of \(MN\) is proportional to the length of \(BC\). Therefore, the perimeter of \(\triangle AMN\) is:
   \[
   AM + MN + AN = 45
   \]

The final answer is \(\boxed{45}\)

Answer: 45


On a plate, there are a total of 58 slices of beigli, a mixture of walnut and poppy seed. Three walnut ones can be chosen from them in the same number of ways as two poppy seed ones and one walnut one. How many poppy seed beigli are on the plate?

Solution:
Solution. Let $x$ be the number of walnut-filled doughnuts, then there are $58-x$ plum-filled doughnuts.

According to the condition, $\binom{x}{3}=\binom{58-x}{2} \cdot x$, which means the number of ways to choose 3 walnut-filled doughnuts is the same as the number of ways to choose 2 plum-filled and 1 walnut-filled doughnut. That is,

$$
\frac{x(x-1)(x-2)}{1 \cdot 2 \cdot 3}=\frac{(58-x)(57-x)}{1 \cdot 2} \cdot x
$$

Simplify by $x \neq 0$, and perform the indicated operations. We arrive at the following quadratic equation: $x^{2}-$ $171 x+4958=0$, from which $x=37$, or 134, but we didn't have that many doughnuts in total.

Therefore, we placed 37 walnut-filled and 21 plum-filled doughnuts on the plate.

()

Tamás Karches (Mosonmagyaróvár, Kossuth L. Gymnasium, 12th grade)

Answer: 21


Let the base edge length of the regular tetrahedron $V-ABC$ be 4, and the side edge length be 8. The section through $A$ intersecting the side edges $VB$ and $VC$ is $\triangle AED$. Then the minimum perimeter of the section $\triangle AED$ is ( ).
(A) $12 \frac{1}{5}$
(B) 11
(C) 12
(D) $11 \frac{1}{5}$

Solution:
5. B.

From the edge $V A$, the surface of the tetrahedron is unfolded as shown in Figure 2. The perimeter of $\triangle A D E$ is minimized if and only if $D$ and $E$ lie on the line segment $A A_{1}$.
Let $\angle B V C=\alpha, \angle A V A_{1}=3 \alpha$. Then $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$.
In $\triangle V B C$, by the cosine rule, $\cos \alpha=\frac{V B^{2}+V C^{2}-B C^{2}}{2 V B \cdot V C}=\frac{7}{8}$. Therefore, $\cos 3 \alpha=\frac{7}{2^{7}}$. In $\triangle A V A_{1}$, by the cosine rule, $A A_{1}=\sqrt{V A^{2}+A_{1} V^{2}-2 V A \cdot A_{1} V \cos 3 \alpha}=11$.

Answer: B


There are two four-digit numbers, Jia and Yi. The common logarithm of Yi is $A+\lg B$, where $A$ and $B$ are natural numbers. The sum of the thousand's and hundred's digits of Jia equals $5B$, and the sum of the unit's and ten's digits of Jia equals the difference between Yi and Jia plus $B$. Find the two numbers.

Solution:
Let the number A be $x y z u$, (where $x, y, z, u$ are all single-digit natural numbers).

From $x+y=5 B$, using the digit domain: $0<x+y \leqslant 18$, i.e., $53 \leqslant 18$. It can be known that $1 \leqslant B \leqslant 3$, thus the number is $B \times 10^{4}$ and $A=3$.

Since the difference between the second number and the first number $=z+u-2 \leqslant 18-2=16$, the thousand's digit $x=B-1$, and the hundred's digit $y=9$. Therefore, $9+(B-1)=5 B, B=2$, so the second number is 2000. Thus, $x=1, y=9$. Also, $(9-z) \times 10+(10-u)$ $=z+u-2$, i.e., $11 z+2 u=102$,
we have $u=\frac{102-11 z}{2}$. By the digit domain of $u$, we get: $1 \leqslant u \leqslant 9$.
So $7 \leqslant z \leqslant 9$.
Since $z$ must be even, we get $z=8, u=7$. Finally, the first number is 1987, and the second number is 2000.

Answer: 1987, 2000


Adam's house number is in exactly one of the following ranges. Which one?
A 123 to 213
B 132 to 231
C 123 to 312
D 231 to 312
E 312 to 321

Solution:
9. E Note that the numbers 123 to 213 are in both option A and option C. Also, numbers 213 to
231 are in both option B and option C, and numbers 231 to 312 are in both C and D. However, numbers 313 to 321 are in option E only. So that must be where Adam's house is.

Answer: E


Let real numbers $x, y$ satisfy the equation
$$
2 x^{2}+3 y^{2}=4 x \text {. }
$$

Then the minimum value of $x+y$ is ( ).
(A) $1+\frac{\sqrt{15}}{3}$
(B) $1-\frac{\sqrt{15}}{3}$
(C) 0
(D) None of the above

Solution:
- 1. B.

Equation (1) can be transformed into
$$
(x-1)^{2}+\frac{y^{2}}{\frac{2}{3}}=1 \text {. }
$$

Let $x+y=t$.
When the line (3) is tangent to the ellipse (2), $t$ takes its extreme values.
At this point, $2 x^{2}+3(x-t)^{2}=4 x$, which simplifies to
$$
5 x^{2}-(6 t+4) x+3 t^{2}=0
$$

It has equal roots.
Thus, $\Delta=(6 t+4)^{2}-60 t^{2}=0$
$\Rightarrow t=1 \pm \frac{\sqrt{15}}{3}$.
Therefore, the minimum value of $x+y$ is $1-\frac{\sqrt{15}}{3}$.

Answer: B


How many small cubes, each having a surface area of $54 \mathrm{~cm}^{2}$, are needed to build a large cube with a surface area of $864 \mathrm{~cm}^{2}$?

Solution:
A small cube has an edge length of $3 \mathrm{~cm}$. A large cube has an edge length of $12 \mathrm{~cm}$. To build the large cube, we need 64 small cubes.

#

Answer: 64


Write the smallest positive integer $n$ that satisfies both of the following properties on the board.
[list]
[*] $n$ is larger than any integer on the board currently.
[*] $n$ cannot be written as the sum of $2$ distinct integers on the board.
[/list]
Find the $100$-th integer that you write on the board. Recall that at the beginning, there are already $4$ integers on the board: $1, 2, 4, 6$.

Solution:
To solve this problem, we need to understand the process of generating the sequence of integers on the board. We start with the integers \(1, 2, 4, 6\) and at each step, we add the smallest positive integer \(n\) that is larger than any integer currently on the board and cannot be written as the sum of two distinct integers already on the board.

1. **Initial Setup:**
   The initial integers on the board are \(1, 2, 4, 6\).

2. **First Step:**
   The next integer \(n\) must be greater than \(6\) and cannot be written as the sum of any two distinct integers from \(\{1, 2, 4, 6\}\).
   - Possible sums: \(1+2=3\), \(1+4=5\), \(1+6=7\), \(2+4=6\), \(2+6=8\), \(4+6=10\).
   - The smallest integer greater than \(6\) that is not in \(\{3, 5, 7, 8, 10\}\) is \(9\).

   So, we add \(9\) to the board: \(\{1, 2, 4, 6, 9\}\).

3. **Second Step:**
   The next integer \(n\) must be greater than \(9\) and cannot be written as the sum of any two distinct integers from \(\{1, 2, 4, 6, 9\}\).
   - Possible sums: \(1+2=3\), \(1+4=5\), \(1+6=7\), \(1+9=10\), \(2+4=6\), \(2+6=8\), \(2+9=11\), \(4+6=10\), \(4+9=13\), \(6+9=15\).
   - The smallest integer greater than \(9\) that is not in \(\{3, 5, 6, 7, 8, 10, 11, 13, 15\}\) is \(12\).

   So, we add \(12\) to the board: \(\{1, 2, 4, 6, 9, 12\}\).

4. **General Pattern:**
   We observe that each new integer added to the board is the smallest integer that cannot be expressed as the sum of any two distinct integers already on the board. This process continues, and the sequence of integers grows.

5. **Finding the 100th Integer:**
   To find the 100th integer, we need to continue this process until we have added 96 more integers to the initial 4 integers. This is a tedious process to do manually, but we can refer to known results or sequences.

   According to the reference provided, this problem is equivalent to a known problem (2015 C4), and the 100th integer in this sequence is given as \(388\).

The final answer is \(\boxed{388}\).

Answer: 388


In a class of 25 children, two are randomly chosen for duty. The probability that both duty students will be boys is $3 / 25$.

How many girls are in the class?

Solution:
Let there be $n$ boys in the class, then the number of ways to choose an ordered pair of duty students from them is $n(n-1)$, the number of ways to choose an ordered pair from the entire class is $25 \cdot 24$. Therefore, the probability that two boys will be chosen is $n(n-1):(25 \cdot 24)=3 / 25$. Hence, $n(n-1)=72=9 \cdot 8$. Therefore, $n=9$, which means there are 9 boys and 16 girls in the class.

## Answer

16 girls.

Answer: 16


Let $ f(t)$ be a continuous function on the interval $ 0 \leq t \leq 1$, and define the two sets of points \[ A_t=\{(t,0): t\in[0,1]\} , B_t=\{(f(t),1): t\in [0,1]\}.\] Show that the union of all segments $ \overline{A_tB_t}$ is Lebesgue-measurable, and find the minimum of its measure with respect to all functions $ f$.

Solution:
1. **Show that the set \( A = \bigcup_{0 \le t \le 1} \overline{A_tB_t} \) is closed:**

   - Let \( P_n \) be a convergent sequence of points in \( A \), such that \( P_n \to P_0 \).
   - By the definition of \( A \), for every \( n \), there exists a \( t_n \) such that \( P_n \in \overline{A_{t_n}B_{t_n}} \).
   - According to the Bolzano-Weierstrass theorem, the sequence \( t_n \) contains a convergent subsequence: \( t_{n_k} \to t_0 \).
   - The distance from the point \( P_{n_k} \) to the segment \( \overline{A_{t_0}B_{t_0}} \) is not greater than \( \max \{|f(t_{n_k}) - f(t_0)|, |t_{n_k} - t_0|\} \), which tends to \( 0 \) by the continuity of \( f(t) \).
   - Since \( A \) contains the limit point of any convergent sequence of its points, \( A \) is closed.

2. **Show that the set \( A \) is Lebesgue-measurable:**

   - Since \( A \) is closed, it is also Lebesgue-measurable.

3. **Determine the minimum measure of \( A \):**

   - Consider the point on the segment \( \overline{A_tB_t} \) that lies on the straight line \( y = c \). The abscissa of this point is \( (1 - c)t + cf(t) \), which is a continuous function of \( t \) due to the continuity of \( f(t) \).
   - Consequently, if two points of the set \( A \) lie on the line \( y = c \), then \( A \) contains the segment that joins these points.

4. **Calculate the measure of \( A \) for different functions \( f(t) \):**

   - If \( f(t) \) is constant, then \( A \) is a triangle of unit base and unit altitude, so it has measure \( \frac{1}{2} \).
   - If the segments \( \overline{A_0B_0} \) and \( \overline{A_1B_1} \) do not intersect, then the trapezoid with vertices \( A_0 \), \( B_0 \), \( A_1 \), and \( B_1 \) is a subset of \( A \), so the measure of \( A \) is not less than \( \frac{1}{2} \).
   - If the segments \( \overline{A_0B_0} \) and \( \overline{A_1B_1} \) intersect at some point \( C \), then the triangles \( A_0CA_1 \) and \( B_0CB_1 \) are subsets of \( A \), so the measure of \( A \) is not smaller than:
     \[
     t(d) = \frac{1}{2}\left( \frac{1}{1 + d} + \frac{d}{1 + d}d \right) = \frac{1}{2}\frac{1 + d^2}{1 + d},
     \]
     where \( d \) denotes the distance from \( B_0 \) to \( B_1 \).

5. **Find the minimum of \( t(d) \):**

   - By a simple calculation, we obtain that the minimum of \( t(d) \) on the positive half-axis is \( \sqrt{2} - 1 \).
   - Thus, the measure of \( A \) cannot be less than \( \sqrt{2} - 1 \).
   - On the other hand, for \( f(t) = (\sqrt{2} - 1)(1 - t) \), the measure of \( A \) is exactly \( \sqrt{2} - 1 \).

\(\blacksquare\)

The final answer is \( \boxed{ \sqrt{2} - 1 } \).

Answer: \sqrt{2} - 1


Solution:
【Key Points】Rapid and Clever Calculation
【Difficulty】Given Order

【Answer】20140309
【Analysis】Calculate in sequence.

Answer: 20140309


Find the smallest positive integer $k$ such that $k!$ ends in at least $43$ zeroes.

Solution:
To find the smallest positive integer \( k \) such that \( k! \) ends in at least 43 zeroes, we need to count the number of trailing zeroes in \( k! \). The number of trailing zeroes in \( k! \) is determined by the number of times 10 is a factor in the numbers from 1 to \( k \). Since 10 is the product of 2 and 5, and there are generally more factors of 2 than 5, we only need to count the number of factors of 5 in \( k! \).

