Find the least positive integer $k$ such that each positive integer $n$ can be written as $n = a_1 \pm a_2 \pm a_3 \pm \cdots \pm a_k$ where $a_1, \dots , a_k$ are simple.

Solution:
To determine if \( k = 9 \) is the least positive integer such that each positive integer \( n \) can be written as \( n = a_1 \pm a_2 \pm a_3 \pm \cdots \pm a_k \) where \( a_1, \dots , a_k \) are simple, we need to explore the construction and verify if a smaller \( k \) can work.

1. **Understanding the Construction for \( k = 9 \)**:
   - Given a positive integer \( n \), we can represent it in its decimal form.
   - Construct nine simple integers \( s_1, s_2, \ldots, s_9 \) such that the \( k \)-th digit in \( s_i \) equals \( 1 \) if and only if the \( k \)-th digit of \( n \) is \( i \) or larger.
   - This ensures that each digit of \( n \) is represented by a combination of the digits of \( s_1, s_2, \ldots, s_9 \).

2. **Example for Clarity**:
   - Consider \( n = 432 \).
   - The simple integers \( s_1, s_2, \ldots, s_9 \) would be:
     \[
     s_1 = 111, \quad s_2 = 111, \quad s_3 = 111, \quad s_4 = 100, \quad s_5 = 000, \quad s_6 = 000, \quad s_7 = 000, \quad s_8 = 000, \quad s_9 = 000
     \]
   - Here, \( s_1, s_2, s_3 \) have all digits as 1 because the digits of \( n \) (4, 3, 2) are all greater than or equal to 1, 2, and 3 respectively. \( s_4 \) has a 1 in the first digit because the first digit of \( n \) is 4, which is greater than or equal to 4.

3. **Checking if \( k = 8 \) or Smaller Can Work**:
   - Suppose \( k = 8 \). We need to check if every positive integer \( n \) can be represented using 8 simple integers.
   - Consider a number \( n \) with a digit 9. To represent this digit, we need at least 9 simple integers because each simple integer can only contribute a 1 or 0 in each digit place.
   - Therefore, \( k = 8 \) is insufficient to represent a digit 9 in \( n \).

4. **Conclusion**:
   - Since \( k = 8 \) is insufficient and \( k = 9 \) works for any positive integer \( n \), \( k = 9 \) is indeed the least positive integer that satisfies the condition.

The final answer is \( \boxed{ k = 9 } \).

Answer: k = 9


Let $A=\{1,2\}$, then the number of mappings from $A$ to $A$ that satisfy $f[f(x)]=f(x)$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4

Solution:
-、1.C.
The functions are $f, g, h$, such that $f(1)=1, f(2)$ $=2 ; g(1)=g(2)=1 ; h(1)=h(2)=2$.

Answer: C


Vlado covered the diagonal of a large square with a side length of $2020 \mathrm{~cm}$ using a row of smaller squares with a side length of $4 \mathrm{~cm}$ cut from green collage paper. The diagonals of the green squares align with the diagonal of the large square, and the intersection of any two consecutive green squares is a square with a side length of 1 cm. Calculate the area of the shape formed by the green squares. Express the result in square decimeters and round to the nearest whole number.

## Result: $\quad 101$

Solution:
## Solution, First Method.

Let's calculate how many squares of side length $4 \mathrm{~cm}$ there are in total.

Start by drawing squares starting from the bottom left corner of the large square, which we will label as $A_{0}$. The top right corner of the first drawn square will be labeled as $A_{1}$, the top right corner of the second drawn square as $A_{2}$, and so on.

From vertex $\mathrm{A}_{0}$ to vertex $\mathrm{A}_{1}$, one can move $4 \mathrm{~cm}$ to the right and $4 \mathrm{~cm}$ up. To reach vertex $A_{2}$, one needs to move another $3 \mathrm{~cm}$ to the right and another $3 \mathrm{~cm}$ up, and to reach vertex $A_{3}$, another $3 \mathrm{~cm}$ to the right and another $3 \mathrm{~cm}$ up. Thus, after the first square, each subsequent square moves to a point that is $3 \mathrm{~cm}$ to the right and $3 \mathrm{~cm}$ up relative to the last point on the diagonal of the large square.

Since adding squares moves equally upwards and to the right, it is sufficient to look at the rightward movements. From the starting point, the sum of all movements must equal the length of the side of the large square, which is $2020 \mathrm{~cm}$. For the first square, the movement is $4 \mathrm{~cm}$, and for each subsequent square, it is $3 \mathrm{~cm}$.

Let $n$ be the total number of squares of side length $4 \mathrm{~cm}$ in this sequence.

Then we have $4 \mathrm{~cm} + (n-1) \cdot 3 \mathrm{~cm} = 2020 \mathrm{~cm}$.

This simplifies to $3n + 1 = 2020$, from which we get $n = 673$.

When calculating the area, we observe that the first square is complete, and all subsequent squares are squares of side length $4 \mathrm{~cm}$ with a square of side length $1 \mathrm{~cm}$ cut out. The area is $16 + 672 \cdot 15 = 10096 \mathrm{~cm}^{2} = 100.96 \mathrm{~dm}^{2}$.

Rounded to the nearest whole number, the area is approximately $101 \mathrm{~dm}^{2}$.

## Solution, Second Method.

From the area of the initial square with side length $2020 \mathrm{~cm}$, we will subtract the parts that are not green. We have two congruent parts, each consisting of rectangles of height $3 \mathrm{~cm}$ and lengths $3 \mathrm{~cm}, 6 \mathrm{~cm}, \ldots, 2016 \mathrm{~cm}$.

The sum of their areas in $\mathrm{cm}^{2}$ is

$$
\begin{gathered}
2 \cdot (3 \cdot 3 + 3 \cdot 6 + \cdots + 3 \cdot 2016) = 2 \cdot 3 \cdot 3 \cdot (1 + 2 + \cdots + 672) = 2 \cdot 3 \cdot 3 \cdot \frac{672 \cdot 673}{2} \\
= 2016 \cdot 2019
\end{gathered}
$$

Thus, the final result in $\mathrm{cm}^{2}$ is:

$$
\begin{aligned}
2020 \cdot 2020 - & 2016 \cdot 2019 = 2020 \cdot 2020 - (2020 - 4) \cdot (2020 - 1) \\
& = 2020 \cdot 2020 - 2020 \cdot 2020 + 4 \cdot 2020 + 1 \cdot 2020 - 4 \\
& = 5 \cdot 2020 - 4 \\
& = 10096
\end{aligned}
$$

The result is $100.96 \mathrm{~dm}^{2}$. Rounded to the nearest whole number, the area is approximately $101 \mathrm{~dm}^{2}$.

Answer: 101


Solution:
## Solution

Answer: $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,-1,1)$ or $(-1,-1,-1)$.

From the first equation $x=z-y-1$. Substituting in the second equation: $2 z^{2}-2 y z+2 y-2 z=$ 0 , so $(\mathrm{z}-1)(\mathrm{z}-\mathrm{y})=0$. Hence $\mathrm{z}=1$ or $\mathrm{y}=\mathrm{z}$. If $\mathrm{z}=1$, then from the first equation $\mathrm{x}+\mathrm{y}=0$, and hence from the last equation, $\mathrm{x}=1, \mathrm{y}=-1$. If $\mathrm{y}=\mathrm{z}$, then $\mathrm{x}=-1$, and hence from the last

equation $\mathrm{y}=\mathrm{z}=-1$.

Answer: (1,-1,1)or(-1,-1,-1)


A white ball is placed into an urn containing two balls, after which one ball is randomly drawn from it. Find the probability that the drawn ball will be white, if all possible assumptions about the initial composition of the balls (by color) are equally likely.

Solution:
Solution. Let $A$ be the event of drawing a white ball. The following hypotheses about the original composition of the balls are possible: $B_{1}$ - no white balls, $B_{2}$ - one white ball, $B_{3}$ - two white balls.

Since there are three hypotheses, and by condition they are equally probable, and the sum of the probabilities of the hypotheses is one (since they form a complete group of events), the probability of each hypothesis is $1 / 3$, i.e., $P\left(B_{1}\right)=P\left(B_{2}\right)=P\left(B_{3}\right)=1 / 3$.

The conditional probability that a white ball will be drawn, given that there were no white balls initially in the urn, $P_{B_{1}}(A)=1 / 3$.

The conditional probability that a white ball will be drawn, given that there was one white ball initially in the urn, $P_{B_{2}}(A)=2 / 3$.

The conditional probability that a white ball will be drawn, given that there were two white balls initially in the urn, $P_{B_{3}}(A)=3 / 3=1$.

The desired probability that a white ball will be drawn is found using the formula for total probability:

$$
\begin{gathered}
P(A)=P\left(B_{1}\right) \cdot P_{B_{1}}(A)+P\left(B_{2}\right) \cdot P_{B_{3}}(A) \dot{+} \\
+P\left(B_{3}\right) \cdot P_{B_{3}}(A)=1 / 3 \cdot 1 / 3+1 / 3 \cdot 2 / 3+1 / 3 \cdot 1=2 / 3
\end{gathered}
$$

Answer: \frac{2}{3}


Find the maximum value of $z+x$, if $(x, y, z, t)$ satisfies the conditions:

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=4 \\
z^{2}+t^{2}=9 \\
x t+y z \geq 6
\end{array}\right.
$$

Solution:
Solution I: From the conditions we have

$$
36=\left(x^{2}+y^{2}\right)\left(z^{2}+t^{2}\right)=(x t+y z)^{2}+(x z-y t)^{2} \geq 36+(x z-y t)^{2}
$$

and this implies $x z-y t=0$.

Now it is clear that

$$
x^{2}+z^{2}+y^{2}+t^{2}=(x+z)^{2}+(y-t)^{2}=13
$$

and the maximum value of $z+x$ is $\sqrt{13}$. It is achieved for $x=\frac{4}{\sqrt{13}}, y=t=\frac{6}{\sqrt{13}}$ and $z=\frac{9}{\sqrt{13}}$.

Answer: \sqrt{13}


Solution:
Answer:
1021
Note that the triangle is a right triangle with right angle at $A$. Therefore, $R^{2}=\frac{(7-2)^{2}+(11-6)^{2}}{4}=\frac{25}{2}=$ $(25)\left(2^{-1}\right) \equiv 1021(\bmod 2017)$. (An equivalent approach works for general triangles; the fact that the triangle is right simply makes the circumradius slightly easier to compute.)

Answer: 1021


There are 10 young men, each with a different weight and height; for any two young men $\mathbf{A}$ and $\mathbf{B}$, if $\mathbf{A}$ is heavier than $\mathbf{B}$, or $\mathbf{A}$ is taller than $\mathbf{B}$, then we say Ã¢ÂÂ$\mathrm{A}$ is not worse than BÃ¢ÂÂ; if a young man is not worse than the other 9 people, he is called a Ã¢ÂÂgreat guyÃ¢ÂÂ. Then, how many Ã¢ÂÂgreat guysÃ¢ÂÂ can there be at most among these 10 people.

Solution: 【Analysis】Let's assume the heights of 10 people are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$, and the weights of 10 people are $10, 9, 8, 7, 6, 5, 4, 3, 2, 1$. For any two people $\mathrm{A}$ and $\mathrm{B}$, either $\mathrm{A}$ is taller than $\mathrm{B}$ and $\mathrm{B}$ is heavier than $\mathrm{A}$, or $\mathrm{A}$ is heavier than $\mathrm{B}$ and $\mathrm{B}$ is taller than $\mathrm{A}$. This means that $\mathrm{A}$ is not worse than $\mathrm{B}$, and at the same time, $\mathrm{B}$ is not worse than $\mathrm{A}$. Therefore, all 10 people can be considered good guys, so the maximum number of good guys can be 10.

Answer: 10


Let $n$ be the product 3659893456789325678 and 342973489379256. Determine the number of digits of $n$.

Solution:
$$
\begin{array}{l}
\text { Let } x=3659893456789325678, y=342973489379256 \\
x=3.7 \times 10^{18}, y=3.4 \times 10^{14} \text { (correct to } 2 \text { sig. fig.) } \\
n=x y=3.7 \times 10^{18} \times 3.4 \times 10^{14}=12.58 \times 10^{32}=1.258 \times 10^{33}
\end{array}
$$

The number of digits of $n$ is 34 .

Answer: 34


Calculate the limit of the function:

$$
\lim _{x \rightarrow 1}\left(1+e^{x}\right)^{\frac{\sin \pi x}{1-x}}
$$

Solution:
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 1}\left(1+e^{x}\right)^{\frac{\sin \pi x}{1-x}}=\left(\lim _{x \rightarrow 1} 1+e^{x}\right)^{\lim _{x \rightarrow 1} \frac{\sin \pi x}{1-x}}= \\
& =\left(1+e^{1}\right)^{\lim _{x \rightarrow 1} \frac{\sin \pi x}{1-x}}=
\end{aligned}
$$

$x=y+1 \Rightarrow y=x-1$

$x \rightarrow 1 \Rightarrow y \rightarrow 0$

We get:

$=(1+e)^{\lim _{y \rightarrow 0} \frac{\sin \pi(y+1)}{1-(y+1)}}=\left(1+e e^{\lim _{y \rightarrow 0} \frac{\sin (\pi y+\pi)}{-y}}=\right.$
$=(1+e)^{\lim _{y \rightarrow 0} \frac{-\sin \pi y}{-y}}=(1+e)^{\lim _{y \rightarrow 0} \frac{\sin \pi y}{y}}=$

Using the substitution of equivalent infinitesimals:

$\sin \pi y \sim \pi y \text{ as } y \rightarrow 0 (\pi y \rightarrow 0)$

We get:

$$
=(1+e)^{\lim _{y \rightarrow 0} \frac{\pi y}{y}}=(1+e)^{\lim _{y \rightarrow 0} \pi}=(1+e)^{\pi}
$$

## Kuznetsov Limits 20-9

Answer: (1+e)^{\pi}


We know: $9=3 \times 3, 16=4 \times 4$, here, $9, 16$ are called "perfect squares". Among the first 300 natural numbers, if we remove all the "perfect squares", what is the sum of the remaining natural numbers?

