Suppose that Ethan has four red chips and two white chips. He selects three chips at random and places them in Urn 1, while the remaining chips are placed in Urn 2. He then lets his brother Josh draw one chip from each urn at random. What is the probability that the chips drawn by Josh are both red?

Solution:
Answer: $\frac{2}{5}$
$\underline{\text { Solution: The possibilities are }}$
\begin{tabular}{ccc}
\hline Urn 1 & Urn 2 & Number of Ways \\
\hline 1R, 2W & $3 \mathrm{R}$ & $\binom{4}{1}\binom{2}{2}=4$ \\
$2 \mathrm{R}, 1 \mathrm{~W}$ & $2 \mathrm{R}, 1 \mathrm{~W}$ & $\binom{4}{2}\binom{2}{1}=12$ \\
$3 \mathrm{R}$ & $1 \mathrm{R}, 2 \mathrm{~W}$ & $\binom{4}{3}\binom{2}{0}=4$ \\
\hline
\end{tabular}

The total number of ways to split the chips into the two urns is twenty. Hence, the probability of getting both red chips is
$$
\frac{4}{20}\left(\frac{1}{3}\right)+\frac{12}{20}\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)+\frac{4}{20}\left(\frac{1}{3}\right)=\frac{2}{5} .
$$

Answer: \frac{2}{5}


Given triangle $ABC$, where angle $B$ is $30^{\circ}, AB=4, BC=6$. The bisector of angle $B$ intersects side $AC$ at point $D$. Find the area of triangle $ABD$.

Solution:
Apply the property of the bisector of a triangle.

## Solution

Note that

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} A B \cdot B C \cdot \sin \angle B=6
$$

By the property of the bisector of a triangle,

$$
\frac{A D}{C D}=\frac{A B}{B C}=\frac{2}{3}
$$

Therefore, $\frac{A D}{A C}=\frac{2}{5}$. Hence,

$$
S_{\triangle \mathrm{ABD}}=\frac{A D}{A C} S_{\Delta \mathrm{ABC}}=\frac{2}{5} \cdot 6=2.4
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-22.jpg?height=277&width=924&top_left_y=2217&top_left_x=567)

Answer

2.4 .

Submit a comment

Answer: 2.4


Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{n}+S_{n}=n$, the sequence $\left\{b_{n}\right\}$ satisfies $b_{1}=a_{1}$, and $b_{n}=a_{n}-a_{n-1}(n \geqslant 2)$.
(1) Find the general term formula for $\left\{a_{n}\right\}$;
(2) Find the general term formula for $\left\{b_{n}\right\}$;
(3) Find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$, $B_{n}$.

Solution:
15. (1) Since $a_{n}+S_{n}=n, a_{n-1}+S_{n-1}=n-1$, subtracting the two equations gives $2 a_{n}-a_{n-1}=1$, then $2\left(a_{n}-\right.$ $1)=\left(a_{n-1}-1\right)$, hence $\left\{a_{n}-1\right\}$ is a geometric sequence, so $a_{n}-1=\left(a_{1}-1\right) \cdot\left(\frac{1}{2}\right)^{n-1}$. And $a_{1}+S_{1}=1$, thus $a_{1}=\frac{1}{2}$, therefore $a_{n}-1=-\left(\frac{1}{2}\right)^{n}$, which means $a_{n}=1-\left(\frac{1}{2}\right)^{n}$;
(2) $b_{n}=a_{n}-a_{n-1}=\left(\frac{1}{2}\right)^{n-1}-\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{n}(n \geqslant 2)$, and $b_{1}=a_{1}=\frac{1}{2}$, hence $b_{n}=$ $\left(\frac{1}{2}\right)^{n}(n \geqslant 1)$
(3) $b_{1}+b_{2}+\cdots+b_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\cdots+\left(a_{n}-a_{n-1}\right)=a_{n}=1-\left(\frac{1}{2}\right)^{n}=B_{n}$.

Answer: 1-(\frac{1}{2})^{n}


The value of the expression $10-10.5 \div[5.2 \times 14.6-(9.2 \times 5.2+5.4 \times 3.7-4.6 \times 1.5)]$ multiplied by 20 is

Solution:
【Answer】186
$$
\begin{aligned}
\text { [Analysis] Original expression } & =10-10.5 \div[5.2 \times(14.6-9.2)-5.4 \times 3.7+4.6 \times 1.5] \\
& =10-10.5 \div(5.4 \times 1.5+4.6 \times 1.5) \\
& =10-10.5 \div 15 \\
& =9.3 \\
& \Rightarrow 9.3 \times 20=186
\end{aligned}
$$

Answer: 186


Let S be a spherical shell radius 1. Find the average straight line distance between two points of S. [In other words S is the set of points (x, y, z) with x^2 + y^2 + z^2 = 1). Solution

Solution: : 4/3. It is sufficient to fix one point and to find the average distance of the other points from it. Take the point as P and the centre as O. Now consider a general point Q on the surface. Let angle POQ = θ. The distance PQ is 2 sin θ/2 and this is the same for all points in a band angular width dθ at the angle θ. The band has radius sin θ. Hence the average distance is 1/4π ∫ 0 π (2π sin θ) (2 sin θ/2) dθ = ∫ 4 sin 2 θ/2 d(sin θ/2) = 4/3. 18th Putnam 1958 © John Scholes jscholes@kalva.demon.co.uk 25 Feb 2002

Answer: \frac{4}{3}


How many rational solutions for $x$ are there to the equation $x^4+(2-p)x^3+(2-2p)x^2+(1-2p)x-p=0$ if $p$ is a prime number?

Solution:
To determine the number of rational solutions for the equation

\[ x^4 + (2 - p)x^3 + (2 - 2p)x^2 + (1 - 2p)x - p = 0 \]

where \( p \) is a prime number, we can use the Rational Root Theorem. The Rational Root Theorem states that any rational solution, expressed as a fraction \(\frac{p}{q}\), must have \( p \) as a factor of the constant term and \( q \) as a factor of the leading coefficient.

1. **Identify the constant term and the leading coefficient:**
   - The constant term is \(-p\).
   - The leading coefficient is \(1\).

2. **List the possible rational roots:**
   - The factors of the constant term \(-p\) are \(\pm 1, \pm p\).
   - The factors of the leading coefficient \(1\) are \(\pm 1\).
   - Therefore, the possible rational roots are \(\pm 1, \pm p\).

3. **Test the possible rational roots:**
   - We need to check if \( x = 1, -1, p, -p \) are solutions to the polynomial equation.

4. **Check \( x = 1 \):**
   \[
   1^4 + (2 - p)1^3 + (2 - 2p)1^2 + (1 - 2p)1 - p = 1 + (2 - p) + (2 - 2p) + (1 - 2p) - p
   \]
   \[
   = 1 + 2 - p + 2 - 2p + 1 - 2p - p = 4 - 6p
   \]
   Since \( 4 - 6p \neq 0 \) for any prime \( p \), \( x = 1 \) is not a solution.