The number of factors of 5 in \( k! \) is given by:
\[
\left\lfloor \frac{k}{5} \right\rfloor + \left\lfloor \frac{k}{5^2} \right\rfloor + \left\lfloor \frac{k}{5^3} \right\rfloor + \cdots
\]

We need this sum to be at least 43. Let's start by testing \( k = 175 \) as an estimate.

1. Calculate the number of factors of 5 in \( 175! \):
\[
\left\lfloor \frac{175}{5} \right\rfloor + \left\lfloor \frac{175}{25} \right\rfloor + \left\lfloor \frac{175}{125} \right\rfloor
\]
\[
= \left\lfloor 35 \right\rfloor + \left\lfloor 7 \right\rfloor + \left\lfloor 1.4 \right\rfloor
\]
\[
= 35 + 7 + 1 = 43
\]

Since \( 175! \) has exactly 43 trailing zeroes, we need to check if there is a smaller \( k \) that also results in at least 43 trailing zeroes.

2. Check \( k = 174 \):
\[
\left\lfloor \frac{174}{5} \right\rfloor + \left\lfloor \frac{174}{25} \right\rfloor + \left\lfloor \frac{174}{125} \right\rfloor
\]
\[
= \left\lfloor 34.8 \right\rfloor + \left\lfloor 6.96 \right\rfloor + \left\lfloor 1.392 \right\rfloor
\]
\[
= 34 + 6 + 1 = 41
\]

Since \( 174! \) has only 41 trailing zeroes, \( k = 174 \) is not sufficient.

3. Check \( k = 175 \) again to confirm:
\[
\left\lfloor \frac{175}{5} \right\rfloor + \left\lfloor \frac{175}{25} \right\rfloor + \left\lfloor \frac{175}{125} \right\rfloor
\]
\[
= 35 + 7 + 1 = 43
\]

Thus, the smallest \( k \) such that \( k! \) ends in at least 43 zeroes is indeed \( k = 175 \).

The final answer is \( \boxed{175} \).

Answer: 175


As shown in Figure 6, fill the ten circles in Figure 6 with the numbers $1,2, \cdots, 10$ respectively, so that the sum of the numbers in any five consecutive adjacent circles is not greater than a certain integer $M$. Find the minimum value of $M$ and complete your filling. ${ }^{[4]}$

Solution:
Let the numbers that satisfy the given conditions be: $a_{1}$, $a_{2}, \cdots, a_{10}$. Then
$$
\begin{array}{l}
a_{1}+a_{2}+a_{3}+a_{4}+a_{5} \leqslant M, \\
a_{2}+a_{3}+a_{4}+a_{5}+a_{6} \leqslant M, \\
\cdots \cdots . \\
a_{10}+a_{1}+a_{2}+a_{3}+a_{4} \leqslant M .
\end{array}
$$

Thus, $5\left(a_{1}+a_{2}+\cdots+a_{10}\right) \leqslant 10 M$
$$
\begin{array}{l}
\Rightarrow \frac{5 \times 10 \times 11}{2} \leqslant 10 M \\
\Rightarrow M \geqslant 27.5 .
\end{array}
$$

Since $M$ is an integer, the minimum value of $M$ is 28.
Divide $1,2, \cdots, 10$ into the following two groups:
$$
\begin{array}{l}
10,7,6,3,2 ; \\
9,8,5,4,1,
\end{array}
$$

Filling them into the figure in sequence yields Figure 7.

Answer: 28


A pyramid has a triangular base with side lengths $20$, $20$, and $24$. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$. The volume of the pyramid is $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.

Solution:
Let the triangular base be $\triangle ABC$, with $\overline {AB} = 24$. We find that the altitude to side $\overline {AB}$ is $16$, so the area of $\triangle ABC$ is $(24*16)/2 = 192$.
Let the fourth vertex of the tetrahedron be $P$, and let the midpoint of $\overline {AB}$ be $M$. Since $P$ is equidistant from $A$, $B$, and $C$, the line through $P$ perpendicular to the plane of $\triangle ABC$ will pass through the circumcenter of $\triangle ABC$, which we will call $O$. Note that $O$ is equidistant from each of $A$, $B$, and $C$. Then,
\[\overline {OM} + \overline {OC} = \overline {CM} = 16\]
Let $\overline {OM} = d$. Then $OC=OA=\sqrt{d^2+12^2}.$
Equation $(1)$:
\[d + \sqrt {d^2 + 144} = 16\]
Squaring both sides, we have
\[d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256\]
\[2d^2 + 2d\sqrt {d^2+144} = 112\]
\[2d(d + \sqrt {d^2+144}) = 112\]
Substituting with equation $(1)$:
\[2d(16) = 112\]
\[d = 7/2\]
We now find that $\sqrt{d^2 + 144} = 25/2$.
Let the distance $\overline {OP} = h$. Using the Pythagorean Theorem on triangle $AOP$, $BOP$, or $COP$ (all three are congruent by SSS):
\[25^2 = h^2 + (25/2)^2\]
\[625 = h^2 + 625/4\]
\[1875/4 = h^2\]
\[25\sqrt {3} / 2 = h\]

Finally, by the formula for volume of a pyramid,
\[V = Bh/3\]
\[V = (192)(25\sqrt{3}/2)/3\]
This simplifies to $V = 800\sqrt {3}$, so $m+n = \boxed {803}$.

NOTE : If you don’t know or remember the formula for the volume of a triangular pyramid, you can derive it using calculus as follows :
Take a small triangular element in the pyramid. We know that it’s area is proportional to the height from the vertex to the base. Hence, we know that $\frac{A_{small element}}{A} = \frac{h^2}{H^2} \implies A_{small element} = \frac{Ah^2}{H^2}$. Now integrate it taking the limits $0$ to $H$

Shortcut
Here is a shortcut for finding the radius $R$ of the circumcenter of $\triangle ABC$.
As before, we find that the foot of the altitude from $P$ lands on the circumcenter of $\triangle ABC$. Let $BC=a$, $AC=b$, and $AB=c$.
Then we write the area of $\triangle ABC$ in two ways:
\[[ABC]= \frac{1}{2} \cdot 24 \cdot 16 = \frac{abc}{4R}\]
Plugging in $20$, $20$, and $24$ for $a$, $b$, and $c$ respectively, and solving for $R$, we obtain $R= \frac{25}{2}=OA=OB=OC$.
Then continue as before to use the Pythagorean Theorem on $\triangle AOP$, find $h$, and find the volume of the pyramid.

Another Shortcut (Extended Law of Sines)
Take the base $\triangle ABC$, where $AB = BC = 20$ and $AC = 24$. Draw an altitude from $B$ to $AC$ that bisects $AC$ at point $D$. Then the altitude has length $\sqrt{20^2 - 12^2} = \sqrt{16^2} = 16$. Next, let $\angle BCA = \theta$. Then from the right triangle $\triangle BDC$, $\sin \theta = 4/5$. From the extended law of sines, the circumradius is $20 \cdot \dfrac{5}{4} \cdot \dfrac{1}{2} = \dfrac{25}{2}$.

Answer: 803


A bored student walks down a hall that contains a row of closed lockers, numbered $1$ to $1024$. He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens?

Solution:
Solution 1
On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of $4$, leaving only lockers $2 \pmod{8}$ and $6 \pmod{8}$. Then he goes ahead and opens all lockers $2 \pmod {8}$, leaving lockers either $6 \pmod {16}$ or $14 \pmod {16}$. He then goes ahead and opens all lockers $14 \pmod {16}$, leaving the lockers either $6 \pmod {32}$ or $22 \pmod {32}$. He then goes ahead and opens all lockers $6 \pmod {32}$, leaving $22 \pmod {64}$ or $54 \pmod {64}$. He then opens $54 \pmod {64}$, leaving $22 \pmod {128}$ or $86 \pmod {128}$. He then opens $22 \pmod {128}$ and leaves $86 \pmod {256}$ and $214 \pmod {256}$. He then opens all $214 \pmod {256}$, so we have $86 \pmod {512}$ and $342 \pmod {512}$, leaving lockers $86, 342, 598$, and $854$, and he is at where he started again. He then opens $86$ and $598$, and then goes back and opens locker number $854$, leaving locker number $\boxed{342}$ untouched. He opens that locker.

Lockers that are $2 \pmod{8}$ and $6 \pmod{8}$ are just lockers that are $2 \pmod{4}$. Edit by [Yiyj1](https://artofproblemsolving.com/wiki/index.php/User:Yiyj1)

Solution 2
We can also solve this with recursion. Let $L_n$ be the last locker he opens given that he started with $2^n$ lockers. Let there be $2^n$ lockers. After he first reaches the end of the hallway, there are $2^{n-1}$ lockers remaining. There is a correspondence between these unopened lockers and if he began with $2^{n-1}$ lockers. The locker $y$ (if he started with $2^{n-1}$ lockers) corresponds to the locker $2^n+2-2y$ (if he started with $2^n$ lockers). It follows that $L_{n} = 2^{n} +2 -2L_{n-1}$ as they are corresponding lockers. We can compute $L_1=2$ and use the recursion to find $L_{10}=\boxed{342}$

Solution 3 (bash)
List all the numbers from $1$ through $1024$, then do the process yourself!!! It will take about 25 minutes (if you don't start to see the pattern), but that's okay, eventually, you will get $\boxed{342}$.
(Note: If you try to do this, first look through all the problems! -Guy)

Edit from EthanSpoon, Doing this with only even numbers will make it faster.
You'll see.
02496

Answer: 342


Portraits of famous scientists hang on a wall. The scientists lived between 1600 and 2008, and none of them lived longer than 80 years. Vasya multiplied the years of birth of these scientists, and Petya multiplied the years of their death. Petya's result is exactly $ 5\over 4$ times greater than  Vasya's result. What minimum number of portraits can be on the wall?

[i]Author: V. Frank[/i]

Solution:
1. Let the years of birth of the scientists be \( a_1, a_2, \ldots, a_n \) and the years of death be \( d_1, d_2, \ldots, d_n \). Given that \( d_i = a_i + b_i \) where \( 1 \leq b_i \leq 80 \) for all \( i \).
2. Vasya's product of the years of birth is \( P_V = a_1 \cdot a_2 \cdot \ldots \cdot a_n \).
3. Petya's product of the years of death is \( P_P = d_1 \cdot d_2 \cdot \ldots \cdot d_n = (a_1 + b_1) \cdot (a_2 + b_2) \cdot \ldots \cdot (a_n + b_n) \).
4. According to the problem, \( P_P = \frac{5}{4} P_V \).

We need to find the minimum number \( n \) such that:
\[
\frac{(a_1 + b_1) \cdot (a_2 + b_2) \cdot \ldots \cdot (a_n + b_n)}{a_1 \cdot a_2 \cdot \ldots \cdot a_n} = \frac{5}{4}
\]

5. This can be rewritten as:
\[
\prod_{i=1}^n \left(1 + \frac{b_i}{a_i}\right) = \frac{5}{4}
\]

6. To find the minimum \( n \), we need to consider the maximum and minimum values of \( \frac{a_i + b_i}{a_i} \). The maximum value occurs when \( a_i \) is at its minimum (1600) and \( b_i \) is at its maximum (80):
\[
\frac{a_i + b_i}{a_i} = \frac{1600 + 80}{1600} = \frac{1680}{1600} = \frac{21}{20}
\]

7. The minimum value occurs when \( a_i \) is at its maximum (2008) and \( b_i \) is at its minimum (1):
\[
\frac{a_i + b_i}{a_i} = \frac{2008 + 1}{2008} = \frac{2009}{2008}
\]

8. We need to find the smallest \( n \) such that:
\[
\left(\frac{21}{20}\right)^k \cdot \left(\frac{2009}{2008}\right)^{n-k} \geq \frac{5}{4}
\]

9. Testing \( n = 4 \):
\[
\left(\frac{21}{20}\right)^4 = \left(1.05\right)^4 \approx 1.21550625
\]
This is less than \( \frac{5}{4} = 1.25 \), so \( n = 4 \) is not sufficient.

10. Testing \( n = 5 \):
\[
\left(\frac{21}{20}\right)^5 = \left(1.05\right)^5 \approx 1.2762815625
\]
This is greater than \( \frac{5}{4} = 1.25 \), so \( n = 5 \) is sufficient.

Therefore, the minimum number of portraits is 5.

The final answer is \( \boxed{5} \).

Answer: 5


Six natural numbers (possibly repeating) are written on the faces of a cube, such that the numbers on adjacent faces differ by more than 1. What is the smallest possible value of the sum of these six numbers?

Solution:
Answer: 18.

Solution: Consider three faces that share a common vertex. The numbers on them differ pairwise by 2, so the smallest possible sum would be for $1+3+5=9$. The same can be said about the remaining three faces.

Thus, the sum cannot be less than 18. We will show that 18 can be achieved - place the number 1 on the top and bottom faces of the cube, 3 on the right and left faces, and 5 on the front and back faces.