Solution:
5.【Solution】The squares not exceeding 300 are:
$1,4,9,16,25,36,49,64,81,100,121,144,169,196,225$,
256, 289, and their sum is 1785
The sum of the first 300 natural numbers is: $1+2+3+\cdots+300=\frac{1+300}{2} \times 300=45150$,
Thus, the sum of the remaining natural numbers is $45150-1785=43365$

Answer: 43365


Calculate the limit of the function:

$\lim _{x \rightarrow \pi} \frac{1+\cos 3 x}{\sin ^{2} 7 x}$

Solution:
## Solution

Substitution:

$x=y+\pi \Rightarrow y=x-\pi$

$x \rightarrow \pi \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow \pi} \frac{1+\cos 3 x}{\sin ^{2} 7 x}=\lim _{y \rightarrow 0} \frac{1+\cos 3(y+\pi)}{\sin ^{2} 7(y+\pi)}= \\
& =\lim _{y \rightarrow 0} \frac{1+\cos (3 y+3 \pi)}{\sin ^{2}(7 y+7 \pi)}=\lim _{y \rightarrow 0} \frac{1+\cos (3 y+\pi)}{\sin ^{2}(7 y+\pi)}= \\
& =\lim _{y \rightarrow 0} \frac{1-\cos 3 y}{(-\sin 7 y)^{2}}=\lim _{y \rightarrow 0} \frac{1-\cos 3 y}{\sin ^{2} 7 y}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\sin 7 y \sim 7 y$, as $y \rightarrow 0(7 y \rightarrow 0)$

$1-\cos 3 y \sim \frac{(3 y)^{2}}{2}$, as $y \rightarrow 0$

We get:

$=\lim _{y \rightarrow 0} \frac{\frac{(3 y)^{2}}{2}}{(7 y)^{2}}=\lim _{y \rightarrow 0} \frac{9 y^{2}}{2 \cdot 49 y^{2}}=\lim _{y \rightarrow 0} \frac{9}{2 \cdot 49}=\frac{9}{98}$

## Problem Kuznetsov Limits 13-3

Answer: \frac{9}{98}


Find the number of matrices that satisfy two conditions:
1) the matrix has the form $\left(\begin{array}{lll}1 & * & * \\ * & 1 & * \\ * & * & 1\end{array}\right)$, where each * can take the value 0 or 1
2) the rows of the matrix do not repeat.

Solution:
Solution: Let $A$ be the set of matrices satisfying condition 1) and $B$ be the subset of $A$ consisting of matrices satisfying condition 2). We need to find the number of elements in the set $B$. Let $A_{ij}, i, j \in\{1,2,3\}, i \neq j$, be the subset of $A$ consisting of matrices in which row $i$ and $j$ are the same. Then $B=A \backslash\left(A_{12} \cup A_{23} \cup A_{13}\right)$ and $|B|=|A|-\left|A_{12} \cup A_{23} \cup A_{13}\right|$. The cardinality $\left|A_{12} \cup A_{23} \cup A_{13}\right|$ is conveniently calculated using the inclusion-exclusion principle:

$\left|A_{12} \cup A_{23} \cup A_{13}\right|=\left|A_{12}\right|+\left|A_{23}\right|+\left|A_{13}\right|-\left|A_{12} \cap A_{23}\right|-\left|A_{13} \cap A_{23}\right|-\left|A_{12} \cap A_{13}\right|+\left|A_{12} \cap A_{23} \cap A_{13}\right|$.

It is easy to calculate the cardinalities of the sets involved in this expression:

$$
\left|A_{12}\right|=\left|A_{23}\right|=\left|A_{13}\right|=2^{3},\left|A_{12} \cap A_{23}\right|=\left|A_{13} \cap A_{23}\right|=\left|A_{12} \cap A_{13}\right|=\left|A_{12} \cap A_{23} \cap A_{13}\right|=1 .
$$

We get $|B|=2^{6}-3 \cdot 2^{3}+3-1=42$.

Answer: 42

Answer: 42


Let $a_{1}, a_{2}, \cdots$ and $b_{1}, b_{2}, \cdots$ be two arithmetic sequences, with their sums of the first $n$ terms being $A_{n}$ and $B_{n}$, respectively. It is known that for all $n \in \mathbf{N}$,
$$
\frac{A_{n}}{B_{n}}=\frac{2 n-1}{3 n+1} \text {. }
$$

Try to write the expression for $\frac{a_{n}}{b_{n}}$ for all $n \in \mathbf{N}$.

Solution:
7. Let the common difference of $a_{1}, a_{2}, \cdots$ be $d_{1}$; the common difference of $b_{1}, b_{2}, \cdots$ be $d_{2}$. Then
$$
\begin{aligned}
A_{n} & =n a_{1}+\frac{n(n-1)}{2} d_{1}, B_{n}=n b_{1}+\frac{n(n-1)}{2} d_{2}, \\
\frac{A_{n}}{B_{n}} & =\frac{2 a_{1}+(n-1) d_{1}}{2 b_{1}+(n-1) d_{2}} .
\end{aligned}
$$

From the given conditions, for all $n \in \mathbf{N}$, we have
$$
\frac{2 a_{1}+(n-1) d_{1}}{2 b_{1}+(n-1) d_{2}}=\frac{2 n-1}{3 n+1} \text {. }
$$

In (1), let $n=1$ to get
$$
\frac{a_{1}}{b_{1}}=\frac{1}{4}, b_{1}=4 a_{1} \text {. }
$$

In (1), let $n \rightarrow+\infty$ to get $\frac{d_{1}}{d_{2}}=\frac{2}{3}, d_{2}=\frac{3}{2} d_{1}$.
In (1), let $n=2$ to get $\frac{2 a_{1}+d_{1}}{2 b_{1}+d_{2}}=\frac{3}{7}$.
Using (2) and (3), the above equation becomes
$$
\frac{2 a_{1}+d_{1}}{8 a_{1}+\frac{3}{2} d_{1}}=\frac{3}{7} \text {, simplifying to } d_{1}=4 a_{1} \text {. }
$$

Thus, $d_{2}=\frac{3}{2} d_{1}=6 a_{1}$,
so $\frac{a_{n}}{b_{n}}=\frac{a_{1}+(n-1) d_{1}}{b_{1}+(n-1) d_{2}}=\frac{a_{1}+(n-1) \cdot 4 a_{1}}{4 a_{1}+(n-1) \cdot 6 a_{1}}$
$$
=\frac{1+4(n-1)}{4+6(n-1)}=\frac{4 n-3}{6 n-2} .
$$

Answer: \frac{4n-3}{6n-2}


Find the general solution of the equation

$$
y^{\prime \prime}+6 y^{\prime}+9 y=14 e^{-3 x} .
$$

Solution:
Solution. 1. Find $\bar{y}$.

The characteristic equation $r^{2}+6 r+9=0$, has roots $r_{1}=r_{2}=-3$. Therefore,

$$
\bar{y}=\left(C_{1}+C_{2} x\right) e^{-3 x}
$$

2. Now find $y^{*}$. Here the right-hand side has the form (3): $n=0$, $P_{0}=14, k=-3$. Since $k=-3$ is a double root of the characteristic equation, the particular solution $y^{*}$ should be sought in the form

$$
y^{*}=A x^{2} e^{-3 x}
$$

where $A$ is a coefficient to be determined. Calculate the derivatives $y^{* \prime}$ and $y^{* \prime \prime}:$

$$
\begin{gathered}
y^{*}=\left(-3 A x^{2}+2 A x\right) e^{-3 x}, \\
y^{*}=\left(9 A x^{2}-12 A x+2 A\right) e^{-3 x} .
\end{gathered}
$$

Substituting the expressions for $y^{*}, y^{*^{\prime}}$ and $y^{* \prime \prime}$ into the given equation, canceling $e^{-3 x}$ from both sides, and combining like terms, we ultimately get $2 A=14$, from which $A=7$. Therefore, the particular solution is:

$$
y^{*}=7 x^{2} e^{-3 x}
$$

Thus, the general solution of the given equation is

$$
y=\bar{y}+y^{*}=\left(C_{1}+C_{2} x\right) e^{-3 x}+7 x^{2} e^{-3 x}
$$

Answer: (C_{1}+C_{2}x)e^{-3x}+7x^{2}e^{-3x}


$\triangle A B C$ has three sides $a, b, c$ satisfying $1 \leqslant a \leqslant 3 \leqslant b \leqslant 5 \leqslant c \leqslant 7$.
Find the perimeter of $\triangle A B C$ when its area is maximized.

Solution:
Solution: Note that
$$
S_{\triangle A B C}=\frac{1}{2} a b \sin C \leqslant \frac{1}{2} a b \leqslant \frac{1}{2} \times 3 \times 5,
$$

equality holds if and only if $a=3, b=5, \angle C=90^{\circ}$. At this time, $c=\sqrt{a^{2}+b^{2}}=\sqrt{34}$ satisfies $5 \leqslant c \leqslant 7$.

Therefore, when $a=3, b=5, c=\sqrt{34}$, the area of $\triangle A B C$ is maximized. At this time, its perimeter is $8+\sqrt{34}$.

Answer: 8+\sqrt{34}


In a new school $40$ percent of the students are freshmen, $30$ percent are sophomores, $20$ percent are juniors, and $10$ percent are seniors. All freshmen are required to take Latin, and $80$ percent of the sophomores, $50$ percent of the juniors, and $20$ percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution: We see that $40\% \cdot 100\% + 30\% \cdot 80\% + 20\% \cdot 50\% + 10\% \cdot 20\% = 76\%$ of students are learning Latin. In addition, $30\% \cdot 80\% = 24\%$ of students are sophomores learning Latin. Thus, our desired probability is $\dfrac{24}{76}=\dfrac{6}{19}$ and our answer is $6+19=\boxed{025}$.

Answer: 25


Point $C$ divides the diameter $A B$ in the ratio $A C: B C=2: 1$. A point $P$ is chosen on the circle. Determine the values that the ratio $\operatorname{tg} \angle P A C: \operatorname{tg} \angle A P C$ can take. In your answer, specify the smallest such value.

Solution:
Answer: $\frac{1}{2}$.

Solution: The angle $\angle A P B=90^{\circ}$, as it subtends the diameter. Drop a perpendicular $C H$ to $A P$, it will be parallel to $P B$. Therefore, $\triangle A P B \propto \triangle A H C$ with a ratio of $\frac{2}{3}$. Hence, $\operatorname{tg} \angle A P C=$ $\frac{H C}{P C}=2 \frac{P B}{A P}=2 \operatorname{tg} \angle P A C$

Answer: \frac{1}{2}


Please write the result of the following calculation as a mixed number: $\frac{0.5 \times 236 \times 59}{119}$

Solution: 1.【Solution】Original expression $\frac{118 \times 59}{119}=\frac{(119-1) \times 59}{119}=59-\frac{59}{119}=58 \frac{60}{119}$

Answer: 58\frac{60}{119}


At least how many groups do we need to divide the first 100 positive integers into so that no group contains two numbers where one is a multiple of the other?

Solution:
Among two powers of 2, the smaller one is a divisor of the larger one, which means that out of the seven powers of 2 less than 100, no two can be placed in the same group. This implies that at least 7 groups are needed.

However, this is sufficient if the numbers $n$ that fall into the group of $2^{k}$ are such that $2^{k} \leqq n<2^{k+1}(0 \leqq k \leqq 6)$. The ratio of any two such numbers—when the larger is divided by the smaller—lies between 1 and 2, and thus cannot be an integer.

Remarks. 1. The above is not the only possible partition. If the numbers that have exactly $i$ prime factors in their prime factorization are placed in the $i$-th group, and 1 is placed in a separate group, we will get exactly 7 suitable groups. The smallest product of 7 factors is $2^{7}$, which is greater than 100, and a number has fewer prime factors in its proper divisors than in the number itself, so indeed no divisibility can hold among the numbers in the same group.

2. By a similar line of reasoning, it can be shown that the first $N$ positive integers can be placed into $1+\left[\log _{2} N\right]$ groups in the required manner, but not into fewer groups.

Answer: 7


One of the angles of the trapezoid is $30^{\circ}$, and the lines containing the lateral sides of the trapezoid intersect at a right angle (Fig. 10.42). Find the length of the shorter lateral side of the trapezoid if its midline is 10 cm, and one of the bases is $8 \mathrm{~cm}$.

Solution:
## Solution.

Given $\angle B C A=30^{\circ}, \angle A B C=90^{\circ}, K M=10$ cm, $D E=8$ cm, $\frac{8+A C}{2}=10, A C=12$ cm (since $K M$ is the midline). Since $\triangle D B E \sim \triangle A B C$, then $\frac{A B}{D B}=\frac{A C}{D E} ; \frac{x+D B}{D B}=\frac{12}{8}$ (where $A D=x$ ), $x=D B \cdot \frac{1}{2}$. In $\triangle D B E, \angle D B E=90^{\circ}, \angle B E D=\angle B C A=30^{\circ}, D B=D E \cdot \sin 30^{\circ}=8 / 2=4$, so, $x=4 \cdot \frac{1}{2}=2$ cm.

Answer: 2 cm.

Answer: 2


The sequence of Fibonacci numbers $F_{1}, F_{2}, \ldots, F_{n}, \ldots$ is such that $F_{1}=F_{2}=1$ and that the n-th term (with $n \geq 3$) is the sum of the two preceding ones (the first terms of the sequence are therefore $F_{1}=1, F_{2}=1, F_{3}=2=1+1, F_{4}=3=2+1, F_{5}=5=3+2$). How many Fibonacci numbers have exactly 2016 digits in their decimal representation?
(A) At least 2 and at most 3
(B) At least 4 and at most 5
(D) At least 8 and at most 9
(E) 10 or more
(C) At least 6 and at most 7

Solution:
9. The answer is $\mathbf{( B )}$. First, observe that (for $n \geq 2$) the relation $F_{n+1}=F_{n}+F_{n-1}$, together with the fact that $F_{n-1} \leq F_{n}$, implies that $F_{n}$ is at least half of $F_{n+1}$. Now let $k \geq 1$ and $F_{n}$ be the smallest Fibonacci number with $k+1$ decimal digits (so $F_{n} \geq 10^{k}$). We have $F_{n-1} \geq \frac{1}{2} \cdot 10^{k}$, and thus $F_{n+1}=F_{n}+F_{n-1} \geq\left(1+\frac{1}{2}\right) \cdot 10^{k}$. Proceeding in the same way, we obtain the inequalities $F_{n+2} \geq\left(\frac{3}{2}+1\right) \cdot 10^{k}, F_{n+3} \geq\left(\frac{5}{2}+\frac{3}{2}\right) 10^{k}, F_{n+4} \geq\left(4+\frac{5}{2}\right) \cdot 10^{k}$, and $F_{n+5} \geq\left(\frac{13}{2}+4\right) \cdot 10^{k}>10^{k+1}$, from which we see that $F_{n+5}$ has at least $k+2$ decimal digits.

On the other hand, by definition we have $F_{n-2} \leq F_{n-1}<10^{k}$, from which we successively obtain $F_{n}<2 \cdot 10^{k}, F_{n+1}<3 \cdot 10^{k}, F_{n+2}<5 \cdot 10^{k}, F_{n+3}<8 \cdot 10^{k}$, so $F_{n+3}$ still has $k+1$ decimal digits. In conclusion, for any number $k \geq 2$, there are at least 4 and at most 5 Fibonacci numbers with $k$ digits.

Answer: B


Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms $S_{n}$ related to $a_{n}$ by $S_{n}=-b a_{n}+1-\frac{1}{(1+b)^{n}}$, where $b$ is a constant independent of $n$, and $b \neq-1$.
(1) Find the expression for $a_{n}$ in terms of $b$ and $n$;
(2) When $b=3$, from which term does the sequence $\left\{a_{n}\right\}$ satisfy $S_{n}-a_{n}>\frac{1}{2}$?