5. **Check \( x = -1 \):**
   \[
   (-1)^4 + (2 - p)(-1)^3 + (2 - 2p)(-1)^2 + (1 - 2p)(-1) - p = 1 - (2 - p) + (2 - 2p) - (1 - 2p) - p
   \]
   \[
   = 1 - 2 + p + 2 - 2p - 1 + 2p - p = 0
   \]
   Therefore, \( x = -1 \) is a solution.

6. **Check \( x = p \):**
   \[
   p^4 + (2 - p)p^3 + (2 - 2p)p^2 + (1 - 2p)p - p = p^4 + 2p^3 - p^4 + 2p^2 - 2p^3 + p - 2p^2 - p
   \]
   \[
   = p^4 - p^4 + 2p^3 - 2p^3 + 2p^2 - 2p^2 + p - p = 0
   \]
   Therefore, \( x = p \) is a solution.

7. **Check \( x = -p \):**
   \[
   (-p)^4 + (2 - p)(-p)^3 + (2 - 2p)(-p)^2 + (1 - 2p)(-p) - p = p^4 - (2 - p)p^3 + (2 - 2p)p^2 - (1 - 2p)p - p
   \]
   \[
   = p^4 - 2p^3 + p^4 + 2p^2 - 2p^3 - p + 2p^2 - p
   \]
   \[
   = p^4 - p^4 - 2p^3 + 2p^3 + 2p^2 - 2p^2 - p + p = 0
   \]
   Therefore, \( x = -p \) is a solution.

Since we have found that \( x = -1 \) and \( x = p \) are solutions, we conclude that there are two rational solutions.

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

Answer: 2


Find the area of the axial section of a cylinder inscribed in a unit cube such that the axis of the cylinder lies on the diagonal of the cube, and each base touches three faces of the cube at their centers.

Solution:
Let the circle of radius $r$ of one of the bases of the cylinder with height $h$ touch the faces $ABCD$, $CC_1D_1D$, and $AA_1D_1D$ of the unit cube $ABCD A_1B_1C_1D_1$ at points $K$, $L$, and $M$ respectively. Since $K$, $L$, and $M$ are the midpoints of the diagonals $AC$, $CD_1$, and $AD_1$ of the corresponding faces of the cube, this circle is inscribed in the equilateral triangle $ACD_1$ with side length $\sqrt{2}$. Therefore,

$$
r = \sqrt{2} \cdot \frac{\sqrt{3}}{6} = \frac{\sqrt{6}}{6}
$$

The circle of the second base of the cylinder is inscribed in the equilateral triangle $BA_1C_1$. The axis of the cylinder passes through the centers of these circles and is perpendicular to the planes $ACD_1$ and $BA_1C_1$. It is known that these planes divide the diagonal of the cube into three equal parts. Therefore,

$$
h = \frac{1}{3} CB_1 = \frac{\sqrt{3}}{3}
$$

Thus, the area of the axial section of the cylinder is

$$
2 r h = 2 \cdot \frac{\sqrt{6}}{6} \cdot \frac{\sqrt{3}}{3} = \frac{\sqrt{2}}{3}
$$

## Answer

$\frac{\sqrt{2}}{3}$.

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


What is the greatest number of circles of radius 1 that can be placed on a plane so that they all intersect a certain fixed unit circle $S$ and no one of them contains the center of $S$ or the center of another circle inside it?

Solution:
50. Fig. 134 shows that 18 circles satisfying the condition of the problem can be placed (six circles in Fig. 134 have centers at the vertices of a regular hexagon inscribed in $S$, the other 12 - at the vertices of squares constructed on the sides of the inscribed hexagon outside it; compare with Fig. 130). The fact that more than 18 circles cannot be placed follows from the result of problem 48: otherwise, the centers of all circles (including the center $O$ of circle $S$) would form a system of 20 or more points enclosed within a circle of radius 2 centered at one of these points $O$ and such that the distance between any two of these points is not less than 1, while according to the theorem of problem 48, such a system of points cannot exist.

![](https://cdn.mathpix.com/cropped/2024_05_21_962b75a91b7185035ab9g-165.jpg?height=587&width=590&top_left_y=510&top_left_x=180)

Fig. 134.

Answer: 18


If $r, s, t, u$ are positive integers and $r^{5}=s^{4}, t^{3}=u^{2}, t-r=19$ and $d=u-s$, find the value of $d$.

Solution:
Let $w=u^{\frac{1}{15}}, v=s^{\frac{1}{15}}$, then $t=u^{\frac{2}{3}}=u^{\frac{10}{15}}=w^{10}, r=s^{\frac{4}{5}}=s^{\frac{12}{15}}=v^{12}$
$$
\begin{array}{l}
t-r=19 \Rightarrow w^{10}-v^{12}=19 \\
\Rightarrow\left(w^{5}+v^{6}\right)\left(w^{5}-v^{6}\right)=19 \times 1
\end{array}
$$
$\because 19$ is a prime number, $w^{5}+v^{6}=19, w^{5}-v^{6}=1$
Solving these equations give $w^{5}=10, v^{6}=9 \Rightarrow w^{5}=10, v^{3}=3$
$$
\begin{array}{l}
u=w^{15}=1000, s=v^{15}=3^{5}=729 \\
d=u-s=1000-243=757
\end{array}
$$

Answer: 757


The function $y=f(x)$ defined on $\mathbf{R}$ has the following properties:
(1) For any $x \in \mathbf{R}$, $f\left(x^{3}\right)=f^{3}(x)$;
(2) For any $x_{1}, x_{2} \in \mathbf{R}, x_{1} \neq x_{2}$, $f\left(x_{1}\right) \neq f\left(x_{2}\right)$. Then the value of $f(0)+f(1)+f(-1)$ is $\qquad$.

Solution: 10. 0. Hint: From $f(0)=f^{3}(0)$, we know $f(0)[1-f(0)][1+f(0)]=0$, therefore, $f(0)=0$ or $f(0)=1$, or $f(0)=-1$; from $f(1)=f^{3}(1)$, similarly $f(1)=0$ or 1 or -1; from $f(-1)=f^{3}(-1)$, similarly $f(-1)=0$ or 1 or -1. But $f(0), f(1), f(-1)$ are pairwise distinct, hence $\{f(0), f(1), f(-1)\}=\{0,1,-1\}$, from which it follows that $f(0)+$ $f(1)+f(-1)=0$.

Answer: 0


How many Dyck paths of length $2 n$ are there?

Solution:
Before starting the proof, note that we can obtain a recurrence relation on the number $C_{n}$ of Dyck paths of length $2 n$. Indeed, consider a Dyck path of length $n$. Then let $0 \leq k < n$ be the largest integer such that the path touches the x-axis at step $2 k$ (by a parity argument, if the path touches the x-axis, it does so at an even abscissa). For a given $k$, to construct a Dyck path of length $2 n$, it is necessary and sufficient to construct one of length $2 k$, then go up, then construct a Dyck path of length $2(n-k-1)$ that does not touch the x-axis, and then go down one step (on the image, the dotted lines should be replaced by Dyck paths). Thus, we show that $C_{n} = \sum_{k=0}^{n-1} C_{k} C_{n-k-1}$.