Answer: 18


The average score of the participants in the Zu Chongzhi Cup Mathematics Invitational at a certain school is 75. Among them, the number of male participants is $80\%$ more than that of female participants, and the average score of female participants is $20\%$ higher than that of male participants. Therefore, the average score of female participants is â points.

Solution: 1. 84

Answer: 84


In the room, there are two light bulbs. When the switch of the first light bulb is turned on, it lights up after 6 seconds and stays on for 5 seconds, then it is off for 6 seconds and on for 5 seconds, and this repeats continuously. When the switch of the second light bulb is turned on, it lights up after 4 seconds and stays on for 3 seconds, then it is off for 4 seconds and on for 3 seconds, and this repeats continuously. Linda turned on both switches at the same time and turned them off after 2021 seconds. How many seconds did both light bulbs shine simultaneously during this time?

Solution:
## Solution.

Every 11 seconds, the first light will repeat a cycle where it will not light for 6 seconds and then light for 5 seconds. The second light will repeat such a cycle every 7 seconds.

The time interval we need to observe as the common period for both lights lasts $7 \cdot 11=77$ seconds.

![](https://cdn.mathpix.com/cropped/2024_05_30_bfe8eca8742272ceb120g-5.jpg?height=42&width=1596&top_left_y=493&top_left_x=184)

![](https://cdn.mathpix.com/cropped/2024_05_30_bfe8eca8742272ceb120g-5.jpg?height=45&width=1619&top_left_y=560&top_left_x=161)$\qquad$
In the first two rectangles, the yellow color represents the seconds during which the lights are on, and the gray color represents the seconds during which the lights are off. The first rectangle shows the state of the first light, and the second shows the state of the second light. In the third rectangle, the seconds when both lights are on simultaneously are marked.

In the observed period of 77 seconds, both lights are on simultaneously for $1+3+1+3+2+2+3=$ 15 seconds. Since $2021=26 \cdot 77+19$, there will be 26 such periods, and then there will be 19 seconds remaining. In these 19 seconds, the lights will be on simultaneously for 2 seconds.

Both lights are on simultaneously for a total of $26 \cdot 15+2=392$ seconds.

Answer: 392


A worker takes care of three machine tools. In one hour, the probabilities that machine tools A, B, and C need attention are $0.9$, $0.8$, and $0.85$, respectively. In one hour, find:
(1) the probability that no machine tool needs attention;
(2) the probability that at least one machine tool needs attention.

Solution:
Let $A$, $B$, and $C$ represent the need for attention for machines A, B, and C, respectively. Then, $\bar{A}$, $\bar{B}$, and $\bar{C}$ represent the lack of need for attention for machines A, B, and C, respectively. According to the problem, $P(A)=0.9$, $P(B)=0.8$, $P(C)=0.85$, and $P(\bar{A})=0.1$, $P(\bar{B})=0.2$, $P(\bar{C})=0.15$.
(1) None of the machines A, B, and C need attention, which means $\bar{A} \cdot \bar{B} \cdot \bar{C}$. Since $\bar{A}$, $\bar{B}$, and $\bar{C}$ are independent, we have $P(\bar{A} \cdot \bar{B} \cdot \bar{C})=P(\bar{A}) \cdot P(\bar{B}) \cdot P(\bar{C})=0.1 \times 0.2 \times 0.15=0.003$.
(2) Let $D$ represent the event that at least one of the machines A, B, and C needs attention. Then, $D=(A \cdot \bar{B} \cdot \bar{C}) \cup (\bar{A} \cdot B \cdot \bar{C}) \cup (\bar{A} \cdot \bar{B} \cdot C) \cup (A \cdot B \cdot \bar{C}) \cup (A \cdot \bar{B} \cdot C) \cup (\bar{A} \cdot B \cdot C) \cup (A \cdot B \cdot C)$. Since $(A \cdot \bar{B} \cdot \bar{C})$, $(\bar{A} \cdot B \cdot \bar{C})$, $(\bar{A} \cdot \bar{B} \cdot C)$, $(A \cdot B \cdot \bar{C})$, $(A \cdot \bar{B} \cdot C)$, $(\bar{A} \cdot B \cdot C)$, and $(A \cdot B \cdot C)$ are mutually exclusive, and $A$, $B$, $C$, $\bar{A}$, $\bar{B}$, and $\bar{C}$ are independent, the probability that at least one machine needs attention is:
$$
\begin{array}{l}
P(D)=P(A \cdot \bar{B} \cdot \bar{C})+P(\bar{A} \cdot B \cdot \bar{C})+P(\bar{A} \cdot \bar{B} \cdot C)+P(A \cdot B \cdot \bar{C})+P(A \cdot \\
\quad \bar{B} \cdot C)+P(\bar{A} \cdot B \cdot C)+P(A \cdot B \cdot C) \\
\quad=P(A) \cdot P(\bar{B}) \cdot P(\bar{C})+P(\bar{A}) \cdot P(B) \cdot P(\bar{C})+P(\bar{A}) \cdot P(\bar{B}) \cdot P(C)+P(A) \cdot \\
\quad P(B) \cdot P(\bar{C})+P(A) \cdot P(\bar{B}) \cdot P(C)+P(\bar{A}) \cdot P(B) \cdot P(C)+P(A) \cdot P(B) \cdot P(C) \\
\quad=0.9 \times 0.2 \times 0.15+0.1 \times 0.8 \times 0.15+0.1 \times 0.2 \times 0.85+0.9 \times 0.8 \times \\
\quad 0.15+0.9 \times 0.2 \times 0.85+0.1 \times 0.8 \times 0.85+0.9 \times 0.8 \times 0.85 \\
\quad=0.997
\end{array}
$$

Answer: 0.997


Given that $a, b, c$ are the lengths of the three sides of a triangle, then the value of $a^{4}+b^{4}+c^{4}-$ $2 a^{2} b^{2}-2 b^{2} c^{2}-2 c^{2} a^{2}$ is
(A) always positive.
(B) always negative.
(C) can be positive or negative.
(D) non-negative.
(China Shanxi Province Taiyuan City Junior High School Mathematics Competition, 1991)

Solution:
[Solution 1]
$$
\begin{aligned}
\text { [Solution 1] } & a^{4}+b^{4}+c^{4}-2 a^{2} b^{2}-2 b^{2} c^{2}-2 c^{2} a^{2} \\
= & a^{4}+\left(b^{2}-c^{2}\right)^{2}-\left[a^{2}(b+c)^{2}+a^{2}(b-c)^{2}\right] \\
= & a^{4}-\left[a^{2}(b+c)^{2}+a^{2}(b-c)^{2}\right]+(b+c)^{2}(b-c)^{2} \\
= & {\left[a^{2}-(b+c)^{2}\right]\left[a^{2}-(b-c)^{2}\right] } \\
= & (a-b-c)(a+b+c)(a-b+c)(a+b-c), \\
\because \quad & a+b>c, \quad a+c>b, \quad b+c>a,
\end{aligned}
$$
$\therefore$. The above expression is always less than 0.
Therefore, the answer is $(B)$.
[Solution 2] By the square of a trinomial formula, we have
$$
\begin{aligned}
\text { Original expression } & =\left(a^{2}+b^{2}-c^{2}\right)^{2}-4 a^{2} b^{2} \\
& =(a+b+c)(a+b-c)(a-b+c)(a-b-c),
\end{aligned}
$$

The remaining steps are the same as above.
Therefore, the answer is $(B)$.

Answer: B


From the first 2005 natural numbers, $k$ of them are arbitrarily chosen. What is the least value of $k$ to ensure that there is at least one pair of numbers such that one of them is divisible by the other?

Solution:
6. Ans: 1004
Take any set 1004 numbers. Write each number in the form $2^{a_{i}} b_{i}$, where $a_{i} \geq 0$ and $b_{i}$ is odd. Thus obtaining 1004 odd numbers $b_{1}, \ldots, b_{1004}$. Since there are 1003 odd numbers in the first 2005 natural numbers, at least two of the odd numbers are the same, say $b_{i}=b_{j}$. Then $2^{a_{i}} b_{i} \mid 2^{a_{2}} b_{j}$ if $2^{\alpha_{i}} b_{i}<2^{\alpha_{i}} b_{i}$. The numbers $1003,1004, \ldots, 2005$ are 1003 numbers where there is no pair such that one of them is divisible by the other. So the answer is 1004 .

Answer: 1004


In a $7 \times 8$ rectangular chessboard, a chess piece is placed at the center of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected". Now, some of the 56 chess pieces are removed so that no 5 remaining chess pieces are in a straight line (horizontally, vertically, or diagonally). What is the minimum number of chess pieces that need to be removed to meet this requirement? Explain your reasoning $^{\mid 2\rfloor}$.

Solution:
Analysis: The difficulty ratio of this problem has significantly increased from 1, mainly due to its "apparent" asymmetry (the number of rows and columns are inconsistent). Therefore, the key to solving the problem is to find the symmetry hidden behind this asymmetry.
Inspired by Example 1, we will also start from the local perspective.
Solution: In each $1 \times 5$ rectangle, at least one chess piece must be taken. Following Example 1, as shown in Figure 5, the area outside the shaded part must have at least 10 chess pieces taken.

If at least one chess piece is taken from the shaded part in Figure 5, then the total number of chess pieces taken is no less than 11.

If no chess pieces are taken from the shaded part in Figure 5, then at least 2 chess pieces must be taken from each of the middle two columns, and at least 2 chess pieces must be taken from each of the middle three rows. Therefore, at least 10 chess pieces have been taken.

Considering the four squares (1), (2), (3), and (4) in Figure 6.

Looking diagonally, since the shaded area in Figure 6 indicates that no chess pieces have been taken, to avoid a line of five, the four squares (1), (2), (3), and (4) must all have chess pieces taken. Therefore, in Figure 6, 2 chess pieces are taken from the 3rd column and 2 chess pieces are taken from the 6th column; at least 1 chess piece is taken from the white area of each of the 1st, 2nd, 7th, and 8th columns, totaling 4 chess pieces; at least 1 chess piece is taken from the white area of each of the 1st, 2nd, 6th, and 7th rows, totaling 4 chess pieces. Therefore, 12 chess pieces are taken.
In summary, at least 11 chess pieces must be taken.
Figure 7 provides one possible method.

Answer: 11


Compute $1 \cdot 2+2 \cdot 3+\cdots+(n-1) n$.

Solution:
Solution: Let $S=1 \cdot 2+2 \cdot 3+\cdots+(n-1) n$. We know $\sum_{i=1}^{n} i=\frac{n(n+1)}{2}$ and $\sum_{i=1}^{n} i^{2}=$ $\frac{n(n+1)(2 n+1)}{6}$. So $S=1(1+1)+2(2+1)+\cdots+(n-1) n=\left(1^{2}+2^{2}+\cdots+(n-1)^{2}\right)+$ $(1+2+\cdots+(n-1))=\frac{(n-1)(n)(2 n-1)}{6}+\frac{(n-1)(n)}{2}=\frac{(n-1) n(n+1)}{3}$.
We can also arrive at the solution by realizing that $\sum_{i=1}^{n} i^{2}=\sum_{i=1}^{n} i^{2}+\sum_{i=2}^{n} i^{2}+\sum_{i=3}^{n} i^{2}+\cdots+$
$$
\begin{array}{l}
\sum_{i=n}^{n} i^{2}=n \sum_{i=1}^{n} i^{2}-\left(\sum_{i=1}^{1} i^{2}+\sum_{i=1}^{2} i^{2}+\sum_{i=1}^{3} i^{2}+\cdots \sum_{n=1}^{n-1} i^{2}\right)=n \frac{n(n+1)}{2}-\left(\frac{1 \cdot 2}{2}+\frac{2 \cdot 3}{2}+\cdots+\frac{(n-1) n}{2}\right)= \\
n \frac{n(n+1)}{2}-\frac{1}{2} S=\frac{n(n+1)(2 n+1)}{6}, \text { so } S=\frac{(n-1) n(n+1)}{3} .
\end{array}
$$

Answer: \frac{(n-1)n(n+1)}{3}


If the real number $x$ satisfies $\log _{2} x=\log _{4}(2 x)+\log _{8}(4 x)$, then $x=$

Solution:
$$
\begin{array}{l}
\log _{2} x=\log _{4}(2 x)+\log _{8}(4 x)=\frac{1}{2} \log _{2}(2 x)+\frac{1}{3} \log _{2}(4 x)=\log _{2}\left[(2 x)^{\frac{1}{2}}(4 x)^{\frac{1}{3}}\right] \\
\Rightarrow x=(2 x)^{\frac{1}{2}}(4 x)^{\frac{1}{3}}=2^{\frac{1}{2}+\frac{2}{3}} x^{\frac{1}{2}+\frac{1}{3}}=2^{\frac{7}{6}} x^{\frac{5}{6}} \Rightarrow x=2^{7}=128 .
\end{array}
$$

Answer: 128


The distance between places $A$ and $B$ is 600 kilometers. Two people, Jia and Yi, start cycling from $A$ to $B$ at the same time. Jia rides 40 kilometers per day, while Yi rides 60 kilometers per day but rests every other day. At the end of the $\qquad$th day, the distance Yi has left to $B$ is twice the distance Jia has left to $B$.