Solution:
11. Analysis (1) $S_{n+1}=-b a_{n+1}+1-\frac{1}{(1+b)^{n+1}}, \quad S_{n}=-b a_{n}+1-\frac{1}{(1+b)^{n}}$ thus $a_{n+1}=-b a_{n+1}+b a_{n}+\frac{b}{(1+b)^{n+1}}$, which means $a_{n+1}=\frac{b}{1+b} a_{n}+\frac{b}{(1+b)^{n+2}}$; let $c_{n}=a_{n}(1+b)^{n+1}$, we get: $c_{n+1}=b c_{n}+b$ If $b=1$, then $a_{n}=\frac{n}{2^{n+1}}$; If $b \neq 1$, then $a_{n}=\frac{b\left(1-b^{n}\right)}{\left(1-b^{2}\right)(1+b)^{n}}$.
(2) When $b=3$, then $a_{n}=\frac{3\left(3^{n}-1\right)}{8 \times 4^{n}}$;

Since $S_{n}-a_{n}=-4 a_{n}+1-\frac{1}{4^{n}}=\frac{1-3^{n+1}}{2 \times 4^{n}}+1$, it is sufficient to have $4^{n}+1>3^{n+1} \quad(*)$.
Also, when $n \geq 4$, $4^{n}=(1+3)^{n} \geq 2+2 n 3^{n-1}+\frac{n(n-1)}{2} 3^{n-2}=2+\frac{n^{2}+11 n}{2} 3^{n-2}>3^{n+1}$, so we only need to consider the cases $n=1,2,3$. Upon inspection, $n=1,2,3$ do not satisfy (*), hence from the 4th term onwards, $S_{n}-a_{n}>\frac{1}{2}$.

Answer: 4


Every day, Xiaoming has to pass through a flat section $A B$, an uphill section $B C$, and a downhill section $C D$ (as shown in the right figure). It is known that $A B: B C: C D=1: 2: 1$, and the speed ratio of Xiaoming on the flat, uphill, and downhill sections is $3: 2: 4$. If the time ratio for Xiaoming to go to school and return home is $\frac{n}{m}$ (where $\mathrm{m}$ and $\mathrm{n}$ are coprime natural numbers), then the value of $\mathrm{m}+\mathrm{n}$ is $\qquad$

Solution:
Answer: 35.
The simple assignment method can calculate the result. Let $\mathrm{AB}$ be 100 meters, then the distance of $\mathrm{BC}$ is 200 meters, and the distance of $\mathrm{CD}$ is 100 meters; assume Xiao Ming's speed on flat ground is 3 meters/second, then his speed on uphill and downhill sections is 2 meters/second and 4 meters/second, respectively. According to the assignment,
the time for Xiao Ming to go to school is: $\frac{100}{3}+\frac{200}{2}+\frac{100}{4}=\frac{475}{3}$
the time for Xiao Ming to return home from school is: $\frac{100}{3}+\frac{200}{4}+\frac{100}{2}=\frac{400}{3}$
Therefore, the time ratio is $\frac{475}{3}: \frac{400}{3}=\frac{19}{16}!$
If you feel that the assignment method is not rigorous enough, then you can set $A B=s$, then $B C=2 s, C D=s$; set $v_{\text {flat }}=3 a$, then $v_{\text {up }}=2 a, v_{\text {down }}=4 a$. You can calculate these expressions to get:
$$
\begin{array}{l}
t_{\text {to school }}=\frac{s}{3 a}+\frac{2 s}{2 a}+\frac{s}{4 a}=\frac{4+12+3}{12} \times \frac{s}{a}=\frac{19}{12} \times \frac{s}{a}, \\
t_{\text {return home }}=\frac{s}{3 a}+\frac{2 s}{4 a}+\frac{s}{2 a}=\frac{4+6+6}{12} \times \frac{s}{a}=\frac{16}{12} \times \frac{s}{a},
\end{array}
$$

Therefore,
$$
\frac{t_{\text {to school }}}{t_{\text {return home }}}=\frac{19}{16}=\frac{n}{m}, \mathrm{~m}+\mathrm{n}=35 \text { . }
$$

Answer: 35


Determine three complex numbers with a modulus of 1 such that both their sum and their product are equal to 1.

Solution:
Solution. Let the required numbers be $z_{1}, z_{2}, z_{3}$. Then $z_{1}+z_{2}+z_{3}=1=z_{1} z_{2} z_{3}$ and

$$
z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}=\frac{1}{z_{3}}+\frac{1}{z_{1}}+\frac{1}{z_{2}}=\frac{\overline{z_{3}}}{\left|z_{3}\right|^{2}}+\frac{\overline{z_{1}}}{\left|z_{1}\right|^{2}}+\frac{\overline{z_{2}}}{\left|z_{2}\right|^{2}}=\overline{z_{1}+z_{2}+z_{3}}=1
$$

According to this, from Vieta's formulas, it follows that the required numbers are solutions to the equation

$$
z^{3}-z^{2}+z-1=0
$$

i.e., these are the numbers $1, i$ and $-i$.

Answer: 1,i,-i


Points $M$ and $N$ are located on the lateral sides $A B$ and $C D$ of trapezoid $A B C D$, respectively, such that $M N \| A D$. It is known that the area of trapezoid $M B C N$ is to the area of trapezoid $A M N D$ as $2: 3$. Find $M N$, if $B C = a, A D = b$.

Solution:
With the help of additional constructions, obtain similar triangles.

## Solution

The first method. Let $P$ be the intersection point of $MN$ with the line passing through point $C$ parallel to $AB$, and $Q$ be the intersection point of $AD$ with the line passing through point $N$ parallel to $AB$. Denote $MN = x; h_1$ and $h_2$ as the heights of the similar triangles $PCN$ and $QND$ (see the left figure).

Let $b > a$. The ratio of the areas of trapezoids $BMNC$ and $MADN$ is $2:3$, so $3(x + a)h_1 = 2(b + x)h_2$, from which $\frac{h_1}{h_2} = \frac{2}{3} \cdot \frac{b + x}{x + a}$.

From the similarity of triangles $CPN$ and $NQD$, it follows that $\frac{h_1}{h_2} = \frac{x - a}{b - x}$. Therefore, $\frac{2}{3} \cdot \frac{b + x}{x + a} = \frac{x - a}{b - x}$. From this, $x = \sqrt{\frac{3a^2 + 2b^2}{5}}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_60e93adedf73503f106cg-29.jpg?height=786&width=1758&top_left_y=1&top_left_x=161)

The second method. Let $O$ be the intersection point of the extensions of the lateral sides $AB$ and $DC$ (see the right figure), $S$ the area of triangle $BOC$.

$MN = x$ is the desired segment. Then $3(S_{MNO} - S) = 2(S_{AOD} - S_{MNO})$, or $3\left(\frac{x^2}{a^2} - 1\right) = 2\left(\frac{b^2}{a^2} - \frac{x^2}{a^2}\right)$.

From this, $x^2 = \frac{1}{2}(3a^2 + 2b^2)$.

## Answer

$\sqrt{\frac{3a^2 + 2b^2}{5}}$

Submit a comment

Answer: \sqrt{\frac{3a^2+2b^2}{5}}


Xiaohong went to the stationery store and bought 4 pencils and 5 notebooks, spending a total of 15 yuan 8 jiao. Xiaoliang bought the same 4 pencils and 7 notebooks, spending a total of 21 yuan 8 jiao. What is the price of 1 notebook?

Solution:
【Answer】Solution: $(21.8-15.8) \div(7-5)$
$$
\begin{array}{l}
=6 \div 2 \\
=3 \text { (yuan) }
\end{array}
$$

Answer: Each exercise book costs 3 yuan.

Answer: 3


Let $S=\{1,2,3, \cdots, 98\}$, find the smallest natural number $n$, such that in any $n$-element subset of $S$, one can always select 10 numbers, and no matter how these 10 numbers are divided into two groups, there is always a number in one group that is coprime with the other four numbers, and a number in the other group that is not coprime with the other four numbers.

Solution:
11. Proof: Take the set $E=\{2 x \mid x \in \mathbf{N}$, and $1 \leqslant x \leqslant 49\}$ of 49 even numbers from $S$. Then, for any $k(k \geqslant 10)$ numbers taken from $E$, it is impossible to select 10 numbers that meet the requirements. Therefore, $n \geqslant 50$.
For any odd number $x$ in $S$, the number of even numbers in $S$ that are not coprime with $x$ is $f(x) \leqslant\left[\frac{49}{3}\right]=16$.
Let $T$ be any 50-element subset of $S$, with $\alpha$ even numbers and $50-\alpha=\beta$ odd numbers. Since $|E|=49$, $T$ must contain at least one odd number.
The set of numbers in $S$ whose smallest prime factor is $k$ (where $k$ is a prime number) is denoted as $R_{k}$, then
$$
\begin{array}{l}
R_{3}=\{3 \times 1,3 \times 3, \cdots, 3 \times 31\},\left|P_{3}\right|=16 ; \\
R_{5}=\{5 \times 1,5 \times 5,5 \times 7,5 \times 11,5 \times 13,5 \times 15,5 \times 19\},\left|P_{5}\right|=7, \\
R_{7}=\{7 \times 1,7 \times 7,7 \times 11,7 \times 13\},\left|P_{7}\right|=4,
\end{array}
$$
$\left|R_{k}\right|=1$ (where $k$ is a prime number greater than 10).
If there exists an odd number $x$ in $T$ such that the number of even numbers in $T$ that are coprime with $x$ is $\geqslant 9$, then $T$ meets the requirements of the problem.
(1) If $\alpha \geqslant 25$,

then since $f(x) \leqslant 16$, for any odd number $a$ in $T$, the number of even numbers in $T$ that are coprime with $a$ is $\geqslant 25-16=9$. The problem is proved.
(2) If $16 \leqslant \alpha \leqslant 24$, then $\beta \geqslant 26$.

Since $\left|\{1\} \cup R_{3} \cup R_{5}\right|=1+16+7=24<\beta$, there must be at least one odd number $a \in R_{7} \cup R_{k}$ in $T$, and $f(a) \leqslant f(7)=\left[\frac{49}{7}\right]=7$. Thus, the number of even numbers in $T$ that are coprime with $a$ is $\geqslant 16-7=9$. The problem is proved.
(3) If $10 \leqslant \alpha \leqslant 15$, then $\beta \geqslant 35$.
Since $\left|\{1\} \cup R_{3} \cup R_{5} \cup R_{7} \cup\{11,13,17,19,23,29\}\right|=34<\beta$, there must be at least one odd number $a \geqslant 31$ in $T$, and $f(a)=1$. Thus, the number of even numbers in $T$ that are coprime with $a$ is $\geqslant 10-1=9$. The problem is proved.
(4) If $\alpha=9$, then $\beta=41$,
Since $\left|\{1\} \cup R_{3} \cup R_{5} \cup R_{7} \cup\{11,13,17,19,23,29,31,37,41,43,47\}\right|=39<\beta$, there must be at least one prime number $a \geqslant 53$ in $T$, and all 9 even numbers in $T$ are coprime with $a$. The problem is proved.
(5) If $\alpha \leqslant 8, \beta \geqslant 42$,
Since $\left|\{1\} \cup R_{3} \cup R_{5} \cup R_{7} \cup\{11,13,17,19,23,29,31,37,41,43,47\}\right|=39<\beta$, there must be at least one prime number $a \geqslant 53$ in $T$, and $a$ is coprime with all other numbers in $T$. The number of odd numbers in $S$ not in $T$ is $50-\beta \leqslant 7$. Taking all the odd numbers of the form $6k-3$ in $T$, there are at least $16-7=9$ such numbers, and they are not coprime. The problem is proved.
In summary, the proposition is true.

Answer: 50


Find all solutions to the following system of equations:

$$
\left\{\begin{aligned}
x+y+z+w & =10 \\
x^{2}+y^{2}+z^{2}+w^{2} & =30 \\
x^{3}+y^{3}+z^{3}+w^{3} & =100 \\
x y z w & =24
\end{aligned}\right.
$$

Solution:
3. By direct verification, we confirm that the set of numbers $(1,2,3,4)$ satisfies the first and fourth, as well as the second and third equations. Since all equations in the system are symmetric with respect to $x, y, z, w$, the other 23 permutations of the numbers $1,2,3,4$ are also solutions to the system. However, the product of the degrees of all four equations is 4!, so there are no other solutions to our system*.

$[$ [

Answer: (1,2,3,4)


Calculate the volumes of the bodies bounded by the surfaces.

$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}+\frac{z^{2}}{16}=1, z=2, z=0
$$

Solution:
## Solution

In the section of the given figure by the plane $z=$ const, there is an ellipse:

$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}=\frac{16-z^{2}}{16}
$$

The area of an ellipse with radii $a$ and $b$ is $\pi \cdot a \cdot b$

By the definition of the radius of an ellipse:

$$
\begin{aligned}
& \frac{x^{2}}{16 \cdot \frac{16-z^{2}}{16}}+\frac{y^{2}}{9 \cdot \frac{16-z^{2}}{16}}=1 \rightarrow a=\sqrt{16-z^{2}} ; b=\frac{3}{4} \sqrt{16-z^{2}} \\
& \Rightarrow S=\pi a b=\pi \sqrt{16-z^{2}} \cdot \frac{3}{4} \sqrt{16-z^{2}}=\frac{3 \pi}{4} \cdot\left(16-z^{2}\right) \\
& V=\int_{0}^{2} S(z) d z=\frac{3 \pi}{4} \int_{0}^{2}\left(16-z^{2}\right) d z=\left.\frac{3 \pi}{4}\left(16 z-\frac{z^{3}}{3}\right)\right|_{0} ^{2}= \\
& =\frac{3 \pi}{4}\left(16 \cdot 2-\frac{2^{3}}{3}-16 \cdot 0+\frac{0^{3}}{3}\right)=\frac{\pi}{4}\left(3 \cdot 32-3 \cdot \frac{8}{3}\right)=\frac{(96-8) \cdot \pi}{4}=22 \pi
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-10$ »

Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals

Ukrainian Banner Network

- Last modified on this page: 11:43, 22 June 2010.
- Content is available under CC-BY-SA 3.0.

Answer: 22\pi


A river flows at a uniform speed, with docks A and B located upstream and downstream, respectively, 200 kilometers apart. Boats A and B depart from docks A and B simultaneously and head towards each other. After meeting, they continue to their respective destinations, immediately turn around, and meet again on their return journey. If the interval between the two meetings is 4 hours, and the still water speeds of boats A and B are $36 \mathrm{~km} / \mathrm{h}$ and $64 \mathrm{~km} / \mathrm{h}$, respectively, then the speed of the current is $\qquad$ $\mathrm{km} / \mathrm{h}$

Solution:
【Answer】 14
【Analysis】 In the context of boats moving in a stream, the time of meeting is independent of the water speed. The first meeting takes $200 \div$ $(36+64)=2$ (hours), and the second meeting takes $2+4=6$ (hours). Together, they cover 3 full distances, which takes $2 \times 3=6$ (hours), which is exactly the same. This indicates that both boats turn around at the same time. Therefore, the downstream speed of $\mathrm{A}$ equals the upstream speed of $B$: $(64-36) \div 2=14(\mathrm{~km} / \mathrm{h})$

Answer: 14


A right-angled trapezoid with area 10 and altitude 4 is divided into two circumscribed trapezoids by a line parallel to its bases. Find their inradii.