![](https://cdn.mathpix.com/cropped/2024_05_10_bca2b404556ff27bef16g-65.jpg?height=128&width=420&top_left_y=1501&top_left_x=1161)

With this relation, we could verify the result by induction. However, we will seek a more direct method to obtain $C_{n}$. Consider the function $f$ as follows: consider a path that is not a Dyck path but ends at $(2 n, 0)$. Let $2 k + 1$ be the smallest abscissa where the path is below the x-axis. The function $f$ associates with this path the path symmetric with respect to the x-axis between the abscissas 0 and $2 k + 1$, followed by the same path translated 2 units upwards (as illustrated below for $n=14$ on a particular path).

![](https://cdn.mathpix.com/cropped/2024_05_10_bca2b404556ff27bef16g-65.jpg?height=429&width=1514&top_left_y=2084&top_left_x=271)

![](https://cdn.mathpix.com/cropped/2024_05_10_bca2b404556ff27bef16g-66.jpg?height=623&width=1516&top_left_y=171&top_left_x=264)

We verify that this function is indeed a bijection between the paths ending at $(2 n, 0)$ that are not Dyck paths and the paths ending at $(2 n, 2)$, by constructing the inverse of $f$. Thus, we have $C_{n} = \binom{2 n}{n} - \binom{2 n}{n+1} = \frac{1}{n+1} \binom{2 n}{n}$. These are the Catalan numbers, which have numerous combinatorial interpretations!

## More Difficult Exercises

Answer: \frac{1}{n+1}\binom{2n}{n}


Let positive integers $a, b$ make $15a + 16b$ and $16a - 15b$ both squares of positive integers. Find the smallest value that the smaller of these two squares can take.

Solution:
Let positive integers $a, b$ be such that $15a + 16b$ and $16a - 15b$ are both squares of positive integers, i.e.,
$$
15a + 16b = r^2, \quad 16a - 15b = s^2 \quad (r, s \in \mathbb{N}).
$$
$$
\text{Thus, } 15^2 a + 16^2 a = 15r^2 + 16s^2,
$$

i.e., $481a = 15r^2 + 16s^2$;
$$
16^2 b + 15^2 b = 16r^2 - 15s^2,
$$

i.e., $481b = 16r^2 - 15s^2$.
Therefore, $15r^2 + 16s^2$ and $16r^2 - 15s^2$ are both multiples of 481. We will prove that $r$ and $s$ are both multiples of 481.

Since $481 = 13 \times 37$, it suffices to prove that $r$ and $s$ are both multiples of 13 and 37.
First, we prove that $r$ and $s$ are both multiples of 13 by contradiction.
Since $16r^2 - 15s^2$ is a multiple of 13, then $13 \nmid r + 13 \nmid s$.
Because $16r^2 \equiv 15s^2 \pmod{13}$. So,
$16r^2 \cdot s^{10} \equiv 15s^{12} \equiv 15 \equiv 2 \pmod{13}$.
Since the left side is a perfect square, taking the sixth power on both sides, we have $\left(\left(4rs^5\right)^2\right)^6 \equiv 1 \pmod{13}$,

thus $2^6 = 1 \pmod{13}$, which is a contradiction.
Next, we prove that $37 \mid r$ and $37 \mid s$ by contradiction.
Assume $37 \nmid r$ and $37 \nmid s$.
Since $37 \mid 15r^2 + 16s^2$ and $37 \mid 16r^2 - 15s^2$, we have
$37 \mid r^2 - 31s^2$, i.e., $r^2 \equiv 31s^2 \pmod{37}$.
Therefore, $r^2 s^{34} = 31s^{36} = 31 \pmod{37}$.
$31^{14} = 1 \pmod{37}$.

But $21^2 \equiv (31^2)^9 \equiv ((-6)^2)^9 \equiv 36^9 \equiv (-1)^9 \equiv -1 \pmod{37}$, which is a contradiction.
Thus, $481 \mid r$ and $481 \mid s$. Clearly, these two perfect squares are both at least $481^2$.

On the other hand, when $a = 481 \times 31$ and $b = 481$, both perfect squares are $481^2$. Therefore, the minimum value sought is $481^2$.

Answer: 481^2


Martin is trying to fill each cell of a rectangular grid with 8 rows and $n$ columns with one of the four letters $\mathrm{P}, \mathrm{O}, \mathrm{F}$, and $\mathrm{M}$ such that for any pair of distinct rows, there is at most one column where the intersections of the two rows are cells with the same letter. What is the largest integer $\mathrm{n}$ for which this is possible?

Solution:
Solution to Exercise 12 In this problem, we are looking for the largest integer satisfying a certain property. Suppose we want to show that the largest integer sought is the integer c. To show that c is indeed the largest integer, we will on the one hand show that if an integer n satisfies the property, then n ≤ c, and on the other hand we will show that we can find a table with exactly c columns.

The statement presents a hypothesis concerning the rows and columns of a table.

We start by looking at the information we have about the rows of this table. There are 8 rows, so there are \(\binom{8}{2} = 28\) pairs of rows.

We then look at the information we have about the columns of the table. Since the hypothesis in the statement mentions identical letters within the same column, we will count how many pairs of identical letters belong to the same column. We fix a column of the table and see how many pairs of identical letters we can form at a minimum. After several trials on a column of size 8, we conjecture that there are always at least 4 pairs of identical letters within the same column. Indeed, in a column there are 8 letters, so a letter appears at least 2 times.

- If each letter appears at most 2 times, then each letter appears exactly 2 times since there are 8 cells in the column and 4 possible letters. We thus have 4 pairs of identical letters in the column.
- If a letter appears exactly 3 times, say the letter P, then we can form 3 pairs of letters P. Among the 5 remaining letters, at least one of the letters O, F, and M appears twice according to the pigeonhole principle, which gives us a fourth pair of identical letters.
- If a letter appears at least 4 times in the column, say the letter P, we can form 4 pairs of letters P.

Thus, in a column, we can always find 4 pairs of identical letters. Therefore, for each column, there are at least 4 pairs of rows such that the intersections with the column have the same letter. But according to the hypothesis in the statement, a pair of rows can be associated with at most one column such that its intersections with the two rows are two cells containing the same letter. We deduce that there are at most \(4 \times n\) pairs of rows. Therefore, \(4n \leq 28\) and \(n \leq 7\).

Conversely, there exists a configuration with the following 7 columns:

| P | P | P | P | P | P | P |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| O | O | O | O | O | P | O |
| F | F | F | F | O | O | P |
| M | M | M | M | P | O | O |
| P | M | F | O | F | F | F |
| O | F | M | P | M | F | M |
| F | P | P | M | M | M | F |
| M | O | O | F | F | M | M |

The largest integer n of columns is therefore n = 7.

Comment from the graders

Many students managed to find the answer n = 7 and provided an example. Most propose a correct argument to show that it is optimal, but sometimes this reasoning is written in a somewhat unclear manner.

Answer: 7


For the linear function $f$, the following holds:

a) If its slope is increased by 1 and the y-intercept by 2, its value at some $x_{0}$ increases by 5.

b) If this $x_{0}$ is decreased by 2, the function $f$ has six times the value at $x_{0}$.

c) The graph of the function $f$ passes through the point $A(4,-3)$.