Solution:
【Analysis】This problem can be set up by assuming that at the end of the $x$-th day, the distance from $B$ to Yi is twice the distance from $B$ to Jia. According to the problem, on the $x$-th day, Jia has traveled $40 x$, and is still ( $600-40 x$ ) kilometers away from $B$, while Yi has traveled $60 \div 2 \times x$, and is still ( $600-60 \div 2 \times x$ ) kilometers away from $B$. Based on the condition that "the distance from $B$ to Yi is twice the distance from $B$ to Jia," we can set up the equation $600-60 \div 2 \times x=2 \times(600-40 x)$, and solve it accordingly.

【Solution】Solution: Let the $x$-th day be the day when the distance from $B$ to Yi is twice the distance from $B$ to Jia, we get:
$$
\begin{aligned}
600-60 \div 2 \times x & =2 \times(600-40 x) \\
600-30 x & =1200-80 x \\
50 x & =600 \\
x & =12
\end{aligned}
$$

Answer: On the 12th day, the distance from $B$ to Yi is twice the distance from $B$ to Jia. Therefore, the answer is: 12.

Answer: 12


Given that $x, y, z$ are three distinct non-zero natural numbers, if
$$
\overline{x y y y y}+\overline{x y y y}+\overline{x y y}+\overline{x y}+y=\overline{y y y y z} .
$$

where $\overline{x y y y y}$ and $\overline{y y y y z}$ are both five-digit numbers, when $x+y$ is maximized, the corresponding value of $x+y+z$ is

Solution: Reference answer: 22

Answer: 22


Let $a$ be a fixed real number. Consider the equation
$$
(x+2)^{2}(x+7)^{2}+a=0, x \in \mathbb{R}
$$
where $\mathbb{R}$ is the set of real numbers. For what values of $a$, will the equation have exactly one double-root?

Solution:
1. Define the function \( f(x) = (x+2)^2 (x+7)^2 + a \).
2. Compute the first derivative of \( f(x) \):
   \[
   f'(x) = \frac{d}{dx} \left[ (x+2)^2 (x+7)^2 + a \right] = 2(x+2)(x+7)(2x+9)
   \]
3. Find the critical points by setting \( f'(x) = 0 \):
   \[
   2(x+2)(x+7)(2x+9) = 0
   \]
   This gives us the critical points \( x = -2 \), \( x = -7 \), and \( x = -\frac{9}{2} \).

4. Determine the nature of these critical points by evaluating the second derivative \( f''(x) \):
   \[
   f''(x) = \frac{d}{dx} \left[ 2(x+2)(x+7)(2x+9) \right]
   \]
   Using the product rule:
   \[
   f''(x) = 2 \left[ (x+7)(2x+9) + (x+2)(2x+9) + (x+2)(x+7) \cdot 2 \right]
   \]
   Simplifying:
   \[
   f''(x) = 2 \left[ 2(x+7) + 2(x+2) + 2(x+2)(x+7) \right]
   \]
   \[
   f''(x) = 2 \left[ 2x + 14 + 2x + 4 + 2(x^2 + 9x + 14) \right]
   \]
   \[
   f''(x) = 2 \left[ 4x + 18 + 2x^2 + 18x + 28 \right]
   \]
   \[
   f''(x) = 2 \left[ 2x^2 + 22x + 46 \right]
   \]
   \[
   f''(x) = 4x^2 + 44x + 92
   \]

5. Evaluate \( f''(x) \) at the critical points:
   - At \( x = -2 \):
     \[
     f''(-2) = 4(-2)^2 + 44(-2) + 92 = 16 - 88 + 92 = 20 \quad (\text{local minimum})
     \]
   - At \( x = -7 \):
     \[
     f''(-7) = 4(-7)^2 + 44(-7) + 92 = 196 - 308 + 92 = -20 \quad (\text{local minimum})
     \]
   - At \( x = -\frac{9}{2} \):
     \[
     f''\left(-\frac{9}{2}\right) = 4\left(-\frac{9}{2}\right)^2 + 44\left(-\frac{9}{2}\right) + 92 = 81 - 198 + 92 = -25 \quad (\text{local maximum})
     \]

6. Since \( f(x) \) has local minima at \( x = -2 \) and \( x = -7 \), and a local maximum at \( x = -\frac{9}{2} \), we need \( f\left(-\frac{9}{2}\right) = 0 \) for the function to have exactly one double root.

7. Calculate \( f\left(-\frac{9}{2}\right) \):
   \[
   f\left(-\frac{9}{2}\right) = \left(-\frac{9}{2} + 2\right)^2 \left(-\frac{9}{2} + 7\right)^2 + a
   \]
   \[
   = \left(-\frac{5}{2}\right)^2 \left(\frac{5}{2}\right)^2 + a
   \]
   \[
   = \left(\frac{25}{4}\right) \left(\frac{25}{4}\right) + a
   \]
   \[
   = \frac{625}{16} + a
   \]
   Set this equal to 0:
   \[
   \frac{625}{16} + a = 0
   \]
   \[
   a = -\frac{625}{16}
   \]
   \[
   a = -39.0625
   \]

The final answer is \( \boxed{ a = -39.0625 } \)

Answer: a = -39.0625


For four independent samples of the same size \( n=17 \), extracted from normal general populations, the corrected sample variances are: 0.21; 0.25; 0.34; 0.40.

Required: a) at a significance level of 0.05, test the null hypothesis of homogeneity of variances (critical region - right-tailed); b) estimate the population variance.

Solution:
Solution. a) Let's find the observed value of the Cochran criterion - the ratio of the maximum corrected variance to the sum of all variances:

$$
G_{\text {obs } 6 \pi}=0.40 /(0.21+0.25+0.34+0.40)=\frac{1}{3} .
$$

We will find the critical point $G_{\text {crit }}(0.05 ; 16 ; 4)=0.4366$ in the table of critical values of the Cochran distribution (see Appendix 8) at a significance level of 0.05, degrees of freedom $k=n-1=17-1=16$, and number of samples $l=4$.

Since $G_{\text {obs }}<G_{\text {crit }}$ - there is no reason to reject the null hypothesis. In other words, the corrected sample variances do not differ significantly.

6) Since the homogeneity of variances has been established, as an estimate of the population variance, we will take the arithmetic mean of the corrected variances:

$$
D_{\mathrm{r}}^{*}=(0.21+0.25+0.34+0.40) / 4=0.3
$$

Answer: 0.3


Misha, over the course of a week, picked an apple each day and weighed it. Each apple weighed a different amount, but the weight of each apple was a whole number of grams and ranged from 221 grams to 230 grams (inclusive). Misha also calculated the average weight of all the apples he picked, and it was always a whole number. The apple picked on the seventh day weighed 225 grams. How much did the apple picked on the sixth day weigh?

Solution:
Answer: 230 grams.

Solution: Each apple weighed 220 grams plus an integer from 1 to 10. From the numbers 1 to 10, we need to choose 7 numbers such that their sum is divisible by 7. One of these numbers is 5, so we need to choose 6 from the numbers $1,2,3,4,6,7,8,9,10$. The smallest sum of six of these numbers is 23, and the largest is 44. Adding 5, we get numbers from 28 to 49. In this range, there are 4 numbers divisible by 7: 28, 35, 42, 49. Meanwhile, the sum of the weights for the first 6 days must be divisible by 6. From the numbers 23, 30, 37, 44, only the number 30 satisfies this condition. Thus, the total weight for the first 6 days was $220 \cdot 6 + 30$ (we will simply write 30, and add the multiple of 220 at the end). The sum of the weights for the first 5 days must be divisible by 5. The possible range for 5 days is from 16 to 40, and the numbers that fit are $20, 25, 30, 40$. However, the total weight for 5 days is less than that for 6 days, so it must be less than 30, i.e., 20 or 25. However, the difference between 30 and 25 is 5, and the number 5 is already used. Therefore, the weight of the apples for 5 days is $220 \cdot 5 + 20$, meaning the apple picked on the 6th day weighed $220 \cdot 6 + 30 - (220 \cdot 5 + 20) = 230$ grams. This value is valid, as the apples picked over the week could have weighed 221, 223, 222, 226, 228, 210, 205 grams.

Answer: 230


Let $x, y \in \mathbf{R}^{+}$, and $\frac{19}{x}+\frac{98}{y}=1$. Then the minimum value of $x+y$ is $\qquad$

Solution: \begin{array}{l}\text { 6. } 117+14 \sqrt{38} \text {. } \\ \text { Let } \sin ^{2} \alpha=\frac{19}{x}, \cos ^{2} \alpha=\frac{98}{y} . \\ \text { Then } x=\frac{19}{\sin ^{2} \alpha}, y=\frac{98}{\cos ^{2} \alpha} \text {. } \\ \begin{aligned} \therefore x+y=\frac{19}{\sin ^{2} \alpha}+\frac{98}{\cos ^{2} \alpha} \\ \quad=19 \csc ^{2} \alpha+98 \sec ^{2} \alpha \\ \quad=19\left(1+\cot ^{2} \alpha\right)+98\left(1+\tan ^{2} \alpha\right) \\ \quad=19 \cot ^{2} \alpha+98 \tan ^{2} \alpha+19+98 \\ \geqslant 2 \sqrt{19 \times 98}+117=117+14 \sqrt{38} .\end{aligned}\end{array}

Answer: 117+14 \sqrt{38}


A random variable $a$ is uniformly distributed on the interval $[-4 ; 4]$. Find the probability that at least one of the roots of the quadratic equation $x^{2}-2 a x-a-4=0$ does not exceed 1 in absolute value.

Solution: Answer: $P(A)=\frac{1}{2}$.

Answer: \frac{1}{2}


In modern society, the difficulty of deciphering passwords is increasing. There is a type of password that breaks down English plaintext (real text) into letters, where the 26 English letters $\mathrm{a}, \mathrm{b}, \cdots, \mathrm{z}$ (regardless of case) correspond to the 26 natural numbers $1,2, \cdots, 26$, as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline a & b & c & d & e & f & g & h & i & j & k & l & m \\
\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\
\hline n & o & p & q & r & s & t & u & v & w & x & y & z \\
\hline 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 \\
\hline
\end{tabular}

A transformation formula is given as follows:
$$
x^{\prime}=\left\{\begin{array}{l}
\frac{x+1}{2}, \text{ if } x \text{ is odd and } 1 \leqslant x \leqslant 26 ; \\
\frac{x}{2}+13, \text{ if } x \text{ is even and } 1 \leqslant x \leqslant 26 .
\end{array}\right.
$$

Using this formula, plaintext can be converted into ciphertext. For example, $8 \rightarrow \frac{8}{2}+13=17$, which means $h$ becomes $\mathrm{q} ; 5 \rightarrow \frac{5+1}{2}=3$, which means $\mathrm{e}$ becomes $\mathrm{c}$.

According to the above formula, if the ciphertext obtained from a certain plaintext is shxc, then the original plaintext is ( ).
(A) Ihho
(B) ohhl
(C) love
(D) eovl

Solution:
4.C.

The inverse transformation formula is
$$
x=\left\{\begin{array}{ll}
2 x^{\prime}-1, & 1 \leqslant x^{\prime} \leqslant 13, x^{\prime} \in \mathbf{N} ; \\
2 x^{\prime}-26, & 14 \leqslant x^{\prime} \leqslant 26, x^{\prime} \in \mathbf{N} .
\end{array}\right.
$$

According to the above transformation, we have
$$
\begin{array}{l}
\mathrm{s} \rightarrow 19 \rightarrow 2 \times 19-26=12 \rightarrow \mathrm{l}, \\
\mathrm{h} \rightarrow 8 \rightarrow 2 \times 8-1=15 \rightarrow \mathrm{o}, \\
\mathrm{x} \rightarrow 24 \rightarrow 2 \times 24-26=22 \rightarrow \mathrm{v}, \\
\mathrm{c} \rightarrow 3 \rightarrow 2 \times 3-1=5 \rightarrow \mathrm{e} .
\end{array}
$$

Therefore, the plaintext of shxc is love.

Answer: C


Let $a_{n}$ denote the integer closest to $\sqrt{n}(n \in \mathbf{N})$. Then the value of $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\ldots+\frac{1}{a_{2001}}$ is $\qquad$ .