Solution:
1. Let $A B C D$ be a right-angled trapezoid with area 10 and altitude $A D=4$. Let the line $M N \| A B, M \in A D, N \in B C$, divides it into two circumscribed trapezoids. Set $A B=a, C D=b$ $(a>b), M N=c, A M=h_{1}$ and $D M=h_{2}$. Then $h_{1}^{2}+(a-c)^{2}=\left(a+c-h_{1}\right)^{2}$ and we get that $h_{1}=\frac{2 a c}{a+c}$.

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-072.jpg?height=427&width=418&top_left_y=376&top_left_x=1204)

Analogously $h_{2}=\frac{2 b c}{b+c}$ which implies that $\frac{h_{1}}{h_{2}}=\frac{a(b+c)}{b(a+c)}$. On the other hand, $\frac{h_{1}}{h_{2}}=\frac{a-c}{c-b}$ and therefore $(a+b)\left(a b-c^{2}\right)=0$. Hence $c^{2}=a b$ and we get that

$$
h_{1}=\frac{2 a \sqrt{b}}{\sqrt{a}+\sqrt{b}}, \quad h_{2}=\frac{2 b \sqrt{a}}{\sqrt{a}+\sqrt{b}}
$$

Then $\sqrt{a b}=2$ and $a+b=5$, i.e., $a$ and $b$ are the roots of the equation $x^{2}-5 x+4=0$. Consequently $a=4, b=1$ and the radii equal $\frac{h_{1}}{2}=\frac{4}{3}$ and $\frac{h_{2}}{2}=\frac{2}{3}$.

Answer: \frac{4}{3},\frac{2}{3}


Given the equation $x^{2}+p x+q=0$. Let's write down the quadratic equation in $y$ whose roots are

$$
y_{1}=\frac{x_{1}+x_{1}^{2}}{1-x_{2}} \quad \text { and } \quad y_{2}=\frac{x_{2}+x_{2}^{2}}{1-x_{1}}
$$

where $x_{1}$ and $x_{2}$ are the roots of the given equation.

Solution:
The desired equation can only be set up if $y_{1}$ and $y_{2}$ are both numbers, that is, their denominators are not 0; in other words, if $x_{1}$ and $x_{2}$ differ from 1; we therefore assume that $x=1$ does not satisfy the given equation, i.e., $1+p+q \neq 0$. The coefficients of the sought equation are essentially given by the sum and product of $y_{1}$ and $y_{2}$. Since the two roots $x_{1}$ and $x_{2}$ can be transformed into each other by swapping, it is hoped that these expressions can be directly converted into expressions involving $x_{1}+x_{2}$ and $x_{1} x_{2}$, and thus can be written in terms of $p$ and $q$ without using the quadratic formula. (Otherwise, the problem - implicitly - clearly wants this.) According to the relationship between the roots and coefficients, if we write the sought equation in the form $a y^{2}+b y+c=0$ and use the fact that $x_{1}+x_{2}=-p, x_{1} x_{2}=q$, then

$$
\begin{aligned}
& \frac{c}{a}=y_{1} y_{2}=\frac{x_{1}\left(1+x_{1}\right) x_{2}\left(1+x_{2}\right)}{\left(1-x_{2}\right)\left(1-x_{1}\right)}=\frac{x_{1} x_{2}\left[1+\left(x_{1}+x_{2}\right)+x_{1} x_{2}\right]}{1-\left(x_{1}+x_{2}\right)+x_{1} x_{2}}=\frac{q(1-p+q)}{1+p+q} \\
& \frac{b}{a}=-\left(y_{1}+y_{2}\right)=-\frac{\left(x_{1}+x_{1}^{2}\right)\left(1-x_{1}\right)+\left(x_{2}+x_{2}^{2}\right)\left(1-x_{2}\right)}{\left(1-x_{2}\right)\left(1-x_{1}\right)}=\frac{x_{1}^{3}+x_{2}^{2}-\left(x_{1}+x_{2}\right)}{1+p+q}= \\
& =\frac{\left(x_{1}+x_{2}\right)^{3}-3 x_{1} x_{2}\left(x_{1}+x_{2}\right)-\left(x_{1}+x_{2}\right)}{1+p+q}=\frac{-p^{3}+3 p q+p}{1+p+q}=\frac{p\left(1+3 q-p^{2}\right)}{1+p+q}
\end{aligned}
$$

thus, for example, with the choice $a=1$, an equation that meets the requirement is:

$$
y^{2}+\frac{p\left(1+3 q-p^{2}\right)}{1+p+q} y+\frac{q(1-p+q)}{1+p+q}=0
$$

or, with the choice $a=1+p+q$ (which ensures $a \neq 0$, so the equation is indeed quadratic):

$$
(1+p+q) y^{2}+p\left(1+3 q-p^{2}\right) y+q(1-p+q)=0
$$

For example, with $p=-1, q=-12$, the roots of the given equation are $x_{1}=-3, x_{2}=4$, and equation (1) is: $-12 y^{2}+36 y+120=0$, whose roots are: $y_{1}=-2, y_{2}=5$, and these, as well as $x_{1}$ and $x_{2}$, indeed satisfy the required relationships.

Kátai Szabolcs (Bp. I., Toldy F. g. I. o. t.)

Answer: (1+p+q)y^{2}+p(1+3q-p^{2})y+q(1-p+q)=0


Find the smallest positive integer $n$ that makes $\frac{n-13}{5n+6}$ a non-zero reducible fraction.

Solution: Solution: Since $(1,5)=1,|1 \times 6-(-13) \times 5|$ $=71 \neq(1,5),(1,71)=1$, the fraction can be simplified by 71. From $\left.n-13=71 m_{1}, n=71 m_{1}+13\left(m_{1} \in Z\right)\right)$, and $n-13>0$, so we take $m_{1}=1$, obtaining the smallest $n=84$.

Answer: 84


Find the smallest number that can be represented as the sum of five, six, and seven consecutive natural numbers.

Solution:
27. The sum of five consecutive natural numbers has the form:

$$
n+(n+1)+(n+2)+(n+3)+(n+\dot{4})=5(n+2)
$$

i.e., this sum is divisible by 5. Similarly, we prove that the sum of seven consecutive natural numbers is divisible by 7, and the sum of six consecutive natural numbers has the form $3(2 n+5)$, i.e., it is an odd number divisible by 3. The smallest odd natural number that is divisible by 3, 5, and 7 is 105. Indeed, $105=19+20+$ $+21+22+23=15+16+17+18+19+20=12+13+14+15+$ $+16+17+18$

Answer: 105


The lateral surface of a cone is unfolded on a plane into a sector with a central angle of $120^{\circ}$ and an area of $S$. Find the volume of the cone.

Solution:
2.19. Let $r$ be the radius of the base of the cone, and $l$ be its slant height. Then the area of the lateral surface $S=\frac{1}{2} l^{2} \cdot \frac{2 \pi}{3}=\frac{\pi l^{2}}{3}$, from which $l=\sqrt{\frac{3 S}{\pi}}$. But $S=\pi r l=\pi r \sqrt{\frac{3 S}{\pi}}$ and, therefore, $r=\sqrt{\frac{S}{3 \pi}}$. The volume of the cone can be found using the formula $V=\frac{1}{3} \pi r^{2} h$, where $h=\sqrt{l^{2}-r^{2}}=\sqrt{\frac{3 S}{\pi}-\frac{S}{3 \pi}}=2 \sqrt{\frac{2 S}{3 \pi}}$. Finally, we get

$$
V=\frac{1}{3} \cdot \frac{\pi S}{3 \pi} \cdot 2 \sqrt{\frac{2 S}{3 \pi}}=\frac{2 S \sqrt{6 \pi S}}{27 \pi}
$$

Answer: $\frac{2 S \sqrt{6 \pi S}}{27 \pi}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-465.jpg?height=392&width=369&top_left_y=1244&top_left_x=46)

Fig. 2.18

Answer: \frac{2S\sqrt{6\piS}}{27\pi}


In an equilateral triangle $ABC$, the midpoint of side $\overline{AB}$ is point $P$. Points $G$ and $H$ are between points $A$ and $P$, and points $I$ and $J$ are between points $P$ and $B$. Points $D, E$, and $F$ divide the length of $\overline{AC}$ into four equal parts. Parallels to side $\overline{AB}$ are drawn through these points. We consider all triangles with one vertex at point $C$ and the other two on one of the constructed parallels to side $\overline{AB}$, including $\overline{AB}$ itself, at the points of intersection with the lines $AC, GC, HC, IC, JC$, or $BC$.

If the total number of such triangles is $x$, and the total number of such triangles that do not contain the centroid $T$ is $y$, determine the ratio

![](https://cdn.mathpix.com/cropped/2024_05_30_1ebddb66d71be8305afcg-09.jpg?height=536&width=585&top_left_y=246&top_left_x=1358)
$x: y$.

Solution:
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_1ebddb66d71be8305afcg-09.jpg?height=751&width=825&top_left_y=932&top_left_x=204)

Let the side $\overline{A B}$ be colored black, the parallel through $D$ be red, the parallel through $E$ be blue, and the parallel through $F$ be green. The centroid $T$ is located on the median $\overline{C P}$ and is between the blue and red parallels because it divides the median in the ratio 2:1 from vertex $C$.

Let's count the triangles whose side opposite vertex $C$ is black.

On the segment $\overline{A B}$, we need to choose two points from the 6 points $A, G, H, I, J$, and $B$.

If we choose point $A$, the second point can be chosen in 5 ways.

If we choose point $G$, the second point can be chosen in 4 ways (excluding point $A$ because triangle $C G A$ is included in the previous case).

If we choose point $H$, the second point can be chosen in 3 ways.

If we choose point $I$, the second point can be chosen in 2 ways.

If we choose point $J$, the second point can be chosen in 1 way.

Thus, there are 15 such triangles.

We have the same number of triangles whose one side is red, blue, or green. Therefore, the total number of triangles is $4 \cdot 15 = 60$.

Now we will calculate how many triangles contain point $T$. Such triangles must have one side either black or red.

If one vertex is at point $A$, the other vertex must be $I, J$, or $B$.

Similarly, if one vertex is at point $G$ or $H$, there are 3 possibilities for the second vertex.

Thus, the number of triangles that contain point $T$ with one black side is $3 + 3 + 3 = 9$, and the same number of triangles have a red side.

In total, 18 triangles contain point $T$, and $60 - 18 = 42$ triangles do not contain point $T$.

The desired ratio is $60: 42 = 10: 7$.

## NATIONAL MATHEMATICS COMPETITION

## 3rd grade - high school - B variant

Poreč, March 29, 2019.

Answer: 10:7


The diagonals of the inscribed quadrilateral $ABCD$ intersect at point $P$, and triangle $APD$ is acute-angled. Points $E$ and $F$ are the midpoints of sides $AB$ and $CD$ respectively. A perpendicular is drawn from point $E$ to line $AC$, and a perpendicular is drawn from point $F$ to line $BD$, these perpendiculars intersect at point $Q$. Find the angle between the lines $PQ$ and $BC$.

Solution:
Answer: $90^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-22.jpg?height=1019&width=993&top_left_y=153&top_left_x=540)

First solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $D P$, respectively, and let $T$ be the point of intersection of the lines $P Q$ and $B C$. Since the angles $\angle B A C$ and $\angle B D C$ subtend the same arc, they are equal. Therefore, the right triangles $A E^{\prime} E$ and $B F^{\prime} F$ are similar. Then, $\frac{E E^{\prime}}{F F^{\prime}}=\frac{A E}{D F}=\frac{B E}{C F}$, the last equality because points $E$ and $F$ are the midpoints of segments $A B$ and $C D$. Moreover, $\angle B E E^{\prime}=\angle C F F^{\prime}$, so triangles $B E E^{\prime}$ and $C F F^{\prime}$ are similar, and in particular, $\angle A B E^{\prime}=\angle D C F^{\prime}$. The angles $\angle A B D$ and $\angle A C D$ subtend the same arc and are therefore equal. Thus, $\angle E^{\prime} B F^{\prime}=\angle A B D-\angle A B E^{\prime}=\angle A C D-\angle D C F^{\prime}=\angle E^{\prime} C F^{\prime}$. Therefore, the quadrilateral $B E^{\prime} F^{\prime} C$ is cyclic and

$$
\angle P B T=\angle C B F^{\prime}=\angle C E^{\prime} F^{\prime}=\angle P E F^{\prime}=\angle P Q F^{\prime}
$$

(the last equality of angles follows from the cyclic nature of the quadrilateral $P E^{\prime} Q F^{\prime}$). It remains to calculate the angles:

$$
\angle P T B=180^{\circ}-\angle P B T-\angle B P T=180^{\circ}-\angle P Q F^{\prime}-\angle Q P F^{\prime}=90^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-23.jpg?height=1128&width=991&top_left_y=156&top_left_x=538)

Second solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $D P$, respectively, and let $K$ and $L$ be the midpoints of segments $A P$ and $D P$, and $T$ be the point of intersection of the lines $P Q$ and $B C$. Let $O$ be the center of the circumcircle of triangle $A P D$, then $O K$ and $O L$ are the perpendicular bisectors of segments $A P$ and $D P$, respectively. Let $\angle B A P=\varphi$ for brevity. Then $\angle C D P=\varphi$, since the angles $\angle B A P$ and $\angle C D P$ subtend the same arc. Therefore, triangles $B A P$ and $C D P$ are similar by two angles and, hence,

$$
\frac{A P}{D P}=\frac{A B}{C D}=\frac{A E}{D F}=\frac{A E \cos \varphi}{D F \cos \varphi}=\frac{A E^{\prime}}{D F^{\prime}}
$$

Thus,

$$
\frac{P E^{\prime}}{P F^{\prime}}=\frac{A P-A E^{\prime}}{D P-D F^{\prime}}=\frac{A P}{D P}=\frac{P K}{P L}
$$

Let the lines $K O$ and $P Q$ intersect at point $K^{\prime}$, and the lines $L O$ and $P Q$ intersect at point $L^{\prime}$. Then triangles $P E^{\prime} Q$ and $P K K^{\prime}$ are similar with the ratio $\frac{P E^{\prime}}{P K}=\frac{P F^{\prime}}{P L}$, and with the same ratio of similarity, triangles $P F^{\prime} Q$ and $P L L^{\prime}$ are similar. Therefore, $P K^{\prime}=P L^{\prime}$ and, hence, points $K^{\prime}, L^{\prime}$, and $O$ coincide. Thus, the line $P Q$ passes through the point $O$. It remains to calculate the angles:

$\angle P T B=180^{\circ}-\angle C B D-\angle B P T=180^{\circ}-\angle C A D-\angle O P D=180^{\circ}-\angle P O L-\angle O P D=90^{\circ}$.