Write the appropriate relationships and calculate the formula for $f(x)$ and the value of $x_{0}$.

Solution:
B1. A linear function has the formula $f(x)=k \cdot x+n$.

At $x_{0}$ it has the value $y_{0}=f\left(x_{0}\right)=k \cdot x_{0}+n$. If the slope $k$ is increased by 1 and the y-intercept $n$ by 2, we get the function $g(x)=(k+1) \cdot x+n+2$. Substituting $x_{0}$, we get $g\left(x_{0}\right)=(k+1) \cdot x_{0}+n+2=$ $k \cdot x_{0}+x_{0}+n+2=y_{0}+x_{0}+2$. Since in this case the value of the function $y_{0}=f\left(x_{0}\right)$ increases by 5, we get $x_{0}+2=5$ and $x_{0}=3$.

At $x_{0}-2=3-2=1$, the function $f(x)$ has the value $f(1)=k \cdot 1+n=k+n$. Considering that this is six times the value of $y_{0}=f\left(x_{0}\right)=f(3)=k \cdot 3+n$, we get the equation $k+n=6(3 k+n)$. This simplifies to $17 k+5 n=0$.

If the graph of the function $f(x)$ passes through the point $A(4,-3)$, then $f(4)=-3$. We get the equation $4 k+n=-3$. Solving the system of equations $17 k+5 n=0$ and $4 k+n=-3$, we get $k=-5$ and $n=17$. The sought linear function thus has the formula $f(x)=-5 x+17$.

Considering the condition in a) for example, $g(x)=(k+1) \cdot x+n+2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots . \ldots$

![](https://cdn.mathpix.com/cropped/2024_06_07_37ea02f5a530d200baf4g-20.jpg?height=57&width=1727&top_left_y=1896&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_37ea02f5a530d200baf4g-20.jpg?height=54&width=1727&top_left_y=1943&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_37ea02f5a530d200baf4g-20.jpg?height=57&width=1727&top_left_y=1996&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_37ea02f5a530d200baf4g-20.jpg?height=57&width=1727&top_left_y=2045&top_left_x=176)

Writing the appropriate equation for condition c) for example, $4 k+n=-3$...................................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_37ea02f5a530d200baf4g-20.jpg?height=48&width=1727&top_left_y=2146&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_37ea02f5a530d200baf4g-20.jpg?height=60&width=1731&top_left_y=2192&top_left_x=174)

Answer: f(x)=-5x+17x_0=3


Calculate the limit of the function:

$\lim _{x \rightarrow 1}\left(\frac{1+\cos \pi x}{\tan^{2} \pi x}\right)^{x^{2}}$

Solution:
## Solution

$\lim _{x \rightarrow 1}\left(\frac{1+\cos \pi x}{\tan^{2} \pi x}\right)^{x^{2}}=\lim _{x \rightarrow 1}\left(\frac{1+1-2 \sin ^{2}\left(\frac{\pi x}{2}\right)}{\tan^{2} \pi x}\right)^{x^{2}}=$

$$
\begin{aligned}
& =\lim _{x \rightarrow 1}\left(\frac{2-2 \sin ^{2}\left(\frac{\pi x}{2}\right)}{\tan^{2} \pi x}\right)^{x^{2}}=\lim _{x \rightarrow 1}\left(\frac{2\left(1-\sin ^{2}\left(\frac{\pi x}{2}\right)\right)}{\tan^{2} \pi x}\right)^{x^{2}}= \\
& =\lim _{x \rightarrow 1}\left(\frac{2 \cos ^{2}\left(\frac{\pi x}{2}\right)}{\tan^{2} \pi x}\right)^{x^{2}}=\left(\lim _{x \rightarrow 1} \frac{2 \cos ^{2}\left(\frac{\pi x}{2}\right)}{\tan^{2} \pi x}\right)^{\lim _{x \rightarrow 1} x^{2}}= \\
& =\left(\lim _{x \rightarrow 1} \frac{2 \cos ^{2}\left(\frac{\pi x}{2}\right)}{\tan^{2} \pi x}\right)^{1^{2}}=\lim _{x \rightarrow 1} \frac{2 \cos ^{2}\left(\frac{\pi x}{2}\right)}{\tan^{2} \pi x}=
\end{aligned}
$$

Substitution:

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

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

We get:

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

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_08a731a792d2ab45f373g-15.jpg?height=146&width=845&top_left_y=1564&top_left_x=152)

$\tan \pi y \sim \pi y$, as $y \rightarrow 0(\pi y \rightarrow 0)$

We get:

$$
=\lim _{y \rightarrow 0} \frac{2\left(\frac{\pi y}{2}\right)^{2}}{(\pi y)^{2}}=\lim _{y \rightarrow 0} \frac{(\pi y)^{2}}{2(\pi y)^{2}}=\lim _{y \rightarrow 0} \frac{1}{2}=\frac{1}{2}
$$

## Problem Kuznetsov Limits 20-30

Answer: \frac{1}{2}


Solution:
9.3280 .

Let $f(x)=(2 x-1)^{8}$, then
$$
\begin{array}{l}
a_{0}=f(0)=1, \\
a_{8}+a_{7}+\cdots+a_{0}=f(1)=1, \\
a_{8}-a_{7}+a_{6}-a_{5}+a_{4}-a_{3}+a_{2}-a_{1}+a_{0} \\
=f(-1)=(-3)^{8}=6561 .
\end{array}
$$

Therefore, $a_{8}+a_{6}+a_{4}+a_{2}$
$$
=\frac{1}{2}(f(1)+f(-1))-f(0)=3280 \text {. }
$$

Answer: 3280


In square $ABCD$ with side length $1$, there are points $P, Q$ on $AB, AD$ respectively. If the perimeter of $\triangle APQ$ is 2, find $\angle PCQ$.

Solution:
2. Take the lines $A B, C D$ as coordinate axes, establish a coordinate system, and solve using the analytical method.

Let $P(a, 0), Q(0, b)$, then $k_{C P}=\frac{1}{1-a}, k_{C Q}=1-b$.
From the perimeter of $\triangle A P Q$ being 2, we get $a b=2(a+b)-2$.
Since $\operatorname{tg} \angle P C Q=\frac{2-(a+b)}{2-(a+b)}=1$, and $\angle P C Q$ is an acute angle, thus $\angle P C Q=45^{\circ}$.

Answer: 45


Discover which positive integers $x$ and $y$ satisfy the equation $x^{4}=y^{2}+71$.

Solution:
The equation can be written in the form $x^{4}-y^{2}=71$. Now, factoring $x^{4}-y^{2}$ we have:

$$
\left(x^{2}-y\right)\left(x^{2}+y\right)=71(*)
$$

Since $x$ and $y$ are integers, each of the factors $\left(x^{2}-y\right)$ and $\left(x^{2}+y\right)$ is also an integer. Therefore, in $\left(^{*}\right)$, we write 71 as the product of two integers. Since 71 is a prime number, it can only be written as the product of integers in the form: $71=1 \times 71$. Thus, we have two cases to consider: $\left(x^{2}-y\right)=1$ and $\left(x^{2}+y\right)=71$, or $\left(x^{2}-y\right)=71$ and $\left(x^{2}+y\right)=1$.