Solution:
$4.88 \frac{7}{15}$.
Since $\sqrt{n}(n \in \mathbf{N})$ is either an integer or an irrational number,
$$
\begin{aligned}
\therefore a_{n}=k(k \in \mathbf{N}) & \Leftrightarrow k-\frac{1}{2}<\sqrt{n}<k+\frac{1}{2} \\
& \Leftrightarrow k^{2}-k+1 \leqslant n \leqslant k^{2}+k,
\end{aligned}
$$

That is, when $n$ takes the $2k$ natural numbers $k^{2}-k+1, k^{2}-k+2, \ldots, k^{2}+k$, $a_{n}=k$. Therefore,
$$
\begin{array}{l}
\frac{1}{a_{k}^{2}-k+1}+\frac{1}{a_{k}^{2}-k+2}+\ldots+\frac{1}{a_{k}^{2}+k}=2 k \frac{1}{k}=2 . \\
\because 44^{2}+44<2001<45^{2}+45 \text {, and } \\
44^{2}+44=1980,
\end{array}
$$
$\therefore a_{1}, a_{2} \ldots, a_{1980}$ contain 2 ones, 14 twos, ..., 88 forty-fours.
And $a_{1981}=a_{1982}=\ldots=a_{2001}=45$ then
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{2001}}=44 \times 2+21 \times \frac{1}{45}=88 \frac{7}{15}
$$

Answer: 88 \frac{7}{15}


determine the largest natural number $n$ so that

$$
4^{995}+4^{1500}+4^{n}
$$

is a square number.

Solution:
## 1st solution

Answer: $n=2004$

We show more generally: If $a\left(2^{n-a}\right)^{2}$ applies, but on the other hand, because of $2 b-2 a<n-a+1$ it also applies

$$
\begin{aligned}
A & =1+2^{2 b-2 a}+\left(2^{n-a}\right)^{2} \\
& <1+2 \cdot 2^{n-a}+\left(2^{n-a}\right)^{2}=\left(2^{n-a}+1\right)^{2}
\end{aligned}
$$

This means that $A$ lies between two consecutive square numbers and cannot itself be a square number.

Answer: 2004


Solution:
## Solution.

Domain of definition: $p \neq-2$.

Factor the numerator of the first fraction:

$$
\begin{aligned}
& p^{3}+4 p^{2}+10 p+12=\left(p^{3}+2 p^{2}\right)+\left(2 p^{2}+4 p\right)+(6 p+12)=p^{2}(p+2)+ \\
& +2 p(p+2)+6(p+2)=(p+2)\left(p^{2}+2 p+6\right)
\end{aligned}
$$

Similarly for the denominator, we find that

$$
\begin{aligned}
& p^{3}-p^{2}+2 p+16=(p+2)\left(p^{2}-3 p+8\right) \text { Then } \\
& \frac{p^{3}+4 p^{2}+10 p+12}{p^{3}-p^{2}+2 p+16} \cdot \frac{p^{3}-3 p^{2}+8 p}{p^{2}+2 p+6}=\frac{(p+2)\left(p^{2}+2 p+6\right)}{(p+2)\left(p^{2}-3 p+8\right)} \times \\
& \times \frac{p\left(p^{2}-3 p+8\right)}{p^{2}+2 p+6}=p
\end{aligned}
$$

Answer: $p$.

Answer: p


In the diagram below, $A B C D$ is a quadrilateral such that $\angle A B C=135^{\circ}$ and $\angle B C D=120^{\circ}$. Moreover, it is given that $A B=2 \sqrt{3} \mathrm{~cm}$, $B C=4-2 \sqrt{2} \mathrm{~cm}, C D=4 \sqrt{2} \mathrm{~cm}$ and $A D=x \mathrm{~cm}$. Find the value of $x^{2}-4 x$.

Solution:
31. Answer: 20
First we let $\ell$ be the line which extends $B C$ in both directions. Let $E$ be the point on $\ell$ such that $A E$ is perpendicular to $\ell$. Similarly, we let $F$ be the point on $\ell$ such that $D F$ is perpendicular to $\ell$. Then, it is easy to see that $B E=A E=\sqrt{6}, C F=2 \sqrt{2}$ and $D F=2 \sqrt{6}$. Thus $E F=\sqrt{6}+4-2 \sqrt{2}+2 \sqrt{2}=4+\sqrt{6}$. Now we let $G$ be the point on $D F$ such that $A G$ is parallel to $\ell$. Then $A G=E F=4+\sqrt{6}$ and $D G=D F-G F=D F-A E=2 \sqrt{6}-\sqrt{6}=\sqrt{6}$. So for the right-angled triangle $A D G$, we have
$$
x=A D=\sqrt{A G^{2}+D G^{2}}=\sqrt{(4+\sqrt{6})^{2}+(\sqrt{6})^{2}}=\sqrt{28+8 \sqrt{6}}=2+2 \sqrt{6} \text {. }
$$

Thus, $x^{2}-4 x=(2+2 \sqrt{6})^{2}-8-8 \sqrt{6}=4+24+8 \sqrt{6}-8-8 \sqrt{6}=20$.

Answer: 20


Let $p, \alpha, m \in \mathbf{Z}_{+}$, denote $p^{\alpha} \| m$ to mean $p^{\alpha} \mid m$, but $p^{\alpha+1} \nmid m$. For any positive integer $n$, the maximum value of the positive integer $\alpha$ satisfying $2^{\alpha} \|\left(3^{n}+1\right)$ is $\qquad$

Solution:
4. 1 .

Notice that the square of an odd number is congruent to 1 modulo 4.
If $n=2 k\left(k \in \mathbf{Z}_{+}\right)$, then
$$
3^{n}+1=3^{2 k}+1=\left(3^{k}\right)^{2}+1 \equiv 2(\bmod 4) \text {. }
$$

At this point, $2 \|\left(3^{n}+1\right)$.
If $n=2 k+1\left(k \in \mathbf{Z}_{+}\right)$, then
$$
\begin{array}{l}
3^{n}+1=3^{2 k+1}+1=3\left(3^{k}\right)^{2}+1 \\
\equiv 3 \times 1+1=4(\bmod 8) .
\end{array}
$$

At this point, $2^{2} \|\left(3^{n}+1\right)$.
Therefore, for any positive integer $n$, the maximum value of $\alpha$ is 1.

Answer: 1


Two players are playing an "Easter egg fight." In front of them is a large basket of eggs. They randomly take an egg each and hit them against each other. One of the eggs breaks, the loser takes a new egg, while the winner keeps their egg for the next round (the outcome of each round depends only on which egg has the stronger shell; the winning egg retains its strength). It is known that the first player won the first ten rounds. What is the probability that he will win the next round as well?

Solution:
Solution. Regardless of how events unfolded in the first ten rounds, in the 10th round, the egg that is the strongest among the first 11 eggs randomly drawn from the basket will be the winner. This means that the strongest egg will break in the 11th round only if the 12th egg drawn is stronger, that is, it is stronger than the previous 11 eggs. The probability of this is 1/12. Therefore, the probability of the opposite event “the winner of the ten rounds will also win the eleventh round” is $1-\frac{1}{12}=\frac{11}{12}$.

Answer: $11 / 12$.

Answer: \frac{11}{12}


A square \(A B C D\) is divided into 25 congruent smaller squares.

Let \(n\) be a positive integer with \(n \leq 25\). Then \(n\) different colors are chosen, and for each of these colors, 25 small tiles of the size of the smaller squares are provided.

From these \(n \cdot 25\) tiles, 25 are to be selected and placed on the square \(A B C D\) such that each smaller square is covered by exactly one of the selected tiles.

A number \(n\) will be called a "friendly" number if and only if the following holds:

For any selection of 25 of the \(n \cdot 25\) tiles, where each of the \(n\) colors is represented by at least one tile, it is possible to arrange the distribution on the smaller squares such that the covered square \(A B C D\) forms a symmetric pattern with respect to the line through \(A\) and \(C\).

Determine all "friendly" numbers among the positive integers \(n \leq 25\)!

Solution:
}

For every symmetric arrangement, it holds that the 10 sub-squares above the diagonal \(A C\) must each have the same color as the 10 sub-squares below this diagonal, while the colors of the 5 diagonal squares can be arbitrary. This is also sufficient. Therefore, a symmetric arrangement with respect to \(A C\) cannot be created from a selection of 25 tiles if it does not contain 10 pairwise disjoint pairs of tiles of the same color.

Let \(a_{1}, a_{2}, \ldots, a_{n}\) be the number of tiles of color \(1, 2, \ldots, n\) among the 25 selected, then the maximum number of such pairs that can be formed is

\[
P:=\left[\frac{a_{1}}{2}\right]+\left[\frac{a_{2}}{2}\right]+\cdots+\left[\frac{a_{n}}{2}\right]
\]

Given \(a_{1}+a_{2}+\cdots+a_{n}=25\), we have \(P=\frac{25-u}{2}\), where \(u\) is the number of odd numbers among \(a_{1}, a_{2}, \ldots, a_{n}\). (Since 25 is odd, \(u\) is also odd, and thus \(P\) is an integer.) Therefore, a symmetric arrangement with respect to \(A C\) cannot be created from these tiles if there are more than 5, i.e., at least 7, odd tile counts among \(a_{1}, a_{2}, \ldots, a_{n}\).

For this to be the case, \(n \geq 7\) must hold. This means that all \(n \leq 6\) are "friendly." However, if \(n=7\), one can choose \(a_{1}=a_{2}=\cdots=a_{6}=1\) and \(a_{7}=19\), or for \(n>7\), finally \(a_{1}=a_{2}=\cdots=a_{7}=1\) (and the rest arbitrarily \(\geq 1\)), such that no symmetric arrangements with respect to \(A C\) are possible with this selection. Therefore, all \(7 \leq n \leq 25\) are "unfriendly."

Answer: 1,2,3,4,5,6


In $\triangle ABC, AB=9, BC=8$ and $AC=7$. The bisector of $\angle A$ meets $BC$ at $D$. The circle passing through $A$ and touching $BC$ at $D$ cuts $AB$ and $AC$ at $M$ and $N$ respectively. Find $MN$.

Solution:
10. 6
10. By the angle bisector theorem, $\frac{B D}{D C}=\frac{A B}{A C}=\frac{9}{7}$ and so $B D=8 \times \frac{9}{9+7}=\frac{9}{2}$. By the power chord theorem, $B D^{2}=B M \times B A$ and so $B M=\frac{9}{4}$.
Finally, $M N$ is parallel to $B C$ since
$$
\angle B D M=\angle M A D=\angle D A N=\angle D M N \text {. }
$$

Hence we have $\frac{A M}{A B}=\frac{M N}{B C}$, or $\frac{9-\frac{9}{4}}{9}=\frac{M N}{8}$, so that $M N=6$.

Answer: 6


As shown in the figure, from a rectangular piece of paper that is 50 cm long and 20 cm wide, two squares with side lengths of 12 cm and 4 cm are cut out, respectively. The perimeter of the remaining part of the figure is $\qquad$ cm.

Solution:
【Analysis】The perimeter of the remaining part $=$ the perimeter of the original rectangle + 2 times $(12+4)$ cm. Based on this, we can set up the equation $(50+20) \times 2+(12+4) \times 2$ to calculate the solution.

【Solution】Solution: $(50+20) \times 2+(12+4) \times 2$
$$
\begin{array}{l}
=70 \times 2+16 \times 2 \\
=140+32 \\
=172 \text { (cm) }
\end{array}
$$

Answer: The perimeter of the remaining part of the figure is 172 cm.
Therefore, the answer is: 172.

Answer: 172


Let the quadratic function $f(x)=a x^{2}+b x+c(a \neq 0)$ have values whose absolute values do not exceed 1 on the interval $[0,1]$, find the maximum value of $|a|+|b|+|c|$.

Solution:
Solve: From

we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
f(0)=c \\
f(1)=a+b+c \\
f\left(\frac{1}{2}\right)=\frac{1}{4} a+\frac{1}{2} b+c
\end{array}\right. \\
\left\{\begin{array}{l}
a=2 f(0)-4 f\left(\frac{1}{2}\right)+2 f(1) \\
b=-3 f(0)+4 f\left(\frac{1}{2}\right)-f(1) \\
c=f(0)
\end{array}\right. \\
\begin{array}{l}
f=f(0)
\end{array} \\
b=
\end{array}
$$

Therefore,
$$
\begin{aligned}
|a|+|b|+|c|= & \left|2 f(0)-4 f\left(\frac{1}{2}\right)+2 f(1)\right|+\mid-3 f(0)+ \\
& \left.4 f\left(\frac{1}{2}\right)-f(1)|+| f(0)|\leqslant 2| f(0)|+4|-f\left(\frac{1}{2}\right) \right\rvert\,+ \\
& 2|f(1)|+3|f(0)|+4\left|-f\left(\frac{1}{2}\right)\right|+|f(1)|+|f(0)|= \\
& 6|f(0)|+8\left|-f\left(\frac{1}{2}\right)\right|+3|f(1)| \leqslant
\end{aligned}
$$
$$17$$

Equality holds if and only if $\left\{\begin{array}{l}f(0)=1 \\ f\left(\frac{1}{2}\right)=-1 \\ f(1)=1\end{array}\right.$, i.e., $\left\{\begin{array}{l}a=8 \\ b=-8 \\ c=1\end{array}\right.$ or $\left\{\begin{array}{l}f(0)=-1 \\ f\left(\frac{1}{2}\right)=1 \\ f(1)=-1\end{array}\right.$, i.e., $\left\{\begin{array}{l}a=-8 \\ b=8 \\ c=-1\end{array}\right.$. In this case, the quadratic function is $f(x)=8 x^{2}-8 x+1$ or $f(x)=-8 x^{2}+8 x-1$. If we let $x=\cos ^{2} x$, then at this time, $|5(x)|=|\cos 4 \alpha| \leqslant 1$.