In the penultimate equality, it was used that the central angle $\angle P O D$ is equal to twice the inscribed angle $\angle C A D$ on one side and to twice the angle $\angle P O L$ on the other side.

![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-24.jpg?height=1128&width=985&top_left_y=156&top_left_x=547)

Third solution. Draw a line $\ell$ from point $P$ perpendicular to side $B C$, let $T$ be its point of intersection with $B C$. Draw lines through points $A$ and $D$ perpendicular to diagonals $A C$ and $B D$, respectively. Let $S$ be their point of intersection. Since $\angle P A S=90^{\circ}=\angle P D S$, points $A, P, D$, and $S$ lie on a circle with diameter $P S$. Then (the equality $\angle D A P=\angle D B C$ follows from the cyclic nature of quadrilateral $A B C D$)

$$
\angle D P S=\angle D A S=90^{\circ}-\angle D A P=90^{\circ}-\angle D B C=\angle B P T
$$

Therefore, point $S$ lies on line $\ell$.

Let $H$ be the orthocenter of triangle $P B C$. Clearly, it also lies on line $\ell$. Then the lines $A S, B H$, and $E Q$ are parallel, and since $A E=E B$, line $E Q$ is the midline of trapezoid $A B H S$. Therefore, $E Q$ intersects line $\ell$ at the midpoint of segment $S H$. Similarly, line $F Q$ also intersects line $\ell$ at the midpoint of segment $S H$. But then lines $E Q, F Q$, and $\ell$ intersect at the midpoint of segment $S H$ and this point is point $Q$. Thus, $Q$ also lies on line $\ell$. In particular, lines $P Q$ and $B C$ intersect at a right angle.

Answer: 90


Find those $x \in[0,2 \pi]$, which satisfy the inequality

$$
2 \cos x \leq|\sqrt{1+\sin 2 x}-\sqrt{1-\sin 2 x}|
$$

Solution:
Solution. First, note that the inequality is meaningful for every value of $x$.

If $\cos x<0$, the left side of the inequality is negative, while the right side is non-negative, so the inequality is satisfied for all $x$ for which $\cos x<0$, i.e., for all $x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$.

Let $\cos x \geq 0$, i.e., $x \in\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]$. In this case, both sides of the inequality are non-negative, so after squaring, the direction of the inequality changes, and we get

$$
\begin{aligned}
& 4 \cos ^{2} x \leq 2-2 \sqrt{\cos ^{2} 2 x} \\
& 2 \cos ^{2} 2 x \leq 1-|\cos 2 x| \\
& |\cos 2 x| \leq-\cos 2 x
\end{aligned}
$$

The last inequality is obviously satisfied for all $x$ for which $\cos 2 x \leq 0$, so taking into account that $x \in\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]$, we get that $x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, \frac{7 \pi}{4}\right]$.

Together with the other solutions (when $\cos x<0$), we get that $x \in\left[\frac{\pi}{4}, \frac{7 \pi}{4}\right]$.

Answer: x\in[\frac{\pi}{4},\frac{7\pi}{4}]


Let $f(n)=3 n^{2}-3 n+1$. Find the last four digits of $f(1)+f(2)+\cdots+f(2010)$.

Solution:
1. 1000
1. Note that $f(n)=n^{3}-(n-1)^{3}$. It follows that
$$
f(1)+f(2)+\cdots+f(2010)=\left(1^{3}-0^{3}\right)+\left(2^{3}-1^{3}\right)+\cdots+\left(2010^{3}-2009^{3}\right)=2010^{3}=201^{3} \times 1000 \text {. }
$$

Since $201^{3}$ has unit digit 1 , it is clear that the answer is 1000 .
Remark. Even without realising $f(n)=n^{3}-(n-1)^{3}$, one could still easily observe that $f(1)+f(2)+\cdots+f(n)=n^{3}$ by computing this sum for some small $n$, which is a natural approach given that the question asks for something related to $f(1)+f(2)+\cdots+f(2010)$.

Answer: 1000


A and B are 36 kilometers apart. Supermen A and B start walking from $\mathbf{A}$ to $\mathbf{B}$ at the same time. Once they reach $\mathbf{B}$, they immediately walk back to $\mathbf{A}$, and upon reaching $\mathbf{A}$, they immediately walk back to $\mathbf{B}$...... They keep moving back and forth between $\mathbf{A}$ and $\mathbf{B}$. If Super A's speed is $2 k$ kilometers/hour, and Super B's speed is $k$ kilometers/hour. Suppose after $p$ hours, the distance between Super A and Super B reaches its maximum for the 2012th time; after $q$ hours, the distance between Super A and Super B reaches its maximum for the 2013th time. If $q-p$ is a positive integer, find: the maximum value of the positive integer $k$

Solution:
【Analysis】As shown in the following diagram, the Lica diagram
It is found that when A is at point A and B is at point B, the distance between the two is the greatest.
The time taken for the first maximum distance between the two is $\frac{36}{k}$ hours.
Thereafter, every $\frac{72}{k}$ hours, the two are at their maximum distance again.
Therefore, $p=\frac{72}{k} \times 2011+\frac{36}{k}$, $q=\frac{72}{k} \times 2012+\frac{36}{k}$,
Thus, $q-p=\frac{72}{k}$,
Hence, the maximum value of $k$ is 72

Answer: 72


How many ways can you assign one of the numbers $1,2,3, \ldots, 10$ to each corner of a cube so that no number is used more than once and so that for each side face the sum of the numbers in the four adjacent corners is odd?

Solution:
## Solution:

We call these sums of four numbers in the corners of a side surface area sums for short. Whether an area sum is odd or not depends only on how many of the numbers in the corners are odd. For each area, therefore, either 1 or 3 adjacent corner numbers must be odd.

Now fix the cube and look at a pair of opposite faces. If both faces have only one adjacent odd corner number, then a total of 6 corner numbers must be even, in contradiction to the fact that only five of the numbers $1,2, \ldots, 10$ are even. If both faces have three adjacent odd corner numbers, then a total of 6 corner numbers must be odd, again a contradiction. Testing the remaining four cases shows that the arrangement of the four odd corner numbers is as follows: One corner and the three adjacent corners have odd numbers, the rest have even numbers.

There are now 8 ways to choose the central odd corner. Subsequently, different numbers from the set $\{1,3,5,7,9\}$ can be assigned pairwise to the four odd corners in $5 \cdot 4 \cdot 3 \cdot 2$ ways. Assign different numbers from the set $\{2,4,6,8,10\}$ in pairs to the four even corners in $5 \cdot 4 \cdot 3 \cdot 2$ ways. The required number of assignments is therefore equal to

$$
8 \cdot(5 \cdot 4 \cdot 3 \cdot 2)^{2}=115200
$$

Answer: 115200


Points $K$ and $M$ are located on the sides $AB$ and $BC$ of triangle $ABC$, such that $BK: KA=1: 4, BM: MC=3: 2$. Lines $MK$ and $AC$ intersect at point $N$.

Find the ratio $AC: CN$.

Solution:
Through point $M$, draw a line parallel to $A B$ until it intersects $A C$ at point $L$. From the similarity of triangles $L M C$ and $A B C$, we find that $A L=3 / 5 A B$, and $L M=2 / 5 A B=1 / 2 A K$. Therefore, $M L$ is the midline of triangle $A N K$, and

$A N=2 A L=6 / 5 A C$.

## Answer

$5: 1$.

Submit a comment

Answer: 5:1


Find all integers n for which the fraction

$$
\frac{n^{3}+2010}{n^{2}+2010}
$$

is equal to an integer.

Solution:
SOLUTION. Fraction

$$
\frac{n^{3}+2010}{n^{2}+2010}=n-\frac{2010(n-1)}{n^{2}+2010}
$$

is an integer precisely when $n^{2}+2010$ is a divisor of the number $2010(n-1)=2 \cdot 3 \cdot 5 \cdot 67(n-1)$.

If $n$ is not a multiple of the prime number 67, then the numbers $n^{2}+2010$ and 67 are coprime, so $n^{2}+2010$ must be a divisor of the number $30(n-1)$. Since $|30(n-1)|<n^{2}+2010$, only $n=1$ satisfies this.

Let $n=67 m$, where $m$ is an integer. Then

$$
\frac{2010(n-1)}{n^{2}+2010}=\frac{30(67 m-1)}{67 m^{2}+30} .
$$

If $m$ is not a multiple of five, the number $67 m^{2}+30$ must be a divisor of the number $6(67 m-1)$. However, for $|m| \leqq 4$ this is not the case, and for $|m| \geqq 6$, $|6(67 m-1)|<67 m^{2}+30$. Therefore, $m=5 k$, where $k$ is an integer. Then

$$
\frac{30(67 m-1)}{67 m^{2}+30}=\frac{6(335 k-1)}{335 k^{2}+6} .
$$

For $|k| \geqq 7$, the absolute value of this fraction is non-zero and less than 1. Of the remaining numbers, $k=0$ and $k=-6$ satisfy the condition.

The number $\frac{n^{3}+2010}{n^{2}+2010}$ is therefore an integer precisely when the integer $n$ is one of the numbers 0, 1, or -2010.

## GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. Find the smallest natural number $n$ for which the quotient $\frac{n^{2}+15 n}{33000}$ is a natural number. [56-B-S-3]

N2. Find all pairs $(p, q)$ of real numbers such that the polynomial $x^{2}+p x q$ is a divisor of the polynomial $x^{4}+p x^{2}+q$. $[56-\mathrm{B}-\mathrm{I}-5]$

Answer: 0,1,-2010


Find the maximum value of $3 \sin \left(x+\frac{\pi}{9}\right)+5 \sin \left(x+\frac{4 \pi}{9}\right)$, where $x$ ranges over all real numbers.

Solution:
31. Answer: 7
$$
\begin{aligned}
& 3 \sin \left(x+\frac{\pi}{9}\right)+5 \sin \left(x+\frac{4 \pi}{9}\right) \\
= & 3 \sin \left(x+\frac{\pi}{9}\right)+5 \sin \left(x+\frac{\pi}{9}+\frac{\pi}{3}\right) \\
= & 3 \sin \left(x+\frac{\pi}{9}\right)+5 \sin \left(x+\frac{\pi}{9}\right) \cos \frac{\pi}{3}+5 \cos \left(x+\frac{\pi}{9}\right) \sin \frac{\pi}{3} \\
= & \frac{11}{2} \sin \left(x+\frac{\pi}{9}\right)+\frac{5 \sqrt{3}}{2} \cos \left(x+\frac{\pi}{9}\right) .
\end{aligned}
$$

Hence the maximum value is $\sqrt{\left(\frac{11}{2}\right)^{2}+\left(\frac{5 \sqrt{3}}{2}\right)^{2}}=7$.

Answer: 7


Find the sum of all integer values of the argument $x$ for which the corresponding values of the function

$$
y=x^{2}+x\left(\log _{2} 20-\log _{5} 8\right)-\log _{2} 5-9 \log _{5} 2
$$

do not exceed 6.

Solution:
Solution. Let $a=\log _{2} 5$. Then the condition of the problem will turn into the inequality

$$
x^{2}+\left(a-\frac{3}{a}+2\right) x-\left(a+\frac{9}{a}+6\right) \leqslant 0
$$

Considering that $a \in(2,3)$, we get $x \in\left[-a-3, \frac{3}{a}+1\right]$. Since $-6<-a-3<-5,2<\frac{3}{a}+1<3$, the integer solutions will be the numbers $-5,-4,-3,-2,-1,0,1,2$.

Answer: $-12 .(\mathrm{C})$

![](https://cdn.mathpix.com/cropped/2024_05_06_a8c31eb257d724803ce6g-06.jpg?height=67&width=901&top_left_y=1848&top_left_x=629)

Answer: -12


Solution:
【Solution】Solution: $928+217+64+8$,
$$
\begin{array}{l}
=928+217+72, \\
=928+72+217, \\
=1000+217, \\
=1217 .
\end{array}
$$

Answer: 1217


Electric lamps are manufactured at three plants. The first plant produces $35 \%$ of the total number of lamps, the second - $50 \%$, and the third $15 \%$. The production of the first plant contains $70 \%$ standard lamps, the second $80 \%$, and the third $90 \%$. The products of all three plants are supplied to the store. What is the probability that

a) a randomly selected lamp is manufactured at the first plant and is standard;

b) a lamp purchased from the store is standard?

Solution:
S o l u t i o n. Let's consider the events:

$B_{1}$ - a randomly selected lamp is manufactured by the first factory;

$B_{2}$ - a randomly selected lamp is manufactured by the second factory;

$B_{3}$ - a randomly selected lamp is manufactured by the third factory;

$C_{1}$ - a randomly selected lamp is manufactured by the first factory and is standard;

$A$ - a lamp purchased from the store is standard.

It should be noted that the first hypothesis - event $B_{1}$, means that a lamp randomly taken from the total number of lamps manufactured by the first factory can be of any quality, i.e., it can be both standard and non-standard. The other two hypotheses of events $B_{2}$ and $B_{3}$ have a similar meaning.

The probabilities of events $B_{1}, B_{2}$, and $B_{3}$ according to formula (1) are: $P\left(B_{1}\right)=0.35 ; P\left(B_{2}\right)=0.5 ; P\left(B_{3}\right)=0.15$. Events $B_{1}, B_{2}$, and $B_{3}$ are mutually exclusive and form a complete group, the sum of the probabilities of these events equals one: $0.35+0.5+0.15=1$.

From the condition, it follows that $P_{B_{1}}(A)=0.7, P_{B_{2}}(A)=0.8, P_{B_{3}}(A)=0.9$.
a) Event $C_{1}$ consists in the fact that a randomly selected lamp, firstly, is manufactured by the first factory, and, secondly, is standard. This means that event $C_{1}$ is the product of two dependent events $B_{1}$ and $A: C_{1}=B_{1} \cdot A$. The probability of event $C_{1}$ will be found using the theorem of multiplication of probabilities of dependent events. Applying formula (12), we get:

$$
P\left(C_{1}\right)=P\left(B_{1}\right) P_{B_{1}}(A)=0.35 \cdot 0.7=0.245
$$

b) Event $A$ represents the sum of the following three mutually exclusive events:

$C_{1}$ - the lamp is manufactured by the first factory and it is standard;

$C_{2}$ - the lamp is manufactured by the second factory and it is standard;

$C_{3}$ - the lamp is manufactured by the third factory and it is standard.

Each of the events $C_{i}(i=1,2,3)$ represents the product of two dependent events $B_{i}$ and $A$.