Let's study each case.
1st case: $\left\{\begin{array}{l}x^{2}-y=1 \\ x^{2}+y=71\end{array}\right.$.

Adding the two equations, we get: $2 x^{2}=72$, which implies $x= \pm 6$. Therefore, $y=( \pm 6)^{2}-1=35$. Since $x, y$ are positive integers, we conclude that the solution in this first case is: $x=6$ and $y=35$.

2nd case: $\left\{\begin{array}{l}x^{2}-y=71 \\ x^{2}+y=1\end{array}\right.$.

If $x^{2}+y=1$, then $x=0$ and $y=1$ or $x=1$ and $y=0$ since $x, y$ are positive integers. On the other hand, it is easy to verify that these values do not satisfy the equation $x^{4}=y^{2}+71$.

Therefore, the solution to the problem is: $x=6$ and $y=35$.

Answer: 6,35


Let $ABC$ be an equilateral triangle with side length $1$. Points $A_1$ and $A_2$ are chosen on side $BC$, points $B_1$ and $B_2$ are chosen on side $CA$, and points $C_1$ and $C_2$ are chosen on side $AB$ such that $BA_1<BA_2$, $CB_1<CB_2$, and $AC_1<AC_2$.
Suppose that the three line segments $B_1C_2$, $C_1A_2$, $A_1B_2$ are concurrent, and the perimeters of triangles $AB_2C_1$, $BC_2A_1$, and $CA_2B_1$ are all equal. Find all possible values of this common perimeter.

[i]Ankan Bhattacharya[/i]

Solution:
1. **Setup and Initial Conditions:**
   - Let $ABC$ be an equilateral triangle with side length $1$.
   - Points $A_1$ and $A_2$ are chosen on side $BC$ such that $BA_1 < BA_2$.
   - Points $B_1$ and $B_2$ are chosen on side $CA$ such that $CB_1 < CB_2$.
   - Points $C_1$ and $C_2$ are chosen on side $AB$ such that $AC_1 < AC_2$.
   - The line segments $B_1C_2$, $C_1A_2$, and $A_1B_2$ are concurrent.
   - The perimeters of triangles $AB_2C_1$, $BC_2A_1$, and $CA_2B_1$ are all equal.

2. **Trisection and Concurrency:**
   - To show that the common perimeter can be $1$, we can trisect each side of the equilateral triangle.
   - Let $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, and $C_2$ trisect the sides $BC$, $CA$, and $AB$ respectively.
   - This means $BA_1 = A_1A_2 = A_2C = \frac{1}{3}$, and similarly for the other points.

3. **Calculation of Perimeters:**
   - The perimeter of $\triangle AB_2C_1$ is:
     \[
     AB_2 + B_2C_1 + C_1A = \frac{2}{3} + \frac{1}{3} + \frac{2}{3} = 1
     \]
   - Similarly, the perimeters of $\triangle BC_2A_1$ and $\triangle CA_2B_1$ are also $1$.

4. **Concurrency at the Center:**
   - The line segments $B_1C_2$, $C_1A_2$, and $A_1B_2$ concur at the centroid of $\triangle ABC$.
   - This is because the points trisecting the sides of an equilateral triangle and the lines connecting these points to the opposite vertices will always meet at the centroid.

5. **Proof of Uniqueness:**
   - Assume $p \neq 1$ and consider the excircles $\omega_A$, $\omega_B$, and $\omega_C$ of triangles $AB_2C_1$, $BC_2A_1$, and $CA_2B_1$ respectively.
   - By tangent segments being equal, if $p$ is the common perimeter:
     \[
     AT_A + AS_A = AB_2 + B_2T_A + AC_1 + C_1S_A = AB_2 + AC_1 + B_2U_A + C_1U_A = AB_2 + AC_1 + B_2C_1 = p
     \]
   - By symmetry, we obtain $AT_A + AS_A = BT_B + BS_B = CT_C + CS_C = p$, so $\omega_A$, $\omega_B$, and $\omega_C$ are congruent.

6. **Contradiction for $p \neq 1$:**
   - Define $B_2'$, $C_1'$, $C_2'$, $A_1'$, $A_2'$, and $B_1'$ such that $B_2'C_1'$ is tangent to the incircle $\omega$ of $\triangle ABC$ and $B_2'C_1' \parallel B_2C_1$.
   - Consider the dilation of factor $\frac{1}{p}$ centered at $A$. This sends $\omega_A \mapsto \omega$, $B_2 \mapsto B_2'$, and $C_1 \mapsto C_2'$.
   - Perform the same dilation at $B$ and $C$.
   - This implies $C_1A_2 \parallel C_1'A_2'$ and similar parallel lines.
   - Let $P$ be the concurrency point of $B_1'C_2'$, $C_1'A_2'$, and $A_1'B_2'$ by Brianchon's theorem.
   - If $p < 1$, then $X$ lies on the strict interior of $\triangle PB_2'C_1'$, $Y$ in $\triangle PC_2'A_1'$, and $Z$ in $\triangle PA_2'B_1'$, leading to a contradiction.
   - If $p > 1$, the same argument applies, leading to a contradiction.

7. **Conclusion:**
   - Therefore, the only possible value for the common perimeter is $1$.

The final answer is $\boxed{1}$.

Answer: 1


For the cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$, draw a line $l$ through vertex $A_{1}$ such that $l$ forms an angle of $60^{\circ}$ with both lines $A C$ and $B C_{1}$. How many such lines $l$ are there?
(A) 1
(B) 2
(C) 3
(D) More than 3

Solution:
3.C.

Obviously, $A D_{1} // B C_{1}$. There are 3 lines passing through point $A$ that form a $60^{\circ}$ angle with both lines $A C$ and $A D_{1}$. Therefore, there are also 3 lines $l$ passing through point $A_{1}$ that satisfy the conditions.

Answer: C


Oleg drew an empty $50 \times 50$ table and wrote a number above each column and to the left of each row. It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 being irrational. Then, in each cell of the table, he wrote the product of the numbers written next to its row and its column ("multiplication table"). What is the maximum number of products in this table that could be rational numbers?

Solution:
Evaluation. Suppose that among the rational numbers there is 0 and it is written at the top side of the table. Let along the left side of the table be written $x$ irrational and $50-x$ rational numbers. Then along the top side are written $50-x$ irrational and $x$ rational numbers. Note that the product of a non-zero rational and an irrational number is irrational. Therefore, in the table, there are at least $x(x-1)+(50-x)^{2} = 2x^{2}-101x+50^{2}$ irrational numbers. The vertex of the parabola $y=2x^{2}-101x+50^{2}$ is at the point 25.25, so the minimum value of this function at an integer point is achieved at $x=25$ and is equal to $25 \cdot 24 + 25^{2} = 1225$. If 0 is replaced by a non-zero rational number, the number of irrational numbers can only increase. Therefore, in the table, there are no more than $2500-1225=1275$ rational numbers.