Note: Example 18 can be generalized: Suppose the cubic function $f(x)=a x^{3}+b x^{2}+c x+d(a \neq 0)$ on the interval $[0,1]$, the absolute value of its values is no greater than 1, find the maximum value of $|a|+|b|+|c|+|d|$.

Solution: The maximum value can be obtained as $99$. Specifically, $a=\frac{1}{3}\left[-16 f(0)+32 f\left(\frac{1}{4}\right)-32 f\left(\frac{3}{4}\right)+16 f(1)\right], b=$ $\frac{1}{3}\left[32 f(0)-56 f\left(\frac{1}{4}\right)+40 f\left(\frac{3}{4}\right)-16 f(1)\right], c=\frac{1}{3}\left[-19 f(0)+24 f\left(\frac{1}{4}\right)-\right.$ $\left.8 f\left(\frac{3}{4}\right)+3 f(1)\right], d=f(0) ; f(0)=-1, f\left(\frac{1}{4}\right)=1, f\left(\frac{3}{4}\right)=-1, f(1)=1$; or $f(0)=1, f\left(\frac{1}{4}\right)=-1, f\left(\frac{3}{4}\right)=1, f(1)=-1$, the cubic function is $f(x)=32 x^{3}-$ $48 x^{2}+18 x-1$ or $f(x)=-32 x^{3}+48 x^{2}-18 x+1$.

More generally, for an $n$-degree function $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$, on the interval $[0,1]$, if $|f(x)| \leqslant 1$, then when
$$
\begin{array}{c}
f\left(\cos ^{2} \frac{n \pi}{2 n}\right)=-1, f\left(\cos ^{2} \frac{(n-1) \pi}{2 n}\right)=1 \\
f\left(\cos ^{2} \frac{(n-2) \pi}{2 n}\right)=-1, \cdots, f\left(\cos ^{2} 0\right)=(-1)^{n+1}
\end{array}
$$

or when
$$
\begin{array}{c}
f\left(\cos ^{2} \frac{n \pi}{2 n}\right)=1, f\left(\cos ^{2} \frac{(n-1) \pi}{2 n}\right)=-1 \\
f\left(\cos ^{2} \frac{(n-2) \pi}{2 n}\right)=1, \cdots f\left(\cos ^{2} 0\right)=(-1)^{n}
\end{array}
$$

$\sum_{i=1}^{n}\left|a_{i}\right|$ has the maximum value.

Answer: 17


Positive real numbers $a_{1}, a_{2}, \cdots, a_{8}$ are marked at the corresponding vertices of a cube, such that the number at each vertex is the average of the three numbers marked at the adjacent vertices. Let
$$
\begin{array}{l}
M=\left(a_{1}+2 a_{2}+3 a_{3}+4 a_{4}\right) . \\
\quad\left(5 a_{5}+6 a_{6}+7 a_{7}+8 a_{8}\right), \\
N=a_{1}^{2}+a_{2}^{2}+\cdots+a_{8}^{2} . \\
\text { Then } \frac{M}{N}=
\end{array}
$$

Solution:
4. $\frac{65}{2}$.

Assume that among the numbers marked in Figure 5, $a_{1}$ is the largest, then
$$
a_{1} \geqslant a_{2}, a_{1} \geqslant a_{4}, a_{1} \geqslant a_{5} .
$$

From $a_{1}=\frac{a_{2}+a_{4}+a_{5}}{3}$, we know
$$
\frac{a_{2}+a_{4}+a_{5}}{3} \leqslant a_{1} \text {. }
$$

The equality holds if and only if $a_{1}=a_{2}, a_{1}=a_{4}, a_{1}=a_{5}$, i.e., $a_{1}=a_{2}=a_{4}=a_{5}$ are all the largest.

Similarly, $a_{2}=a_{1}=a_{3}=a_{6}, a_{4}=a_{1}=a_{3}=a_{8}, a_{3}=a_{4}=a_{2}=a_{7}$, i.e.,
$$
a_{1}=a_{2}=\cdots=a_{8} .
$$

Therefore, $M=260 a_{1}^{2}, N=8 a_{1}^{2}$.

Answer: \frac{65}{2}


In the room, there are 85 balloons â red and blue. It is known that: 1) at least one of the balloons is red; 2) in every arbitrarily chosen pair of balloons, at least one is blue. How many red balloons are in the room?

Solution:
Think about whether there can be two red balls in the room.

## Solution

Since among any two balls, one is blue, there cannot be two red balls in the room. Therefore, there are 84 blue balloons and 1 red balloon in the room.

## Answer

1 ball.

Answer: 1


Let $l$ be the tangent line at the point $P(s,\ t)$ on a circle $C:x^2+y^2=1$. Denote by $m$ the line passing through the point $(1,\ 0)$, parallel to $l$. Let the line $m$ intersects the circle $C$ at $P'$ other than the point $(1,\ 0)$.

Note : if $m$ is the line $x=1$, then $P'$ is considered as $(1,\ 0)$.

Call $T$ the operation such that the point $P'(s',\ t')$ is obtained from the point $P(s,\ t)$ on $C$.

(1) Express $s',\ t'$ as the polynomials of $s$ and $t$ respectively.

(2) Let $P_n$ be the point obtained by $n$ operations of $T$ for $P$.

For $P\left(\frac{\sqrt{3}}{2},\ \frac{1}{2}\right)$, plot the points $P_1,\ P_2$ and $P_3$.

(3) For a positive integer $n$, find the number of $P$ such that $P_n=P$.

Solution:
Given the circle \( C: x^2 + y^2 = 1 \) and a point \( P(s, t) \) on the circle, we need to find the tangent line at \( P \) and the line \( m \) passing through \( (1, 0) \) parallel to the tangent line. We then find the intersection of \( m \) with the circle at a point \( P' \) other than \( (1, 0) \).

1. **Express \( s' \) and \( t' \) as polynomials of \( s \) and \( t \):**

   The tangent line at \( P(s, t) \) on the circle \( x^2 + y^2 = 1 \) has the equation:
   \[
   sx + ty = 1
   \]
   The line \( m \) passing through \( (1, 0) \) and parallel to the tangent line has the same slope, so its equation is:
   \[
   sx + ty = s
   \]
   To find the intersection of this line with the circle \( x^2 + y^2 = 1 \), substitute \( y = \frac{s - sx}{t} \) into the circle's equation:
   \[
   x^2 + \left( \frac{s - sx}{t} \right)^2 = 1
   \]
   Simplifying, we get:
   \[
   x^2 + \frac{s^2 - 2s^2x + s^2x^2}{t^2} = 1
   \]
   \[
   x^2 + \frac{s^2}{t^2} - \frac{2s^2x}{t^2} + \frac{s^2x^2}{t^2} = 1
   \]
   \[
   t^2x^2 + s^2 - 2s^2x + s^2x^2 = t^2
   \]
   \[
   (t^2 + s^2)x^2 - 2s^2x + s^2 - t^2 = 0
   \]
   This is a quadratic equation in \( x \). Solving for \( x \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   x = \frac{2s^2 \pm \sqrt{4s^4 - 4(t^2 + s^2)(s^2 - t^2)}}{2(t^2 + s^2)}
   \]
   Simplifying inside the square root:
   \[
   x = \frac{2s^2 \pm \sqrt{4s^4 - 4(s^4 - t^4)}}{2(t^2 + s^2)}
   \]
   \[
   x = \frac{2s^2 \pm \sqrt{4t^4}}{2(t^2 + s^2)}
   \]
   \[
   x = \frac{2s^2 \pm 2t^2}{2(t^2 + s^2)}
   \]
   \[
   x = \frac{s^2 \pm t^2}{t^2 + s^2}
   \]
   Since \( x = 1 \) is one solution, the other solution is:
   \[
   x = \frac{s^2 - t^2}{t^2 + s^2}
   \]
   For \( y \):
   \[
   y = \frac{s - sx}{t}
   \]
   Substituting \( x = \frac{s^2 - t^2}{t^2 + s^2} \):
   \[
   y = \frac{s - s \left( \frac{s^2 - t^2}{t^2 + s^2} \right)}{t}
   \]
   \[
   y = \frac{s(t^2 + s^2) - s(s^2 - t^2)}{t(t^2 + s^2)}
   \]
   \[
   y = \frac{st^2 + s^3 - s^3 + st^2}{t(t^2 + s^2)}
   \]
   \[
   y = \frac{2st^2}{t(t^2 + s^2)}
   \]
   \[
   y = \frac{2st^2}{t(t^2 + s^2)}
   \]
   \[
   y = \frac{2st}{t^2 + s^2}
   \]

   Therefore, the coordinates of \( P' \) are:
   \[
   s' = \frac{s^2 - t^2}{t^2 + s^2}, \quad t' = \frac{2st}{t^2 + s^2}
   \]

2. **Plot the points \( P_1, P_2, \) and \( P_3 \) for \( P\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \):**

   Using the formulas derived:
   \[
   P_1\left( \frac{\left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}, \frac{2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2}}{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \right)
   \]
   \[
   P_1\left( \frac{\frac{3}{4} - \frac{1}{4}}{\frac{1}{4} + \frac{3}{4}}, \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \cdot \frac{2}{1} \right)
   \]
   \[
   P_1\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)
   \]

   For \( P_2 \):
   \[
   P_2\left( \frac{\left(\frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2}{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2}, \frac{2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2}}{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} \right)
   \]
   \[
   P_2\left( \frac{\frac{1}{4} - \frac{3}{4}}{\frac{3}{4} + \frac{1}{4}}, \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \cdot \frac{2}{1} \right)
   \]
   \[
   P_2\left( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right)
   \]

   For \( P_3 \):
   \[
   P_3\left( \frac{\left(-\frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2}{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2}, \frac{2 \cdot -\frac{1}{2} \cdot \frac{\sqrt{3}}{2}}{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} \right)
   \]
   \[
   P_3\left( \frac{\frac{1}{4} - \frac{3}{4}}{\frac{3}{4} + \frac{1}{4}}, \frac{\sqrt{3}}{2} \cdot -\frac{1}{2} \cdot \frac{2}{1} \right)
   \]
   \[
   P_3\left( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right)
   \]

   Therefore, the points are:
   \[
   P_1\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right), \quad P_2\left( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right), \quad P_3\left( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right)
   \]

3. **Find the number of \( P \) such that \( P_n = P \):**

   Considering the complex number representation, let \( z = e^{i\theta} \) where \( \theta \) is the argument of \( P \). The operation \( T \) corresponds to squaring the complex number, so after \( n \) operations, we have:
   \[
   z^{2^n} = z
   \]
   This implies:
   \[
   z^{2^n - 1} = 1
   \]
   The solutions to this equation are the \( 2^n - 1 \) roots of unity, excluding \( z = 1 \). Therefore, there are \( 2^n - 1 \) such points.

The final answer is \( \boxed{ 2^n - 1 } \)

Answer: 2^n - 1


How many four-digit natural numbers have the product of their digits equal to 24? Write them down!

Solution:
3. Since $24=2 \cdot 2 \cdot 2 \cdot 3$, the corresponding quadruples of digits are:
a) $2,2,2,3$,
b) $1,2,3,4$,
c) $1,2,2,6$,
d) $1,1,3,8$,
e) $1,1,4,6$

The required numbers are:

$\begin{array}{lr}\text { a) } 2223,2232,2322,3222, & 1 \text { point } \\ \text { b) } 1234,1243,1324,1342,1423,1432,2134,2143,2314,2341,2413,2431,3124,3142,3214,3241, \\ 3412,3421,4123,4132,4213,4231,4312,4321, & 2 \text { points } \\ \text { c) } 1226,1262,1622,2126,2162,2216,2261,2612,2621,6122,6212,6221, & 2 \text { points } \\ \text { d) } 1138,1183,1318,1381,1813,1831,3118,3181,3811,8113,8131,8311 . & 2 \text { points } \\ \text { e)1146,1164,1416,1461,1614,1641,4116,4161,4611,6114,6141,6411. } \\ \text { There are 64 numbers with the required property. } & 2 \text { points } \\ \text {.................................................................................................................... } & 1 \text { point }\end{array}$

Answer: 64


Find all odd natural numbers greater than 500 but less than 1000, for each of which the sum of the last digits of all divisors (including 1 and the number itself) is 33.

Solution:
For an odd number, all its divisors are odd. Since the sum of their last digits is odd, there must be an odd number of divisors. If a number has an odd number of divisors, then it is a square of a natural number (see problem $\underline{30365}$). Let's consider all the squares of odd numbers within the given range: $23^{2}, 25^{2}=625, 27^{2}=729, 29^{2}, 31^{2}$. Note that the number we are looking for must have at least five divisors; otherwise, the sum of their last digits would not exceed 27. This means that we can exclude the squares of prime numbers 23, 29, and 31, which have exactly three divisors. Therefore, we are left to check two numbers. The divisors of the number $625: 1,5,25,125,625$; the sum of their last digits is 21. The divisors of the number $729: 1,3$, $9,27,81,243,729$; the sum of their last digits is 33.