Thus, $A=B_{1} \cdot A+B_{2} \cdot A+B_{3} \cdot A$. Applying formula (16), we get

$$
\begin{aligned}
& P(A)=P\left(B_{1}\right) P_{B_{1}}(A)+P\left(B_{2}\right) P_{B_{2}}(A)+P\left(B_{3}\right) P_{B_{3}}(A)= \\
& =0.35 \cdot 0.7+0.5 \cdot 0.8+0.15 \cdot 0.9=0.245+0.4+0.135=0.78
\end{aligned}
$$

Answer: 0.78


If in the previous task we also want to know whether the counterfeit coin is lighter or heavier than the others, how many coins is the maximum number from which three measurements are still sufficient to determine the answer to this question?

Solution:
In this measurement procedure, the counterfeit coin only did not end up in the pan in one case: if we found balance in every measurement. In all other cases, we always knew whether the counterfeit coin was lighter or heavier than the others. Thus, with three measurements, we can always obtain this more precise answer from 12 coins.

It would be more tedious to show that with 13 coins (not only with the above but with any other measurement procedure), there can always be cases where it is not possible to answer this more precise question. We are afraid you might get bored. If anyone is still curious about this, ask, and we will write it down for them.

There were those who did not find the most efficient procedures, solving the problem with 4 to 7 measurements, or with 11 to 5 coins (depending on how incompetent they were).

Answer: 12


Today's date is written as: 22.11.2015. Name the last past date that is written with the same set of digits.

Solution: Solution. In 2015, there are no earlier such dates. The previous suitable year is 2012, the last month in it - 12, the largest possible number - 15. Answer: 15.12.2012.

Answer: 15.12.2012


As shown in Figure 1, $A$ is the right vertex of the hyperbola $\frac{x^{2}}{4}-y^{2}=1$. Two perpendicular lines passing through $A$ intersect the right branch of the hyperbola at points $M$ and $N$, respectively. Is the line $MN$ guaranteed to pass through a fixed point on the $x$-axis? If such a fixed point does not exist, please explain the reason; if there is such a fixed point $P$, try to find the coordinates of this fixed point $P$.

Solution:
10. Clearly, the right vertex $A(2,0)$.

Translate the $y$-axis to the right by 2 units, making $A$ the origin of the new rectangular coordinate system. In the new coordinate system, the equation of the hyperbola is
$$
\frac{\left(x^{\prime}+2\right)^{2}}{4}-y^{2}=1 \text {, }
$$

which simplifies to $4 y^{2}-x^{\prime 2}-4 x^{\prime}=0$.
If $M N \perp x$-axis, then the slope of $A M$ is 1, i.e., $l_{A M}: y=x^{\prime}$.
Substituting into equation (1) yields $M\left(\frac{4}{3}, \frac{4}{3}\right)$.
Consequently, $N\left(\frac{4}{3},-\frac{4}{3}\right)$.
Thus, the intersection point of $M N$ with the $x$-axis is $P\left(\frac{4}{3}, 0\right)$.
If $M N$ is not perpendicular to the $x$-axis, let
$l_{M N}: y=k x^{\prime}+t(t \neq 0)$.
Then $\frac{y-k x^{\prime}}{t}=1$.
Thus, equation (1) can be rewritten as
$$
4 y^{2}-x^{\prime 2}-4 x^{\prime} \cdot \frac{y-k x^{\prime}}{t}=0 \text {, }
$$

which simplifies to $4 t\left(\frac{y}{x^{\prime}}\right)^{2}-4\left(\frac{y}{x^{\prime}}\right)+4 k-t=0$.
The two roots $k_{1} 、 k_{2}$ of the quadratic equation in $\frac{y}{x^{\prime}}$ are the slopes of $A M$ and $A N$.
Since $A M \perp A N$, we have
$$
k_{1} k_{2}=\frac{4 k-t}{4 t}=-1 \text {. }
$$

Solving for $t$ gives $t=-\frac{4}{3} k$.
Thus, $l_{M N}: y=k x-\frac{4}{3} k$, which also passes through the point $P\left(\frac{4}{3}, 0\right)$.
In summary, in the new coordinate system, the line $M N$ passes through the fixed point $P\left(\frac{4}{3}, 0\right)$ on the $x$-axis, i.e., in the original coordinate system, the line $M N$ passes through the fixed point $P\left(\frac{10}{3}, 0\right)$ on the $x$-axis.

Answer: \left(\frac{10}{3}, 0\right)


At the beginning of the winter, there were at least 66 students registered in a ski class. After the class started, eleven boys transferred into this class and thirteen girls transferred out. The ratio of boys to girls in the class was then $1: 1$. Which of the following is not a possible ratio of boys to girls before the transfers?
(A) $4: 7$
(B) $1: 2$
(C) $9: 13$
(D) $5: 11$
(E) $3: 5$

Solution:
Solution 1

Let $b$ represent the number of boys initially registered in the class.

Let $g$ represent the number of girls initially registered in the class.

When 11 boys transferred into the class, the number of boys in the class was $b+11$.

When 13 girls transferred out of the class, the number of girls in the class was $g-13$.

The ratio of boys to girls in the class at this point was $1: 1$.

That is, the number of boys in the class was equal to the number of girls in the class, or $b+11=g-13$ and so $g=b+24$.

Since there were at least 66 students initially registered in the class, then $b+g \geq 66$.

Substituting $g=b+24, b+g$ becomes $b+(b+24)=2 b+24$, and so $2 b+24 \geq 66$ or $2 b \geq 42$, so $b \geq 21$.

At this point, we may use the conditions that $g=b+24$ and $b \geq 21$ to determine which of the 5 answers given is not possible.

Each of the 5 answers represents a possible ratio of $b: g$ or $b:(b+24)$ (since $g=b+24)$.

We check whether $b:(b+24)$ can equal each of the given ratios while satisfying the condition that $b \geq 21$.

In (A), we have $b:(b+24)=4: 7$ or $\frac{b}{b+24}=\frac{4}{7}$ or $7 b=4 b+96$ and then $3 b=96$, so $b=32$.

In (B), we have $b:(b+24)=1: 2$ or $\frac{b}{b+24}=\frac{1}{2}$ or $2 b=b+24$, so $b=24$.

In (C), we have $b:(b+24)=9: 13$ or $\frac{b}{b+24}=\frac{9}{13}$ or $13 b=9 b+216$ and then $4 b=216$, so $b=54$.

In (D), we have $b:(b+24)=5: 11$ or $\frac{b}{b+24}=\frac{5}{11}$ or $11 b=5 b+120$ and then $6 b=120$, so $b=20$.

In (E), we have $b:(b+24)=3: 5$ or $\frac{b}{b+24}=\frac{3}{5}$ or $5 b=3 b+72$ and then $2 b=72$, so $b=36$.

Thus, the only ratio that does not satisfy the condition that $b \geq 21$ is $5: 11$ (since $b=20$ ).

## Solution 2

As in Solution 1, we establish the conditions that $g=b+24$ and $b \geq 21$.

Again, the ratio $b: g$ or $\frac{b}{g}$ then becomes $\frac{b}{b+24}$ (since $g=b+24$ ).

Since $b \geq 21$, then $b$ can equal $21,22,23,24, \ldots$, but the smallest possible value for $b$ is 21 .

When $b=21, \frac{b}{b+24}$ becomes $\frac{21}{21+24}=\frac{21}{45}=0.4 \overline{6}$.

When $b=22, \frac{b}{b+24}$ becomes $\frac{22}{22+24}=\frac{22}{46} \approx 0.478$.

When $b=23, \frac{b}{b+24}$ becomes $\frac{23}{23+24}=\frac{23}{47} \approx 0.489$.

When $b=24, \frac{b}{b+24}$ becomes $\frac{24}{24+24}=\frac{24}{48}=0.5$.

As $b$ continues to increase, the value of the ratio $\frac{b}{b+24}$ or $\frac{b}{g}$ continues to increase.

Can you verify this for yourself?

Thus, the smallest possible value of the ratio $\frac{b}{g}$ is $\frac{21}{45}$ or $0.4 \overline{6}$.

Comparing this ratio to the 5 answers given, we determine that $\frac{5}{11}=0 . \overline{45}<0.4 \overline{6}=\frac{21}{45}$, which is not possible.

(You may also confirm that each of the other 4 answers given are all greater than $\frac{21}{45}$ and are obtainable, as in Solution 1.)

Therefore, the only ratio of boys to girls which is not possible is (D).

ANSWER: (D)

Answer: D


Given two sets of real numbers
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}, B=\left\{b_{1}, b_{2}, \cdots, b_{50}\right\} \text {. }
$$

If the mapping $f$ from $A$ to $B$ makes each element in $B$ have a preimage, and
$$
f\left(a_{1}\right) \leqslant f\left(a_{2}\right) \leqslant \cdots \leqslant f\left(a_{100}\right),
$$

then the number of such mappings is ( ).
(A) $\mathrm{C}_{100}^{50}$
(B) $\mathrm{C}_{99}^{48}$
(C) $\mathrm{C}_{100}^{49}$
(D) $\mathrm{C}_{99}^{49}$

Solution:
Let's assume $b_{1}<b_{2}<\cdots<b_{50}$.
Since each element in set $B$ has a preimage, let the set of preimages of $b_{i}$ be $A_{i}(i=1,2, \cdots, 50)$, with the number of elements being $x_{i}$. Then
$$
x_{1}+x_{2}+\cdots+x_{50}=100 \text {. }
$$

Thus, the problem is transformed into finding the number of positive integer solutions to the indeterminate equation (1), which has $\mathrm{C}_{99}^{49}$ solutions. Therefore, the correct choice is (D).

Answer: D


What is the greatest number of acute angles that can occur in a convex polygon?

Solution: 5. The maximum number of acute angles in a convex polygon is three. Indeed, the sum of all exterior angles of an $n$-sided polygon is always $4 d$. If any $n$-sided polygon ($n \geqslant 4$) had four acute interior angles, then the exterior angles supplementary to these interior angles would be obtuse, and their sum would be greater than $4 d$; therefore, the sum of all exterior angles would also be greater than $4 d$.

Answer: 3


Let $a, b_{1}, b_{2}, \cdots, b_{n}, c_{1}, c_{2}, \cdots, c_{n} \in \mathbf{R}$, such that
$$
x^{2 n}+a \sum_{i=1}^{2 n-1} x^{i}+1=\prod_{i=1}^{n}\left(x^{2}+b_{i} x+c_{i}\right)
$$

holds for any real number $x$. Find the values of $c_{1}, c_{2}, \cdots, c_{n}$.

Solution:
Proof of a lemma first.
Lemma Let $p(z)=z^{2 n}+a \sum_{i=1}^{2 n-1} z^{i}+1$. Then the polynomial $p(z)$ has at least $n-1$ complex roots on the upper unit circle.
Proof In fact, if $p(z)=0$, then
$$
-a=\frac{\left(z^{2 n}+1\right)(z-1)}{z^{2 n}-z} \text {. }
$$

Let $z=\omega^{2}$.
Thus, $-a=\frac{\left(\omega^{4 n}+1\right)\left(\omega^{2}-1\right)}{\omega^{4 n}-\omega^{2}}$
$$
=\frac{\left(\omega^{2 n}+\omega^{-2 n}\right)\left(\omega-\omega^{-1}\right)}{\omega^{2 n-1}-\omega^{-(2 n-1)}} \text {. }
$$

Take $\omega=\mathrm{e}^{\mathrm{i} \theta}$, we get
$$
-a=\frac{2 \cos 2 n \theta \cdot \sin \theta}{\sin (2 n-1) \theta}=\frac{\sin (2 n+1) \theta}{\sin (2 n-1) \theta}-1 .
$$

It suffices to prove: for each real number $a$, the equation about $\theta$
$$
f(\theta)=\sin (2 n+1) \theta+(a-1) \sin (2 n-1) \theta=0
$$

has at least $n-1$ solutions in the interval $\left(0, \frac{\pi}{2}\right)$.
If $a=1$, then when $\theta_{k}=\frac{k \pi}{2 n+1}(k=1,2, \cdots, n)$, we have $f\left(\theta_{k}\right)=0$.
If $a \neq 1$, for $\theta_{k}=\frac{k \pi}{2 n+1}(k=1,2, \cdots, n)$, note that,
$$
\begin{array}{l}
(k-1) \pi<\theta_{k}<k \pi, \\
f\left(\theta_{k}\right) f\left(\theta_{k+1}\right)<0,
\end{array}
$$

i.e., $f(\theta)=0$ has at least one solution in the interval $\left(\theta_{k}, \theta_{k+1}\right)$.

In summary, $p(z)=0$ has at least $n-1$ roots in the interval $(0, \pi)$.
Back to the original problem.
Let the $n-1$ roots of $p(z)=0$ on the upper unit circle be $z_{1}, z_{2}, \cdots, z_{n-1}$. Then their conjugate complex numbers $\overline{z_{1}}, \overline{z_{2}}, \cdots$, $\overline{z_{n-1}}$ are also its roots.
Thus, $x^{2}+b_{k} x+c_{k}=\left(x-z_{k}\right)\left(x-\overline{z_{k}}\right)$.
Therefore, $c_{k}=z_{k} \overline{z_{k}}=1(k=1,2, \cdots, n-1)$.
Since $c_{1} c_{2} \cdots c_{n}=1$, we have $c_{n}=1$, i.e., $c_{1}=c_{2}=\cdots=c_{n}=1$.

Answer: c_{1}=c_{2}=\cdots=c_{n}=1


There are three numbers $a, b, c$, $a \times b=24, a \times c=36, b \times c=54$, then $a+b+c=$

Solution:
【Solution】Solution: Since, $(a \times b) \times(a \times c) \div(b \times c)=24 \times 36 \div 54=16$, that is $a^{2}=16$,
so $a=4$,
$$
\begin{array}{l}
b=24 \div a=6, \\
c=36 \div a=9, \\
a+b+c=4+6+9=19 ;
\end{array}
$$

Therefore, the answer is: 19 .

Answer: 19


What is the maximum number of angles less than $150^{\circ}$ that a non-crossing 2017-sided polygon can have, all of whose angles are strictly less than $180^{\circ}$?

Solution:
Answer: 11. If the angles are $180-x_{i}$, then we have $x_{i}>0$ and $x_{1}+\cdots+x_{2017}=360$, so at most 11 of the $x_{i}$ can be greater than 30. We can verify that conversely there exists such a polygon.

## 5 High School Eliminatory: Solutions

Answer: 11


In a regular hexagon $ABCDEF$ with an area of 135 square centimeters, there is a rhombus $AMDN$ composed of two equilateral triangles. Then, the area of the remaining part after cutting out this rhombus is ( ) square centimeters.
(A) 60
(B) 65
(C) 70
(D) 75

Solution:
2. D.