Example, when there are 1275 rational numbers. Along the left side, place the numbers 1, 2, ..., 24, 25,

$\sqrt{2}, 2\sqrt{2}, \ldots, 25\sqrt{2}$, and along the top - the numbers $0, 26, 27, \ldots, 49, 26\sqrt{2}, 27\sqrt{2}, \ldots, 50\sqrt{2}$. Then

the irrational numbers will be only $25 \cdot 24 + 25^{2} = 1225$ products of non-zero rational and

irrational numbers.

## Answer

1275 products.

## $\underline{\text{Irrational Equations}}$ Methods for solving problems with parameters Problem $\underline{76453}$ Topics: Difficulty: $4-$ Solve the equation $\sqrt{a-\sqrt{a+x}}=x$.

## Solution

Clearly, $a \geq 0, x \geq 0$. Let $y=\sqrt{a+x} \geq 0$, then $x=\sqrt{a-y}$. Squaring and subtracting, we get $x+y=y^{2}-x^{2}$, that is,

$x+y=0$ or $y-x=1$.

$x+y$ can only be zero if $x=y=0$, in which case $a=0$.

If $y=x+1$, then $(x+1)^{2}=x+a, x^{2}+x+1-a=0$. For $a<1$ this equation has no non-negative roots. For $a \geq 1$ the only non-negative root is $\frac{\sqrt{4a-3}-1}{2}$.

## Answer

For $a=0$, $x=0$; for $a \geq 1$, $x=\frac{\sqrt{4a-3}-1}{2}$; for $a<0$ and $0<a<1$, there are no solutions.

Answer: 1275


Find all pairs $(x, y)$ of real numbers such that $|x|+|y|=1340$ and $x^{3}+y^{3}+2010 x y=670^{3}$.

Solution:
Solution. Answer: $(-670 ;-670),(1005 ;-335),(-335 ; 1005)$.

To prove this, let $z=-670$. We have

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

Thus either $x+y+z=0$, or $x=y=z$. In the latter case we get $x=y=-670$, which satisfies both the equations. In the former case we get $x+y=670$. Then at least one of $x, y$ is positive, but not both, as from the second equation we would get $x+y=1340$. If $x>0 \geq y$, we get $x-y=1340$, which together with $x+y=670$ yields $x=1005, y=-335$. If $y>0 \geq x$ we get similarly $x=-335, y=1005$.

Answer: (-670,-670),(1005,-335),(-335,1005)


In the triangle ABC, the length of the altitude from A is not less than BC, and the length of the altitude from B is not less than AC. Find the angles.

Solution: Let k be twice the area of the triangle. Then k≥BC 2 , k≥AC 2 and k≤AC.BC, with equality in the last case only if AC is perpendicular to BC. Hence AC and BC have equal lengths and are perpendicular. So the angles are 90, 45, 45. 4th ASU 1964 (C) John Scholes jscholes@kalva.demon.co.uk 22 Sep 1998

Answer: 90,45,45


Compute the perimeter of the triangle that has area $3-\sqrt{3}$ and angles $45^\circ$, $60^\circ$, and $75^\circ$.

Solution:
1. Let the triangle be denoted \( \triangle ABC \), with \( \angle A = 45^\circ \), \( \angle B = 60^\circ \), and \( \angle C = 75^\circ \).

2. Drop an altitude \( CD \) from \( C \) to \( AB \). This splits \( \triangle ABC \) into two right triangles: \( \triangle ACD \) and \( \triangle BCD \).

3. Note that \( \triangle ACD \) is a \( 45^\circ-45^\circ-90^\circ \) triangle, and \( \triangle BCD \) is a \( 30^\circ-60^\circ-90^\circ \) triangle.

4. In \( \triangle ACD \), since it is a \( 45^\circ-45^\circ-90^\circ \) triangle, we have:
   \[
   AD = CD
   \]

5. In \( \triangle BCD \), since it is a \( 30^\circ-60^\circ-90^\circ \) triangle, we have:
   \[
   BD = \frac{CD}{\sqrt{3}}
   \]

6. Therefore, the length of \( AB \) is:
   \[
   AB = AD + BD = CD + \frac{CD}{\sqrt{3}} = CD \left(1 + \frac{1}{\sqrt{3}}\right)
   \]

7. The area of \( \triangle ABC \) is given by:
   \[
   \text{Area} = \frac{1}{2} \cdot AB \cdot CD
   \]

8. Substituting the given area \( 3 - \sqrt{3} \) and the expression for \( AB \):
   \[
   3 - \sqrt{3} = \frac{1}{2} \cdot CD \left(1 + \frac{1}{\sqrt{3}}\right) \cdot CD
   \]

9. Simplifying the expression:
   \[
   3 - \sqrt{3} = \frac{1}{2} \cdot CD^2 \left(1 + \frac{1}{\sqrt{3}}\right)
   \]
   \[
   3 - \sqrt{3} = \frac{1}{2} \cdot CD^2 \left(\frac{\sqrt{3} + 1}{\sqrt{3}}\right)
   \]
   \[
   3 - \sqrt{3} = \frac{1}{2} \cdot CD^2 \cdot \frac{\sqrt{3} + 1}{\sqrt{3}}
   \]
   \[
   3 - \sqrt{3} = \frac{CD^2 (\sqrt{3} + 1)}{2\sqrt{3}}
   \]

10. Solving for \( CD^2 \):
    \[
    CD^2 = \frac{2\sqrt{3}(3 - \sqrt{3})}{\sqrt{3} + 1}
    \]

11. Simplifying further:
    \[
    CD^2 = \frac{2\sqrt{3}(3 - \sqrt{3})(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)}
    \]
    \[
    CD^2 = \frac{2\sqrt{3}(3\sqrt{3} - 3 - 3 + \sqrt{3})}{2}
    \]
    \[
    CD^2 = \frac{2\sqrt{3}(2\sqrt{3} - 6)}{2}
    \]
    \[
    CD^2 = 2\sqrt{3} - 6
    \]
    \[
    CD = \sqrt{2\sqrt{3} - 6}
    \]

12. Since \( \triangle ACD \) is a \( 45^\circ-45^\circ-90^\circ \) triangle:
    \[
    AC = CD \sqrt{2} = \sqrt{2(2\sqrt{3} - 6)}
    \]

13. Since \( \triangle BCD \) is a \( 30^\circ-60^\circ-90^\circ \) triangle:
    \[
    BC = 2 \cdot \frac{CD}{\sqrt{3}} = 2 \cdot \frac{\sqrt{2\sqrt{3} - 6}}{\sqrt{3}}
    \]

14. The perimeter of \( \triangle ABC \) is:
    \[
    AB + AC + BC = CD \left(1 + \frac{1}{\sqrt{3}}\right) + \sqrt{2(2\sqrt{3} - 6)} + 2 \cdot \frac{\sqrt{2\sqrt{3} - 6}}{\sqrt{3}}
    \]

15. Simplifying the expression for the perimeter:
    \[
    \boxed{3\sqrt{2} + 2\sqrt{3} - \sqrt{6}}
    \]

Answer: 3\sqrt{2} + 2\sqrt{3} - \sqrt{6}


Find the minimum value of the sum

$$
\left|x-1^{2}\right|+\left|x-2^{2}\right|+\left|x-3^{2}\right|+\ldots+\left|x-10^{2}\right|
$$