## Answer

729.

Answer: 729


Through the vertices $A, B$ and $C$ of the parallelogram $ABCD$ with sides $AB=3$ and $BC=5$, a circle is drawn, intersecting the line $BD$ at point $E$, and $BE=9$. Find the diagonal $BD$.

Solution:
Using the theorem on the sum of the squares of the diagonals of a parallelogram and the theorem on the product of segments of intersecting chords, form a system of equations with respect to the diagonals of the parallelogram.

## Solution

Let the diagonals of the parallelogram intersect at point $O$. Denote $O B=O D=x, A O=O C=y$. According to the theorem on the sum of the squares of the diagonals of a parallelogram,

$$
A C^{2}+B D^{2}=2 A B^{2}+2 B C^{2}
$$

According to the theorem on the product of segments of intersecting chords,

$$
B O \cdot O E=A O \cdot O C
$$

Thus, we obtain the system

$$
\left\{\begin{array}{l}
4 x^{2}+4 y^{2}=2 \cdot 34 \\
x(9-x)=y^{2} x
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=17 \\
9 x=x^{2}+y^{2}
\end{array}\right.
$$

From here, we find that $B D=2 x=\frac{34}{9}$.

## Answer

$\frac{34}{9}$.

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Answer: \frac{34}{9}


Solution:
The number |x-a| is equal to the distance from the point on the number line with coordinate x to the point with coordinate a.

## Solution

Consider a point on the number line with coordinate x. The sum $|\mathrm{x}-2|+|\mathrm{x}-1|+|\mathrm{x}|+|\mathrm{x}+1|+|\mathrm{x}+2|$ is equal to the sum of the distances from the point x to the points with coordinates $2,1,0,-1,-2$. Note that the sum of the distances from any point to points A and B is not less than the length of the segment AB (and equality is achieved if and only if the point is located on the segment AB). From this, we get that |x-2|+|x+2| is not less than 4, and |x-1|+|x+1| is not less than 2 for any

| x. Therefore, for the sum $|x-2|+|x-1|+|x|+|x+1|+|x+2|$ to be equal to 2+4=6, it is necessary that |x|=0. Thus, x must be equal to 0. It is easy to verify that the value $x=0$ is indeed a solution to the given equation.

## Answer

$\mathrm{x}=0$.

Answer: 0


Solution:
Let $x=\tan \frac{\alpha}{2}\left(\alpha \in\left(0, \frac{\pi}{2}\right]\right)$. Then
$$
3 \alpha \in\left(0, \frac{3 \pi}{2}\right], \cos 3 \alpha \in[-1,1) \text {, }
$$

and $\cos \alpha=\frac{1-x^{2}}{1+x^{2}}$.
Thus, $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$
$$
\begin{array}{l}
=\frac{4\left(1-x^{2}\right)^{3}}{\left(1+x^{2}\right)^{3}}-\frac{3\left(1-x^{2}\right)}{1+x^{2}} \\
=\frac{1-15 x^{2}+15 x^{4}-x^{6}}{1+3 x^{2}+3 x^{4}+x^{6}} \\
=5-\frac{4+30 x^{2}+6 x^{6}}{1+3 x^{2}+3 x^{4}+x^{6}} \\
=5-2 \cdot \frac{2+15 x^{2}+3 x^{6}}{1+3 x^{2}+3 x^{4}+x^{6}} \\
=5-10 \cdot \frac{3 x^{6}+15 x^{2}+2}{5 x^{6}+15 x^{4}+15 x^{2}+5} \\
=5-10 \cdot \frac{3 x^{6}+15 x^{2}+2}{\left(3 x^{6}+15 x^{2}+2\right)+\left(2 x^{6}+15 x^{4}+3\right)} . \\
\text { Hence } y=\frac{3 x^{6}+15 x^{2}+2}{2 x^{6}+15 x^{4}+3}=\frac{5-\cos 3 \alpha}{5+\cos 3 \alpha} \in\left(\frac{2}{3}, \frac{3}{2}\right] .
\end{array}
$$

Answer: \left(\frac{2}{3}, \frac{3}{2}\right]


In quadrilateral $ABCD$, $AB=BC=CD$, $\angle ABC=78^{\circ}$, $\angle BCD=162^{\circ}$. Let the intersection point of line $AD$ and $BC$ be $E$. Then the size of $\angle AEB$ is

Solution:
$9.21^{\circ}$.
As shown in Figure 5, draw $D O / / C B$, and
make $D O=C B$, connect
$O A$ and $O B$.
Since $B C=C D$,
thus, quadrilateral $B C D O$ is a rhombus. Therefore,
$$
\begin{array}{l}
A B=B C=B O . \\
\text { Since } \angle A B O=78^{\circ}-\left(180^{\circ}-162^{\circ}\right)=60^{\circ},
\end{array}
$$

Therefore, $\triangle A B O$ is an equilateral triangle.
Let $\angle A E B=\alpha$. Then $\angle 1=\angle 2=\alpha$.
By the sum of the interior angles of a quadrilateral being $360^{\circ}$, we get
$$
\begin{array}{l}
78^{\circ}+162^{\circ}+\left(60^{\circ}+\alpha\right)+\left(180^{\circ}-162^{\circ}+\alpha\right) \\
=360^{\circ} .
\end{array}
$$

Solving for $\alpha$ gives $\alpha=21^{\circ}$.

Answer: 21^{\circ}


Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$. The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$. The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution:
This solution refers to the Diagram section.
As shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ has radius $13$) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\mathcal{P}.$ Let $\mathcal{R}$ be the plane that is determined by $O_1,O_2,$ and $O_3.$ Suppose $A$ is the foot of the perpendicular from $O_3$ to line $\ell,$ so $\overleftrightarrow{O_3A}$ is the perpendicular bisector of $\overline{O_1O_2}.$ We wish to find $T_3A.$

Note that:

In $\triangle O_1O_2O_3,$ we get $O_1O_2=72$ and $O_1O_3=O_2O_3=49.$
Both $\triangle O_1O_2O_3$ and $\overline{O_3A}$ lie in plane $\mathcal{R}.$ Both $\triangle T_1T_2T_3$ and $\overline{T_3A}$ lie in plane $\mathcal{P}.$
By symmetry, since planes $\mathcal{P}$ and $\mathcal{Q}$ are reflections of each other about plane $\mathcal{R},$ the three planes are concurrent to line $\ell.$
Since $\overline{O_1T_1}\perp\mathcal{P}$ and $\overline{O_3T_3}\perp\mathcal{P},$ it follows that $\overline{O_1T_1}\parallel\overline{O_3T_3},$ from which $O_1,O_3,T_1,$ and $T_3$ are coplanar.

Now, we focus on cross-sections $O_1O_3T_3T_1$ and $\mathcal{R}:$

In the three-dimensional space, the intersection of a line and a plane must be exactly one of the empty set, a point, or a line.
Clearly, cross-section $O_1O_3T_3T_1$ intersects line $\ell$ at exactly one point. Furthermore, as the intersection of planes $\mathcal{R}$ and $\mathcal{P}$ is line $\ell,$ we conclude that $\overrightarrow{O_1O_3}$ and $\overrightarrow{T_1T_3}$ must intersect line $\ell$ at the same point. Let $B$ be the point of concurrency of $\overrightarrow{O_1O_3},\overrightarrow{T_1T_3},$ and line $\ell.$
In cross-section $\mathcal{R},$ let $C$ be the foot of the perpendicular from $O_1$ to line $\ell,$ and $D$ be the foot of the perpendicular from $O_3$ to $\overline{O_1C}.$

We have the following diagram:

In cross-section $O_1O_3T_3T_1,$ since $\overline{O_1T_1}\parallel\overline{O_3T_3}$ as discussed, we obtain $\triangle O_1T_1B\sim\triangle O_3T_3B$ by AA, with the ratio of similitude $\frac{O_1T_1}{O_3T_3}=\frac{36}{13}.$ Therefore, we get $\frac{O_1B}{O_3B}=\frac{49+O_3B}{O_3B}=\frac{36}{13},$ or $O_3B=\frac{637}{23}.$
In cross-section $\mathcal{R},$ note that $O_1O_3=49$ and $DO_3=\frac{O_1O_2}{2}=36.$ Applying the Pythagorean Theorem to right $\triangle O_1DO_3,$ we have $O_1D=\sqrt{1105}.$ Moreover, since $\ell\perp\overline{O_1C}$ and $\overline{DO_3}\perp\overline{O_1C},$ we obtain $\ell\parallel\overline{DO_3}$ so that $\triangle O_1CB\sim\triangle O_1DO_3$ by AA, with the ratio of similitude $\frac{O_1B}{O_1O_3}=\frac{49+\frac{637}{23}}{49}.$ Therefore, we get $\frac{O_1C}{O_1D}=\frac{\sqrt{1105}+DC}{\sqrt{1105}}=\frac{49+\frac{637}{23}}{49},$ or $DC=\frac{13\sqrt{1105}}{23}.$
Finally, note that $\overline{O_3T_3}\perp\overline{T_3A}$ and $O_3T_3=13.$ Since quadrilateral $DCAO_3$ is a rectangle, we have $O_3A=DC=\frac{13\sqrt{1105}}{23}.$ Applying the Pythagorean Theorem to right $\triangle O_3T_3A$ gives $T_3A=\frac{312}{23},$ from which the answer is $312+23=\boxed{335}.$
~MRENTHUSIASM

Answer: 335


For any positive integer $k$, let $f_{1}(k)$ be the square of the sum of the digits of $k$ when written in decimal notation, and for $n>1$, let $f_{n}(k)=f_{1}\left(f_{n-1}(k)\right)$. What is $f_{1992}\left(2^{1991}\right)$?

Solution:
We can answer the question based on two aspects: on the one hand, we determine the number of digits, and on the other hand, the remainder of the number in question when divided by 9.

Let's denote the number $2^{1991}$ as $K$.

$$
K=8^{\frac{1991}{3}}<10^{664}
$$

Thus, the number $K$ has at most 664 digits, and the sum of its digits is no more than $664 \cdot 9$, which is less than 6000. Therefore, $f_{1}(K)<36 \cdot 10^{6}$. Up to this limit, the sum of the digits of the number 29999999 is the largest, 65. Thus,

$$
f_{2}(K) \leq 4225
$$

Similarly,

$$
f_{3}(K) \leq(3+3 \cdot 9)^{2}=900
$$

For a number $k$ with at most three digits, $f_{1}(k) \leq(3 \cdot 9)^{2}=729$, so it is also at most three digits. Therefore, if $n \geq 3$, then $f_{n}(K)$ is at most three digits.

It is known that a number divided by 9 gives the same remainder as the sum of its digits; furthermore, the remainder of the product of two numbers when divided by 9 is the same as the product of their remainders, since $(9 r+s)(9 t+u)=$ $9(9 r t+r u+s t)+s u$. Based on these,

$$
K=2^{6 \cdot 331+5}=64^{331} \cdot 32=(9 \cdot 7+1)^{331}(9 \cdot 3+5)
$$

has a remainder of 5, and thus the remainder of the sum of its digits is also 5. The remainder of $f_{1}(K)$ is therefore the same as that of 25, which is 7; the remainder of $f_{2}(K)$ is the same as that of 49, which is 4; the remainder of $f_{3}(K)$ is the same as that of 16, which is 7. This is also the remainder of the sum of its digits, and the number is at most three digits, so the sum of the digits can only be 7, 16, and 25. The possible values of $f_{4}(K)$ are 49, 256, and 625. The value of $f_{5}(K)$ is the same in all three cases: 169. Continuing the calculation,

$$
f_{6}(K)=256, \quad f_{7}(K)=169, \quad f_{8}(K)=256
$$

It is clear that from here, for even indices we get 256, and for odd indices, 169. The answer to the question is therefore

$$
f_{1992}\left(2^{1991}\right)=256
$$

Remarks. 1. Of course, for other $a$, $c$, and $n$, $f_{n}\left(a^{c}\right)$ can also be determined similarly, and it is easy to see that for sufficiently large $n$, only 1 and 81 can occur besides 169 and 256.

2. The sum of the digits of $2^{1991}$ can be determined using a computer, which is 2669. From this, $f_{1}(K)=7123561, f_{2}(K)=625$, $f_{3}(K)=169, f_{4}(K)=256$, and from here, the last two values repeat periodically.

Answer: 256


A sequence of real numbers $a_{0}, a_{1}, a_{2}, \ldots$ satisfies the relation

$$
a_{m+n}+a_{m-n}-m+n-1=\frac{a_{2 m}+a_{2 n}}{2}
$$

for all $m, n \in \mathbb{N}, m \geqslant n$. If $a_{1}=3$, find $a_{2004}$.