As shown in Figure 4, connect $A D$, and let $A D=2 h$. Then the regular hexagon is composed of six equilateral triangles with side length $h$. It is easy to know that the area of the equilateral $\triangle C O B$ with side length $h$ is $\frac{\sqrt{3}}{4} h^{2}$.
Therefore, $S_{\text {regular hexagon }}=6 \times \frac{\sqrt{3}}{4} h^{2}=\frac{3 \sqrt{3}}{2} h^{2}=135$ $\Rightarrow h^{2}=30 \sqrt{3}$.
Let the side length of the rhombus $A M=a$.
Then $h=\frac{\sqrt{3}}{2} a \Rightarrow a^{2}=\frac{4}{3} h^{2}$.
Thus, the area of the rhombus $A M D N$ is
$$
\begin{array}{l}
2 \times \frac{\sqrt{3}}{4} a^{2}=\frac{\sqrt{3}}{2} \times \frac{4}{3} h^{2}=\frac{2}{\sqrt{3}} \times 30 \sqrt{3} \\
=60 \text { (square centimeters). }
\end{array}
$$

Therefore, the area of the remaining part after cutting out this rhombus is 75 square centimeters.

Answer: D


Insert two numbers between 2015 and 131 so that the four numbers are arranged in descending order, and the difference between any two adjacent numbers is equal. The sum of the two inserted numbers is

Solution:
$$
2146
$$

【Solution】According to the analysis, after inserting two numbers, the sequence forms an arithmetic sequence. Using the properties of an arithmetic sequence, we can find the sum of the two numbers. The sum of the middle two numbers $=2015+131=2146$. Therefore, the answer is: 2146.

Answer: 2146


Three circles each with a radius of 1 are placed such that each circle touches the other two circles, but none of the circles overlap. What is the exact value of the radius of the smallest circle that will enclose all three circles?

Solution:
Suppose that the three small circles have centres $A, B$ and $C$.

For the larger circle to be as small as possible, these three small circles must just touch the larger circle (which has centre $O$ ) at $P, Q$ and $S$, respectively.

Note that $A P=B Q=C S=1$.

![](https://cdn.mathpix.com/cropped/2024_04_30_68642ad613c12af1e439g-13.jpg?height=398&width=393&top_left_y=1666&top_left_x=931)

Let $r$ be the radius of the larger circle. Then $O P=O Q=O S=r$.

Since the larger circle is tangent to each of the smaller circles at $P, Q$ and $S$, this means that the circles share a common tangent at each of these points.

Consider the point $P$. Since $A P$ is perpendicular to the tangent to the smaller circle at $P$ and $O P$ is perpendicular to the tangent to the larger circle at $P$ and this tangent is common, then $A P$ and $O P$ must be parallel. Since $A P$ and $O P$ both pass through $P$, then $A P$ lies along $O P$. Since $A P=1$ and $O P=r$, then $O A=r-1$.

In a similar way, we can find that $O B=O C=r-1$.

Now since the circles with centres $A$ and $B$ just touch, then the distance $A B$ equals the sum of the radii of the circles, or $A B=2$.

Similarly, $A C=B C=2$.

Consider $\triangle A O B$. We have $O A=O B=r-1$ and $A B=2$.

By symmetry, $\angle A O B=\angle B O C=\angle C O A=\frac{1}{3}\left(360^{\circ}\right)=120^{\circ}$.

Let $M$ be the midpoint of $A B$. Thus, $A M=M B=1$.

![](https://cdn.mathpix.com/cropped/2024_04_30_68642ad613c12af1e439g-14.jpg?height=192&width=417&top_left_y=419&top_left_x=911)

Since $\triangle A O B$ is isosceles, then $O M$ is perpendicular to $A B$ and bisects $\angle A O B$.

This means that $\triangle A O M$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Therefore, $\frac{A O}{A M}=\frac{2}{\sqrt{3}}$ and so $\frac{r-1}{1}=\frac{2}{\sqrt{3}}$ or $r=1+\frac{2}{\sqrt{3}}$.

Thus, the radius of the smallest circle enclosing the other three circles is $1+\frac{2}{\sqrt{3}}$

ANSWER: $1+\frac{2}{\sqrt{3}}$

Answer: 1+\frac{2}{\sqrt{3}}


Determine the number of $8$-tuples of nonnegative integers $(a_1,a_2,a_3,a_4,b_1,b_2,b_3,b_4)$ satisfying $0\le a_k\le k$, for each $k = 1,2,3,4$, and $a_1 + a_2 + a_3 + a_4 + 2b_1 + 3b_2 + 4b_3 + 5b_4 = 19$.

Solution:
1. **Define new variables**: Let \( c_k = a_k + (k+1)b_k \). This transformation helps us simplify the problem by combining \( a_k \) and \( b_k \) into a single variable \( c_k \).

2. **Determine \( a_k \) and \( b_k \) from \( c_k \)**: Note that there is only one way to write \( c_k \) in the given form:
   \[
   b_k = \left\lfloor \frac{c_k}{k+1} \right\rfloor
   \]
   and
   \[
   a_k = c_k - (k+1) \left\lfloor \frac{c_k}{k+1} \right\rfloor.
   \]
   This means \( b_k \) is the number of complete times \( k+1 \) goes into \( c_k \), and \( a_k \) is the remainder of \( c_k \) upon division by \( k+1 \).

3. **Rewrite the original equation**: The original equation \( a_1 + a_2 + a_3 + a_4 + 2b_1 + 3b_2 + 4b_3 + 5b_4 = 19 \) can be rewritten using \( c_k \):
   \[
   c_1 + c_2 + c_3 + c_4 = 19.
   \]

4. **Count the number of solutions**: We need to find the number of nonnegative integer solutions to the equation \( c_1 + c_2 + c_3 + c_4 = 19 \). This is a classic "stars and bars" problem, where we distribute 19 indistinguishable stars (units) into 4 distinguishable bins (variables \( c_1, c_2, c_3, c_4 \)).

5. **Apply the stars and bars theorem**: The number of nonnegative integer solutions to \( c_1 + c_2 + c_3 + c_4 = 19 \) is given by:
   \[
   \binom{19 + 4 - 1}{4 - 1} = \binom{22}{3}.
   \]

6. **Calculate the binomial coefficient**:
   \[
   \binom{22}{3} = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 1540.
   \]

Conclusion:
\[
\boxed{1540}
\]

Answer: 1540


Triangle ABC has integer side lengths and perimeter 7. Determine all possible lengths of side $\mathrm{AB}$.

Solution:
Solution: By the triangle inequality, the largest possible length of any side is 3 . Moreover, if all side lengths were less than 3 , the perimeter would be less than 7 . Thus, at least one side has length 3 . This leaves the other sides has $(2,2)$ or $(1,3)$. Thus, $\mathrm{AB}$ can be 1,2 , or 3 .
Answer: $1,2,3$.

Answer: 1,2,3


If the line $y=k x+1$ intersects the circle $x^{2}+y^{2}+k x+m y-4=0$ at points $P$ and $Q$, and points $P$ and $Q$ are symmetric with respect to the line $x+y=0$, then the area of the plane region represented by the system of inequalities
$$
\left\{\begin{array}{l}
k x-y+1 \geqslant 0, \\
k x-m y \leqslant 0, \\
y \geqslant 0
\end{array}\right.
$$

is ( ).
(A) 2
(B) 1
(C) $\frac{1}{2}$
(D) $\frac{1}{4}$

Solution:
2.D.

From the fact that points $P$ and $Q$ are symmetric with respect to the line $x+y=0$, we know that the line $y=kx+1$ is perpendicular to $x+y=0$, and the center $\left(-\frac{k}{2},-\frac{m}{2}\right)$ of the circle lies on the line $x+y=0$, which gives $k=1$, and $\left(-\frac{k}{2}\right)+\left(-\frac{m}{2}\right)=0$. Solving these equations simultaneously, we get $k=1, m=-1$. Therefore, the system of inequalities $\left\{\begin{array}{l}x-y+1 \geqslant 0, \\ x+y \leqslant 0, \\ y \geqslant 0\end{array}\right.$ represents the plane region of the isosceles right triangle $\triangle ABO$ in Figure 3, with an area of $S=\frac{1}{4}$.

Answer: D


The sum of the digits in the decimal representation of a natural number $n$ is 100, and the sum of the digits of the number $44 n$ is 800. What is the sum of the digits of the number $3 n$?

Solution:
Note that $44 n$ is the sum of 4 instances of the number $n$ and 4 instances of the number $10 n$.

If we add these numbers digit by digit, then in each digit place, there will be the sum of the quadrupled digit from the same place in the number $n$ and the quadrupled digit from the next place. If no carries occur in this process, then each digit of the number $n$ is added 8 times, and the sum of the digits in all places is 800. With carries, the sum of the digits obviously decreases (since 10 is subtracted from one place and only 1 is added to another). Therefore, in the situation described in the problem, no carries occur. This means, in particular, that any digit of the number $n$ does not exceed 2. Then, when multiplying $n$ by 3, each of its digits is simply multiplied by 3, and, therefore, the sum of the digits is also multiplied by 3.

Therefore, the sum of the digits of the number 3n is 300.

## Answer

300.00

Answer: 300


Among 2000 indistinguishable balls, half are aluminum with a mass of 10 g, and the rest are duralumin with a mass of 9.9 g. It is required to separate the balls into two piles such that the masses of the piles are different, but the number of balls in them is the same. What is the smallest number of weighings on a balance scale without weights that can achieve this?

Solution:
Let's compare the mass of 667 balls with the mass of another 667 balls. If the masses of these two piles are not equal, the required condition is met.

Suppose the specified masses are equal. Then the mass of 666 balls that did not participate in the weighing is not equal to the mass of any 666 balls lying on one of the pans of the scales.

Indeed, if each of the weighed piles contains exactly $k$ aluminum balls, then among any 666 balls of any of these piles, the number of aluminum balls is $k$ or $k-1$. Meanwhile, among the 666 balls that did not participate in the weighing, there are exactly $1000-2k$ aluminum balls. It remains to note that neither the equality $k = 1000 - 2k$, nor the equality $k - 1 = 1000 - 2k$ can be satisfied for any integer $k$.

## Answer

In one weighing.

Answer: 1


At the ceremony marking the City Day, a tribune was set up where the spectators were properly arranged in 12 rows and a certain number of columns. Each spectator has either a cap or a hat on their head. The mayor noticed that in each of the 12 rows of the tribune, there are exactly 6 spectators with caps, and in each column of the tribune, there are exactly 5 spectators with hats. There are a total of 5 empty seats on the tribune.

a) How many spectators are on the tribune?

b) Show an example of a seating arrangement that meets the conditions of the problem.

Solution:
## Solution.

a) Let $s$ be the number of columns. The total number of seats is then $12s$.

On the other hand, the total number of seats is equal to the sum of the number of spectators with caps, which is 72, the number of spectators with hats, which is $5s$, and the number of empty seats, which is 5.

Thus, we have $12s = 72 + 5s + 5$, or $7s = 77$, i.e., $s = 11$.

There are 11 columns on the stand, and $12 \cdot 11 - 5 = 127$ spectators.

b) An example of a seating arrangement that meets the required conditions:

| K | K | K | K | K | K | S | S | S | S | S |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| S | K | K | K | K | K | K | S | S | S | S |
| S | S | K | K | K | K | K | K | S | S | S |
| S | S | S | K | K | K | K | K | K | S | S |
| S | S | S | S | K | K | K | K | K | K | S |
| S | S | S | S | S | K | K | K | K | K | K |
| K | S | S | S | S | S | K | K | K | K | K |
| K | K | S | S | S | S | S | K | K | K | K |
| K | K | K | S | S | S | S | S | K | K | K |
| K | K | K | K | S | S | S | S | S | K | K |
| K | K | K | K | K | S | S | S | S | S | K |
| K | K | K | K | K | K | S | S | S | S | S |

We use K to denote spectators with caps, S to denote spectators with hats, and the rest are empty seats.

Notice that after arranging the first eleven rows as shown above, in the last row, we can arrange the spectators with caps on any 6 seats, while the rest remain empty.

Answer: 127


We want to plant our seedlings in such a way that as many seedlings go into each row as there are rows. In our first attempt, 39 seedlings turned out to be excessive. If we increase the number of rows (and thus the number of trees per row) by 1, we find that 50 seedlings are missing. How many seedlings do we have available?

Solution:
I. solution: If we plant $x$ seedlings in each row, and there are $x$ rows, then $x^{2}$ seedlings are planted, leaving 39 remaining. If we plant $x+1$ seedlings in each row, and there are $x+1$ rows, then $(x+1)^{2}$ seedlings are planted, leaving 50 spaces empty. Thus,

$$
x^{2}+39=(x+1)^{2}-50, \quad \text { from which } \quad x=44 .
$$

Since we have $x^{2}+39$ seedlings, this means $44^{2}+39=1975$ seedlings are available.

György Komlóssy (Szolnok, Verseghy g. I. o. t.)

![](https://cdn.mathpix.com/cropped/2024_05_02_2965b58b76ff0a9c9995g-1.jpg?height=436&width=422&top_left_y=481&top_left_x=838)

II solution: In the first planting attempt, we would plant the seedlings in $x$ rows, and in the second attempt, in $x+1$ rows. Therefore, on the shaded area in the diagram, 89 seedlings must be placed. Since $2 x+1$ seedlings fit in this area, $2 x+1=89$, $x=44$.

Gábor Fenyó (Bp. V., Eötvös g. I. o. t.)

III. solution: If the number of seedlings is $x$, then in the first attempt, $\sqrt{x-39}$ seedlings are planted in each row, and in the second attempt, $\sqrt{x+50}$ seedlings are planted in each row. However, the number of rows in the second planting is 1 more than in the first, so

$$
\sqrt{x-39}+1=\sqrt{x+50}
$$

which means

$$
\begin{aligned}
& x-39+2 \sqrt{x-39}+1=x+50 \\
& \sqrt{x-39}=44 \\
& x=44^{2}+39=1975
\end{aligned}
$$

Árpád Bokor (Szombathely, Nagy L, g. II. o. t.)

Answer: 1975


Through the vertex $A$ of a right circular cone, a section of maximum area is drawn. Its area is twice the area of the section passing through the axis of the cone. Find the angle at the vertex of the axial section of the cone.

Solution:
2.28. Consider an arbitrary section passing through the vertex $A$. This section is a triangle $A B C$, and its sides $A B$ and $A C$ are the generators of the cone, i.e., they have a constant length. Therefore, the area of the section is proportional to the sine of the angle $B A C$. The angle $B A C$ varies from $0^{\circ}$ to $\varphi$,
where $\varphi$ is the angle at the vertex of the axial section of the cone. If $\varphi \leqslant 90^{\circ}$, then the axial section has the maximum area, and if $\varphi > 90^{\circ}$, then the section with a right angle at the vertex $A$ has the maximum area. Thus, from the problem statement, it follows that $\boldsymbol{\operatorname { s i n }} \varphi=0.5$ and $\varphi > 90^{\circ}$, i.e., $\varphi=120^{\circ}$.

Answer: 120


The number of tetrahedra with vertices at the vertices of the cube $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is

Solution: 658 Prompt: From the 8 vertices of a cube, any four points can be chosen, with $\mathrm{C}_{8}^{4}=70$ (ways) of selection. Removing the 6 faces and 6 diagonal planes, we get 58 tetrahedra.