Solution: 4. Grouping the terms, we get $\left(\left|x-1^{2}\right|+\left|x-10^{2}\right|\right)+\left(\left|x-2^{2}\right|+\left|x-9^{2}\right|\right)+\left(\left|x-3^{2}\right|+\left|x-8^{2}\right|\right)+\left(\left|x-4^{2}\right|+\left|x-7^{2}\right|\right)+\left(\left|x-5^{2}\right|+\left|x-6^{2}\right|\right)$. It is known that the pair $|x-a|+|x-b|$ takes the minimum value when $a \leq x \leq b$ and this value is equal to $|b-a|$. Therefore, the minimum value of the first bracket $\left(\left|x-1^{2}\right|+\left|x-10^{2}\right|\right)$ will be $10^{2}-1^{2}$, the second bracket respectively $9^{2}-2^{2}$, the third $8^{2}-3^{2}$, and so on. However, the intervals $5^{2} \leq x \leq 6^{2}, 4^{2} \leq x \leq 7^{2}, \ldots$ are nested within each other, since the numbers $1^{2}<2^{2}<3^{2} \ldots<10^{2}$. Therefore, the interval $\left[1,10^{2}\right]$ contains all other intervals, on each of which all terms of the form $|x-a|+|x-b|$ simultaneously take the minimum value. Then our expression takes the minimum value, equal to $10^{2}-1^{2}+9^{2}-2^{2}+8^{2}-3^{2}+7^{2}-4^{2}+6^{2}-5^{2}$, (i.e., the sum of the lengths of the nested intervals). We find this sum: $11 \cdot 9+11 \cdot 7+11 \cdot 5+11 \cdot 3+11 \cdot 1=11(9+7+5+3+1)=$ $=11 \cdot 25=275$ Answer: $\{275\}$

Answer: 275


In the sequence $\left\{a_{n}\right\}$, for $1 \leqslant n \leqslant 5$, we have $a_{n}=n^{2}$, and for all positive integers $n$, we have
$$
a_{n+5}+a_{n+1}=a_{n+4}+a_{n} \text {. }
$$

Then $a_{2023}=$ $\qquad$

Solution:
3. 17 .

For all positive integers $n$, we have
$$
a_{n+5}+a_{n+1}=a_{n+4}+a_{n}=\cdots=a_{5}+a_{1}=26 \text {. }
$$

Then $a_{n}=26-a_{n+4}=26-\left(26-a_{n+8}\right)=a_{n+8}$, which means $\left\{a_{n}\right\}$ is a sequence with a period of 8.
Therefore, $a_{2023}=a_{7}=26-a_{3}=26-9=17$.

Answer: 17


In a box, there are 3 red, 4 gold, and 5 silver stars. Randomly, one star is taken from the box and hung on the Christmas tree. What is the probability that a red star will end up on the top of the tree, there will be no more red stars on the tree, and there will be exactly 3 gold stars, if a total of 6 stars are taken from the box?

Solution:
Solution: There are a total of $3+4+5=12$ toys. It is necessary for a red toy to be on top - the probability is $\frac{3}{12}$. From the remaining 11, 3 golden, 2 silver, and 0 red toys are chosen, so the probability is $\frac{C_{4}^{3} \cdot C_{5}^{2} \cdot C_{2}^{0}}{C_{11}^{5}}$.

The desired probability is obtained by multiplying the two obtained: $\frac{3 \cdot C_{4}^{3} \cdot C_{5}^{2} \cdot C_{2}^{0}}{12 \cdot C_{11}^{5}}$

We will calculate these expressions separately: $C_{2}^{0}=1, \quad C_{5}^{2}=\frac{5 \cdot 4}{1 \cdot 2}=10$, $C_{4}^{3}=4, C_{11}^{5}=\frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}=462$.

Substitute the obtained results into the answer: $\frac{3 \cdot 4 \cdot 10 \cdot 1}{12 \cdot 462}=\frac{5}{231}$.

Answer: $\frac{5}{231}$.

Answer: \frac{5}{231}


The range of the function $y=\frac{x^{2}-1}{x^{2}+2}$ is $\qquad$ .

Solution:
From the given, we have $x^{2}=\frac{2 y+1}{1-y} \geqslant 0$, solving this yields $-\frac{1}{2} \leqslant y<1$, therefore, the range of the function $y=\frac{x^{2}-1}{x^{2}+2}$ is $\left[-\frac{1}{2}, 1\right)$.

Solution Guidance Construct an inequality that the dependent variable satisfies, and obtain the range of the function by solving the inequality. This is another important method for studying the range of a function, or rather, this is an application of the "construction method" in solving problems related to the range of a function.

Answer: [-\frac{1}{2},1)


An integer $n$ is called a combi-number if every pair of different digits from all possible digits $0$ to $9$ appear next to each other at least once in the number. Thus, in a combi-number, the digits 3 and 5 appear next to each other somewhere. It does not matter whether they appear in the order 35 or 53. We agree that a combi-number does not start with the digit 0.

What is the smallest number of digits a combi-number can consist of?

Solution:
B2. 50 Consider a random combi-number. First, we will prove that it consists of at least 50 digits. Then, we will show that there actually exists a combi-number with exactly 50 digits.

Note that if we swap the same digits everywhere in a combi-number, we still get a combi-number: for example, you can replace all threes with ones and vice versa. This way, a combi-number never needs to start with a zero.

Suppose the digit $c$ is not in the first or the last position in the combi-number. Every time $c$ appears in the combi-number, it forms a pair with both its left and right neighbor. Since $c$ must appear in 9 pairs, $c$ must therefore appear at least 5 times in the combi-number.

Suppose $c$ is the first or the last digit, but not both. Then $c$ must also appear 5 times in the combi-number, because the $c$ at the beginning (or at the end) is only in one pair. If the first and the last digit of the combi-number are the same digit $c$, then you need $c$ six times.

In total, we see that each digit from $0$ to $9$ must appear at least five times in the combi-number. Therefore, the combi-number must have at least 50 digits. It can also have exactly 50 digits, for example with the number

$$
98765432109753196307418529517394864208406283726150 .
$$

The way we created this number is as follows. We made all possible sequences of digits with a constant jump. So 98765432109 with a jump of 1, 975319 and 864208 with a jump of 2, and so on. We then concatenated these sequences in a convenient way. Of course, there are many other ways to create a 50-digit combi-number.

Answer: 50


Solution: 267. $\sqrt{x^{2}-x}=2-x ; \quad x^{2}-x=4+x^{2}-4 x ; \quad 3 x=4$; $x=\frac{4}{3}$

Answer: \frac{4}{3}


For each positive integer $n$, let $S(n)$ be the sum of the digits of $n$. Determines the smallest positive integer $a$ such that there are infinite positive integers $n$ for which you have $S (n) -S (n + a) = 2018$.