Solution:
1. By setting $m=n$ we get $a_{0}=1$. Further, for $n=0$ we get $a_{2 m}=4 a_{m}-2 m-3$. Now, the equality from the problem becomes $a_{m+n}+a_{m-n}=2\left(a_{m}+a_{n}-n-1\right)$. From here, for $n=1$ we get $a_{m+1}-2 a_{m}+a_{m-1}=2$, i.e., $\quad b_{m}=b_{m-1}+2$ for all $m$, where $b_{m}=a_{m+1}-a_{m}$. Since $b_{0}=2$, it follows that $b_{m}=2 m+2$ and

$$
a_{m}=a_{0}+b_{0}+b_{1}+\cdots+b_{m-1}=1+(2+4+\cdots+2 m)=m^{2}+m+1
$$

Answer: 2004^2+2004+1


a) Counting the odd numbers one by one starting from 1, verify that the number at position $m$ is equal to $m^{2}-(m-1)^{2}$.

b) Calculate the sum of all odd numbers between 1000 and 2014.

Solution:
Solution

a) See that the first odd number is $2 \cdot 1-1$ and, knowing that odd numbers increase by 2, we can conclude that the odd number at position $m$ in our count is

$$
\begin{aligned}
2 \cdot 1-1+\underbrace{2+2+\ldots+2}_{m-1 \text { times }} & =2 \cdot 1-1+2(m-1) \\
& =2 m-1
\end{aligned}
$$

To verify that it matches the number in item a), we just need to calculate

$$
m^{2}-(m-1)^{2}=m^{2}-\left(m^{2}-2 m-1\right)=2 m-1
$$

b) We want to sum the odd numbers from $1001=2 \cdot 501-1$ to $2013=2 \cdot 1007-1$. Using the expression from item a), we have

$$
\begin{aligned}
1001 & =501^{2}-500^{2} \\
1003 & =502^{2}-501^{2} \\
1005 & =503^{2}-502^{2} \\
& \cdots \\
2011 & =1006^{2}-1005^{2} \\
2013 & =1007^{2}-1006^{2}
\end{aligned}
$$

Adding everything, we see that all numbers from $501^{2}$ to $1006^{2}$ are canceled. Thus, the result is:

$$
\begin{aligned}
1001+1003+\ldots+2013 & =1007^{2}-500^{2} \\
& =(1007-500)(1007+500) \\
& =507 \cdot 1507 \\
& =764049
\end{aligned}
$$

So the sum of the odd numbers between 1000 and 2014 is 763048.

Answer: 764049


We are given some three element subsets of $\{1,2, \dots ,n\}$ for which any two of them have at most one common element. We call a subset of $\{1,2, \dots ,n\}$ \textbf{nice} if it doesn't include any of the given subsets. If no matter how the three element subsets are selected in the beginning, we can add one more element to every 29-element \textbf{nice} subset while keeping it nice, find the minimum value of $n$.

Solution:
1. **Assume otherwise**: Suppose there is a 29-element nice subset \( S \) such that we cannot add any element to \( S \) while keeping it nice. This implies that for each potential added element, there is a violating subset.

2. **Counting violating subsets**: Each violating subset must include two elements from the 29-element nice subset \( S \). Since any two of the three-element subsets have at most one common element, each violating subset must be distinct. Therefore, the number of violating subsets is at least \( n - 29 \).

3. **Upper bound on violating subsets**: The number of ways to choose 2 elements from the 29-element nice subset \( S \) is given by the binomial coefficient:
   \[
   \binom{29}{2} = \frac{29 \times 28}{2} = 406
   \]
   Therefore, the number of violating subsets is at most 406.

4. **Inequality**: Combining the above, we get:
   \[
   n - 29 \leq 406 \implies n \leq 435
   \]

5. **Claim**: There is such an example for \( n = 435 \).

6. **Construction**: Consider the set of pairs of elements of \( \{1, 2, \ldots, 30\} \). Create a three-element subset for every triple \( \{a, b\}, \{b, c\}, \{c, a\} \). These subsets satisfy the condition that any two of them have at most one common element.

7. **29-element nice subset**: The 29-element nice subset \( \{\{1, 2\}, \{1, 3\}, \ldots, \{1, 30\}\} \) cannot have any other pairs added without violating the condition. This is because any additional pair would form a three-element subset with two pairs already in the nice subset.

8. **Minimum value of \( n \)**: Hence, the minimum value of \( n \) is:
   \[
   \binom{30}{2} + 1 = 435 + 1 = 436
   \]

The final answer is \(\boxed{436}\)

Answer: 436


There are three villages $A$, $B$, and $C$ forming a triangle (as shown in Figure 5). The ratio of the number of primary school students in villages $A$, $B$, and $C$ is $1: 2: 3$. A primary school needs to be established. Where should the school be located to minimize the total distance $S$ traveled by the students to school?

Solution:
Solution: Let the primary school be located at point $P$, and the number of students from villages $A$, $B$, and $C$ be $a$, $2a$, and $3a$ respectively. Then,
$$
\begin{array}{l}
S=a P A+2 a P B+3 a P C \\
=a(P A+P C)+2 a(P B+P C) \\
\geqslant a A C+2 a B C .
\end{array}
$$

Equality holds if and only if $P=C$.
Therefore, setting the primary school in village $C$ can minimize the total distance $S$ traveled by the schoolchildren.

Answer: P=C


The popular gear-changing bicycles nowadays have several gears with different numbers of teeth installed on the drive shaft and the rear shaft. By connecting different combinations of gears with a chain, several different speeds are achieved through various gear ratios.
âHopeâ brand gear-changing bicycles have three gears on the drive shaft, with the number of teeth being $48,36,24$; the rear shaft has four gears, with the number of teeth being $36,24,16,12$. Question: How many different speeds does this type of gear-changing bicycle have?

Solution:
5.【Solution】Calculate all gear ratios and list them in a table:
\begin{tabular}{|c|cccc|}
\hline Main & 36 & 24 & 16 & 12 \\
\hline 48 & $\frac{4}{3}$ & 2 & 3 & 4 \\
\hline 36 & 1 & $\frac{3}{2}$ & $\frac{9}{4}$ & 3 \\
\hline 24 & $\frac{2}{3}$ & 1 & $\frac{3}{2}$ & 2 \\
\hline
\end{tabular}

There are 4 pairs of identical gear ratios: $1, \frac{3}{2}, 2, 3$. Removing the repeated gear ratios, 8 different ratios remain, so there are 8 different speeds in total.

Answer: 8


Given the sides of $\triangle ABC$ are $a, b, c$ and satisfy: $a+b \leqslant 2c, b+c \leqslant 3a$.

Then the range of $\frac{c}{a}$ is $\qquad$

Solution:
5. $\left(\frac{2}{3}, 2\right)$.

From the problem, we know that $a+b+c \leqslant 3 c, a+b+c \leqslant 4 a$.
Let $s=a+b+c=1$. Then
$c \geqslant \frac{1}{3}, a \geqslant \frac{1}{4}, a+c>\frac{1}{2}$.
If $c>a \Rightarrow \max \left\{\frac{1}{3}, a\right\} \leqslant c \leqslant \frac{1}{2}$,
If $c<a \Rightarrow \frac{1}{3} \leqslant c \leqslant \max \left\{\frac{1}{2}, a\right\}$,
If $c=a \Rightarrow \frac{c}{a}=1$.
In summary, $\frac{c}{a} \in\left(\frac{2}{3}, 2\right)$.

Answer: \left(\frac{2}{3}, 2\right)


Suppose $P(x)$ is a monic polynomial of degree $2023$ such that $P(k) = k^{2023}P(1-\frac{1}{k})$ for every positive integer $1 \leq k \leq 2023$. Then $P(-1) = \frac{a}{b}$ where $a$ and $b$ are relatively prime integers. Compute the unique integer $0 \leq n < 2027$ such that $bn-a$ is divisible by the prime $2027$.

Solution:
Given the monic polynomial \( P(x) \) of degree 2023 such that \( P(k) = k^{2023} P\left(1 - \frac{1}{k}\right) \) for every positive integer \( 1 \leq k \leq 2023 \), we need to find \( P(-1) \) in the form \( \frac{a}{b} \) where \( a \) and \( b \) are relatively prime integers. Then, we need to compute the unique integer \( 0 \leq n < 2027 \) such that \( bn - a \) is divisible by the prime 2027.

1. **Understanding the Functional Equation:**
   \[
   P(k) = k^{2023} P\left(1 - \frac{1}{k}\right)
   \]
   This implies that if \( k \) is a root of \( P(x) \), then \( 1 - \frac{1}{k} \) is also a root.

2. **Analyzing the Polynomial:**
   Since \( P(x) \) is monic and of degree 2023, we can write:
   \[
   P(x) = x^{2023} + a_{2022} x^{2022} + \cdots + a_1 x + a_0
   \]

3. **Using the Functional Equation:**
   Substitute \( k = 1 \):
   \[
   P(1) = 1^{2023} P(0) = P(0)
   \]
   This implies \( P(1) = P(0) \).

4. **Substitute \( k = 2 \):**
   \[
   P(2) = 2^{2023} P\left(1 - \frac{1}{2}\right) = 2^{2023} P\left(\frac{1}{2}\right)
   \]

5. **Substitute \( k = -1 \):**
   \[
   P(-1) = (-1)^{2023} P\left(1 - \frac{1}{-1}\right) = -P(2)
   \]
   Since \( P(-1) = -P(2) \), we need to find \( P(2) \).

6. **Finding \( P(2) \):**
   From the previous step:
   \[
   P(2) = 2^{2023} P\left(\frac{1}{2}\right)
   \]
   Substitute \( k = \frac{1}{2} \):
   \[
   P\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{2023} P\left(1 - 2\right) = \left(\frac{1}{2}\right)^{2023} P(-1)
   \]
   Therefore:
   \[
   P(2) = 2^{2023} \left(\frac{1}{2}\right)^{2023} P(-1) = P(-1)
   \]
   This implies:
   \[
   P(-1) = -P(2) = -P(-1)
   \]
   Hence:
   \[
   2P(-1) = 0 \implies P(-1) = 0
   \]

7. **Conclusion:**
   Since \( P(-1) = 0 \), we have \( a = 0 \) and \( b = 1 \). We need to find \( n \) such that \( bn - a \) is divisible by 2027:
   \[
   1n - 0 \equiv 0 \pmod{2027}
   \]
   Therefore:
   \[
   n \equiv 0 \pmod{2027}
   \]

The final answer is \( \boxed{0} \).

Answer: 0


How many divisors does the integer

$$
N=a^{\alpha} \cdot b^{\beta} \cdot c^{\gamma} \ldots l^{\lambda}
$$

have, where $a, b, c, \ldots l$ are prime numbers and $\alpha, \beta, \gamma, \ldots \lambda$ are any integers?

Solution:
50. Generalizing the method used in the answer to the previous question, we get for the total number of divisors the expression:

$$
(\alpha+1)(\beta+1)(\gamma+1) \cdots(\lambda+1)
$$

Answer: (\alpha+1)(\beta+1)(\gamma+1)\cdots(\lambda+1)


The sum of all four-digit natural numbers divisible by 45, where the tens digit is three times the hundreds digit, minus the sum of all three-digit natural numbers divisible by 18, where the hundreds digit is the largest one-digit prime number.

Solution:
2. The sum of all four-digit natural numbers divisible by 45, where the tens digit is three times the hundreds digit, decreased by the sum of all three-digit natural numbers divisible by 18, where the hundreds digit is the largest one-digit prime number.

## Solution.

A number is divisible by 45 if it is divisible by 9 and 5. A number is divisible by 9 if the sum of its digits is divisible by 9, and by 5 if its last digit is 0 or 5.

Since the tens digit is three times the hundreds digit, the possible combinations of these two digits are:

hundreds digit 0, tens digit 0

hundreds digit 1, tens digit 3

hundreds digit 2, tens digit 6

hundreds digit 3, tens digit 9

$1 \text{ POINT}$

The units digit of the desired four-digit numbers can be 0 or 5.

If the units digit is 0, then (due to the rule of divisibility by 9) we have the following solutions: $9000, 5130, 1260$, and 6390.

If the units digit is 5, then (due to the rule of divisibility by 9) we have the following solutions: 4005, 9135, 5265, and 1395.

The sum of all four-digit numbers that satisfy the desired property is:

$9000 + 5130 + 1260 + 6390 + 4005 + 9135 + 5265 + 1395 = 41580$.

A number is divisible by 18 if it is divisible by 9 and 2. A number is divisible by 9 if the sum of its digits is divisible by 9, and by 2 if its last digit is 0, 2, 4, 6, or 8.

Since the hundreds digit is the largest one-digit prime number, the hundreds digit is 7.

The sum of all three-digit numbers that satisfy the desired property is:

$720 + 702 + 792 + 774 + 756 + 738 = 4482$.

Note. If the student, when determining the four-digit numbers, failed to notice that the tens and hundreds digits can both be zero, i.e., omitted the numbers 9000 and 4005, the sum is 28575, the difference is 24093, and the task should be graded with a maximum of 7 POINTS.

Answer: 37098