Answer: 58


Draw in oblique parallel projection four plane-faced bodies, each with exactly 6 vertices, where the first has exactly 5, the second exactly 6, the third exactly 7, and the fourth exactly 8 faces!

Determine the number of all edges for each of these bodies!

Solution:
}

For example, the bodies shown in the figure could be drawn with 9, 10, 11, and 12 edges respectively.
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1312.jpg?height=266&width=1266&top_left_y=798&top_left_x=402)

Answer: 9,10,11,12


Solve the equation in natural numbers $\mathrm{n}$ and $\mathrm{m}$

$(n+1)!(m+1)!=(n+m)!$

Solution:
Answer:

$\{(2 ; 4),(4 ; 2)\}$

## Solution

Obviously, $\mathrm{n}>1$ and $\mathrm{m}>1$ (since when $\mathrm{n}=1$ we get $2(m+1)!=(1+m)!$ and similarly when $\mathrm{m}=1$).

Divide the equation by $(n+1)!$, we get $1 \cdot 2 \cdot 3 \cdot \ldots \cdot(m+1)=(n+2)(n+3) \ldots(n+m)$

Here on the left side there are $\mathrm{m}+1$ factors, on the right side there are $\mathrm{m}-1$.

When $\mathrm{n}>3$ we get $(1 \cdot 2 \cdot 3) \cdot 4 \cdot \ldots \cdot(m+1)=(n+2)(n+3) \ldots(n+m)$

Or $6 \cdot 4 \cdot \ldots \cdot(m+1)=(n+2)(n+3) \ldots(n+m)$

Here we have the same number of factors, and when $\mathrm{n}>3$ we have

$6 \leq n+2$

$4<n+3$

$5<n+4$

$\cdots$

$(m+1)<(n+m)$

Therefore, the equality is impossible. Thus, if (n, m) is a solution, then $1<\mathrm{n}<4$. By the symmetry of the equation, if ($\mathrm{n}, \mathrm{m}$) is a solution, then $1<\mathrm{m}<4$. (These statements are not necessarily true simultaneously).

Now let $\mathrm{n}=2$, then $(1 \cdot 2 \cdot 3) \cdot(1 \cdot 2 \cdot \ldots \cdot(m+1))=1 \cdot 2 \cdot 3 \cdot \ldots \cdot(2+m)$

Or after simplification $(1 \cdot 2 \cdot 3)=(2+m)$

From this, $\mathrm{m}=4$

Now let $\mathrm{n}=3$, then $(1 \cdot 2 \cdot 3 \cdot 4) \cdot(1 \cdot 2 \cdot \ldots \cdot(m+1))=1 \cdot 2 \cdot 3 \cdot \ldots \cdot(3+m)$

Or after simplification $(1 \cdot 2 \cdot 3 \cdot 4)=(2+m)(3+m)$

There are no natural solutions

Answer: (2,4),(4,2)


Solution:
6. $x=1$.

Let $f(t)=2^{t}+3^{t}+t$, then the original equation becomes $f\left(x^{2}-2 x\right)=-\frac{1}{6}=f(-1)$. Since $f(t)$ is monotonically increasing, we have $x^{2}-2 x=-1$, which gives $x=1$.

Answer: 1


The mean (average) of the four integers $78,83,82$, and $x$ is 80 . Which one of the following statements is true?

(A) $x$ is 2 greater than the mean

(B) $x$ is 1 less than the mean

(C) $x$ is 2 less than the mean

(D) $x$ is 3 less than the mean

(E) $x$ is equal to the mean

Solution:
Solution 1

Since 78 is 2 less than 80 and 82 is 2 greater than 80 , the mean of 78 and 82 is 80 .

Since the mean of all four integers is 80 , then the mean of 83 and $x$ must also equal 80 .

The integer 83 is 3 greater than 80 , and so $x$ must be 3 less than 80 .

That is, $x=80-3=77$.

(We may check that the mean of $78,83,82$, and 77 is indeed 80 .)

Solution 2

Since the mean of the four integers is 80 , then the sum of the four integers is $4 \times 80=320$.

Since the sum of 78,83 and 82 is 243 , then $x=320-243=77$.

Therefore, $x$ is 77 which is 3 less than the mean 80 .

ANSWER: (D)

Answer: D


Elements are non-empty sets $S$ of natural numbers that satisfy âif $x \in S$, then $14-x \in S$â. The number of such sets $S$ is $\qquad$ .

Solution: 127 127. From $x, 14-x \in \mathbf{N}$, we know $1 \leqslant x \leqslant 13$. Pairing $1,2, \cdots, 13$ into $\{1,13\}$, $\{2,12\},\{3,11\},\{4,10\},\{5,9\},\{6,8\},\{7\}$. $S$ is the union of some of these 7 sets. Therefore, the number of $S$ is $2^{7}-1=127$.

Answer: 127


Xiao Ming has three identical custom dice, each face of the dice is written with an integer, and the sum of the numbers on opposite faces is equal. When the three dice are thrown at the same time, the numbers on the top faces are 57, 67, and 78, and the numbers on the bottom faces are all prime numbers. Then the sum of the six numbers on each die is $\qquad$

Answer: 240


Determine the maximal number of $120^\circ$ angles in a $7$-gon inscribed in a circle such that all sides of the $7$-gon are of different length.

Solution:
1. **Understanding the Problem:**
   We need to determine the maximal number of \(120^\circ\) angles in a 7-gon inscribed in a circle, where all sides are of different lengths.

2. **Initial Claim:**
   We claim that the maximum number of \(120^\circ\) angles is 2.

3. **Setup:**
   Consider a cyclic heptagon \(ABCDEFG\). If two adjacent angles are equal, then the sides are symmetric with respect to their perpendicular bisector, contradicting our original condition of different-length sides.

4. **Case Analysis:**
   - Suppose there are 3 non-adjacent angles equal to \(120^\circ\), say \(\angle B = \angle D = \angle F = 120^\circ\).
   - Let \(O\) be the circumcenter of the circle.

5. **Angle Calculation:**
   - Since \(\angle B = 120^\circ\), the arc subtended by \(\angle B\) is \(240^\circ\) (since the central angle is twice the inscribed angle).
   - Similarly, \(\angle D = 120^\circ\) implies the arc subtended by \(\angle D\) is \(240^\circ\).
   - \(\angle F = 120^\circ\) implies the arc subtended by \(\angle F\) is \(240^\circ\).

6. **Contradiction:**
   - If \(\angle B = \angle D = \angle F = 120^\circ\), then the central angles \(\angle COA = \angle COE = \angle EOG = 120^\circ\).
   - This implies that the sum of these central angles is \(360^\circ\), which is the total angle around point \(O\).
   - However, this configuration would force \(\angle GOA = 0^\circ\), which is impossible in a geometric figure.

7. **Conclusion:**
   - Therefore, having 3 non-adjacent \(120^\circ\) angles leads to a contradiction.
   - Hence, the maximum number of \(120^\circ\) angles in such a 7-gon is 2.

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

Answer: 2


Solution:
Answer: $\frac{5}{14}$
Solution: Note that $-\frac{-x-2}{-x}=3 \Longrightarrow x=-\frac{1}{2}$. Hence $f(3)=\frac{-\frac{1}{2}-2}{2\left(-\frac{1}{2}\right)-6}=\frac{5}{14}$.

Answer: \frac{5}{14}


As shown in Figure 1, the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ are perpendicular to each other. If $AB=3, BC$ $=4, CD=5$, then the length of $AD$ is ( ).
(A) $3 \sqrt{2}$
(B) 4
(C) $2 \sqrt{3}$
(D) $4 \sqrt{2}$

Solution:
4.A.

As shown in Figure 6, let the four segments into which the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ are divided be $a, b, c, d$. By the Pythagorean theorem, we have
$$
a^{2}+b^{2}=3^{2}, b^{2}+c^{2}=4^{2}, c^{2}+d^{2}=5^{2} \text {. }
$$

Then $a^{2}+d^{2}=18=AD^{2}$. Therefore, $AD=3 \sqrt{2}$.

Answer: A


Let $k$ be a positive real number. Suppose that the set of real numbers $x$ such that $x^2+k|x| \leq 2019$ is an interval of length $6$. Compute $k$.

Solution:
1. **Define the function and analyze its properties:**
   Let \( f(x) = x^2 + k|x| \). Since \( f(x) = f(-x) \), the function is symmetric about the y-axis. This symmetry implies that if \( x \) is a solution, then \( -x \) is also a solution.

2. **Determine the interval:**
   Given that the set of real numbers \( x \) such that \( x^2 + k|x| \leq 2019 \) forms an interval of length 6, we can denote this interval as \([-a, a]\) where \( 2a = 6 \). Therefore, \( a = 3 \).

3. **Evaluate the function at the endpoints of the interval:**
   Since the interval is \([-3, 3]\), we need to have:
   \[
   f(3) = 2019 \quad \text{and} \quad f(-3) = 2019
   \]
   Because of the symmetry, it suffices to consider \( f(3) \):
   \[
   f(3) = 3^2 + k|3| = 9 + 3k
   \]
   Setting this equal to 2019, we get:
   \[
   9 + 3k = 2019
   \]

4. **Solve for \( k \):**
   \[
   3k = 2010
   \]
   \[
   k = \frac{2010}{3} = 670
   \]

The final answer is \( \boxed{670} \).

Answer: 670


For a given positive integer $n(\ge 2)$, find maximum positive integer $A$ such that a polynomial with integer coefficients $P(x)$ with degree $n$ that satisfies the condition below exists.

$\bullet P(1), P(2), \cdots P(A)$ is a multiple of $A$

$\bullet P(0) = 0$ and the coefficient of the first term of $P(x)$ is $1$, In other words, $P(x)$ has the following form:
$$P(x) = c_nx^n + c_{n-1}x^{n-1}+\dots+c_2x^2+x \ \ (c_n \neq 0)$$

Solution:
1. **Claim**: The maximum positive integer \( A \) such that a polynomial \( P(x) \) with integer coefficients and degree \( n \) satisfies the given conditions is \( n! \).

2. **Constructing a Polynomial**: Consider the polynomial \( P(x) = x(x+1)(x+2)\cdots(x+n-1) \). This polynomial has the following properties:
   - It is a monic polynomial of degree \( n \).
   - The coefficient of the linear term is 1.
   - \( P(0) = 0 \).

3. **Verifying the Conditions**:
   - For \( P(i) \) where \( i = 1, 2, \ldots, n \):
     \[
     P(i) = i(i+1)(i+2)\cdots(i+n-1)
     \]
     This can be rewritten using the factorial notation:
     \[
     P(i) = \frac{(i+n-1)!}{(i-1)!}
     \]
     Since \( (i+n-1)! \) is divisible by \( n! \), it follows that \( P(i) \) is a multiple of \( n! \) for \( i = 1, 2, \ldots, n \).

4. **Upper Bound for \( A \)**:
   - We need to show that \( A \leq n! \). Consider the \( n \)-th finite difference of \( P(x) \), denoted by \( \Delta^n P(x) \).
   - For a polynomial \( P(x) \) of degree \( n \), the \( n \)-th finite difference \( \Delta^n P(x) \) is a constant and equals \( n! \).
   - By the given condition, \( P(z) \) is a multiple of \( A \) for all positive integers \( z \). Therefore, \( A \) must divide \( \Delta^n P(0) \), which is \( n! \).

5. **Conclusion**: Since \( A \) must divide \( n! \) and we have constructed a polynomial where \( P(i) \) is a multiple of \( n! \) for \( i = 1, 2, \ldots, n \), the maximum value of \( A \) is \( n! \).

\(\blacksquare\)

The final answer is \( \boxed{ n! } \)

Answer: n!


In Rt $\triangle ABC$, it is known that $\angle A=20^{\circ}, \angle B=90^{\circ}, AD$ is the bisector of $\angle BAC$, point $E$ is on side $AB$, and lines $CE$ and $DE$ are connected. If $\angle DCE=30^{\circ}$, find the degree measure of $\angle ADE$.

Solution:
Solve As shown in Figure 2, construct $\angle A E F=20^{\circ}, E F$ intersects $A D$ at point $F$. Take point $G$ on side $A C$ such that $A G=A E$, and connect $F G$, $E G$, and $C F$.
Since $A D$ bisects $\angle E A G$, by its symmetry we know
$$
F E=F G,
$$

and
$$
\begin{array}{l}
\angle G F D=\angle E F D \\
=\angle A E F+\angle E A F \\
=20^{\circ}+10^{\circ}=30^{\circ} .
\end{array}
$$

Therefore, $\triangle F E G$ is an equilateral triangle.
Thus, $G F=G E$.
It is easy to see that $\angle A G E=80^{\circ}, \angle G C E=40^{\circ}$.
Then $\angle G E C=\angle A G E-\angle G C E$
$$
=40^{\circ}=\angle G C E \text {. }
$$

Therefore, $G C=G E=G F$.
Thus, $\angle G C F=\frac{1}{2} \angle A G F=\frac{1}{2} \times 20^{\circ}=10^{\circ}$.
Hence $\angle F C E=\angle G C E-\angle G C F=30^{\circ}$.
Since $\angle E F D=\angle E C D=30^{\circ}$, points $E$, $F$, $C$, and $D$ are concyclic.
Therefore, $\angle A D E=\angle F C E=30^{\circ}$.
(Wan Xiren, Hunan Library Training Building Wanxi School, 410011)

Answer: 30^{\circ}


$1 \times 1=1, 2 \times 2=4, 3 \times 3=9, 4 \times 4=16, \ldots \ldots$, the products $1, 4, 9, 16, \ldots \ldots$ in the expressions are called perfect squares. Among the 100 natural numbers from 1 to 100, the sum of all non-perfect square numbers is $\qquad$

Solution: Reference answer: 4665

Answer: 4665


Solution:
## Solution.

Domain of definition: $0<x \neq 1$

Switch to base 2. We have $\frac{1}{\log _{2} x}-\frac{1}{2} \log _{2} x+\frac{7}{6}=0 \Leftrightarrow$ $\Leftrightarrow 3 \log _{2}^{2} x-7 \log _{2} x-6=0$. Solving this equation as a quadratic in terms of $\log _{2} x$, we find $\left(\log _{2} x\right)_{1}=-\frac{2}{3}$ or $\left(\log _{2} x\right)_{2}=3$, from which $x_{1}=\frac{1}{\sqrt[3]{4}}, x_{2}=8$.

Answer: $\quad \frac{1}{\sqrt[3]{4}} ; 8$.

Answer: \frac{1}{\sqrt[3]{4}};8


How many six-digit numbers exist where each digit in an even position is one greater than the digit to its left (digits are numbered from left to right)?

Solution: 9. \(8 \cdot 9 \cdot 9=648\). Hint. The six-digit numbers in question are determined by the choice of the first, third, and fifth digits. (Kvant, 1972, No. 5, p. 81; MSU, 1971)

Answer: 648