Solution:
1. **Lemma: \( S(n) \equiv n \pmod{9} \)**

   **Proof:**
   Consider a positive integer \( n \) expressed in its decimal form:
   \[
   n = 10^x a_1 + 10^{x-1} a_2 + \cdots + 10 a_{x-1} + a_x
   \]
   where \( a_i \) are the digits of \( n \) and \( 1 \leq a_i \leq 9 \). We can rewrite \( n \) as:
   \[
   n = (10^x - 1)a_1 + (10^{x-1} - 1)a_2 + \cdots + (10 - 1)a_{x-1} + a_1 + a_2 + \cdots + a_x
   \]
   Since \( 10^k \equiv 1 \pmod{9} \) for any integer \( k \), we have:
   \[
   n \equiv a_1 + a_2 + \cdots + a_x \pmod{9}
   \]
   Therefore, \( S(n) \equiv n \pmod{9} \), proving our lemma.

2. **Determine \( a \) such that \( S(n) - S(n + a) = 2018 \)**

   From the lemma, we know:
   \[
   S(n) \equiv n \pmod{9} \quad \text{and} \quad S(n + a) \equiv n + a \pmod{9}
   \]
   Therefore:
   \[
   S(n) - S(n + a) \equiv n - (n + a) \equiv -a \pmod{9}
   \]
   Given \( S(n) - S(n + a) = 2018 \), we have:
   \[
   2018 \equiv -a \pmod{9}
   \]
   Calculating \( 2018 \mod 9 \):
   \[
   2018 \div 9 = 224 \text{ remainder } 2 \quad \Rightarrow \quad 2018 \equiv 2 \pmod{9}
   \]
   Thus:
   \[
   -a \equiv 2 \pmod{9} \quad \Rightarrow \quad a \equiv -2 \pmod{9} \quad \Rightarrow \quad a \equiv 7 \pmod{9}
   \]
   The smallest positive integer \( a \) that satisfies \( a \equiv 7 \pmod{9} \) is \( a = 7 \).

3. **Verification:**

   Let \( n = 1\underbrace{000\cdots0}_{k} \underbrace{99\dots9}_{225} \), for any \( 1 \leq k \) with \( k \in \mathbb{Z} \), i.e., \( n = 10^{k+225} + 10^{226} - 1 \).

   Then:
   \[
   n + a = n + 7 = 1\underbrace{000\cdots0}_{k-1} 1\underbrace{000\dots0}_{224}6
   \]
   This can be written as:
   \[
   n + 7 = 10^{k+225} + 10^{226} + 6
   \]

   Calculating the sum of the digits:
   \[
   S(n) = 1 + 225 \cdot 9 = 226
   \]
   \[
   S(n + 7) = 8
   \]
   Therefore:
   \[
   S(n) - S(n + 7) = 226 - 8 = 218
   \]
   This does not match the required difference of 2018. Hence, we need to reconsider the construction of \( n \).

   Instead, consider \( n = 10^k - 1 \) for large \( k \). Then:
   \[
   S(n) = 9k
   \]
   \[
   n + 7 = 10^k - 1 + 7 = 10^k + 6
   \]
   \[
   S(n + 7) = 1 + 6 = 7
   \]
   Therefore:
   \[
   S(n) - S(n + 7) = 9k - 7
   \]
   Setting \( 9k - 7 = 2018 \):
   \[
   9k = 2025 \quad \Rightarrow \quad k = 225
   \]
   Thus, for \( k = 225 \), we have:
   \[
   S(n) - S(n + 7) = 2018
   \]
   This confirms that \( a = 7 \) satisfies the condition.

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

Answer: 7


Observe the pattern in the following expressions:
$$
\begin{array}{l}
1 \times 2+2 \times 3=2 \times 2 \times 2 \\
2 \times 3+3 \times 4=2 \times 3 \times 3 \\
3 \times 4+4 \times 5=2 \times 4 \times 4 \\
4 \times 5+5 \times 6=2 \times 5 \times 5
\end{array}
$$

Use the pattern to calculate: $75 \times 222+76 \times 225-25 \times 14 \times 15-25 \times 15 \times 16=$ $\qquad$

Solution: Reference answer: 22500

Answer: 22500


$488 \square$ is a four-digit number, the math teacher says: âI fill in this blank with 3 digits in sequence, the resulting 3 four-digit numbers can be successively divisible by $9,11,7$.Ã¢ÂÂ What is the sum of the 3 digits the math teacher filled in?

Solution: Reference answer: 17

Answer: 17


Given in $\triangle ABC$, $AC \geqslant AB$, side $BC$ is divided into $n$ ($n$ is an odd number) equal parts. Let $\alpha$ represent the angle subtended at point $A$ by the segment containing the midpoint of side $BC$, $h$ be the altitude from $A$ to side $BC$, and $BC=a$. If $\tan \alpha=\frac{4 n h}{\left(n^{2}-1\right) a}$, then $\angle BAC=$ $\qquad$

Solution:
3. $90^{\circ}$.

Draw a perpendicular from point $A$ to $BC$, with the foot of the perpendicular being $D$. Let the segment containing the midpoint of $BC$ be $B' C'$. Then
$$
B B'=\frac{n-1}{2 n} a, B C'=\frac{n+1}{2 n} a .
$$

Let $\overrightarrow{B D}=x \overrightarrow{B C}(x \in \mathbf{R})$. Then
$$
D B'=\left(\frac{n-1}{2 n}-x\right) a, D C'=\left(\frac{n+1}{2 n}-x\right) a \text {. }
$$

Thus, $\tan \angle C' A D=\frac{\left(\frac{n+1}{2 n}-x\right) a}{h}$,
$$
\tan \angle B' A D=\frac{\left(\frac{n-1}{2 n}-x\right) a}{h} \text {. }
$$

Therefore, $\tan \alpha=\tan \angle C' A B'$
$$
\begin{array}{l}
=\frac{\left[\left(\frac{n+1}{2 n}-x\right)-\left(\frac{n-1}{2 n}-x\right)\right] a h}{h^{2}+\left(\frac{n+1}{2 n}-x\right)\left(\frac{n-1}{2 n}-x\right) a^{2}} \\
=\frac{4 n h a}{\left(n^{2}-1\right) a^{2}+4 n^{2}\left[h^{2}-x(1-x) a^{2}\right]} .
\end{array}
$$

Given $\tan \alpha=\frac{4 n h}{\left(n^{2}-1\right) a}$, we know
$$
h^{2}=x a \cdot(1-x) a \text {. }
$$

Thus, $A D^{2}=B D \cdot C D$, and $D$ lies on the segment $BC$.
By the inverse of the projection theorem, $\angle B A C=90^{\circ}$.

Answer: 90^{\circ}


In the known sequence $1,4,8,10,16,19,21,25,30,43$, the number of arrays where the sum of consecutive numbers is divisible by 11 is $\qquad$.

Solution:
7
3.【Analysis and Solution】Since the problem is about divisibility by 11, we can first subtract multiples of 11 from each term to make the numbers smaller and easier to handle, thus obtaining the following sequence:
$$
1,4,-3,-1,5,-3,-1,3,-3,-1 \text {. }
$$

Let $S_{n}$ be the sum of the first $n$ terms, then
$$
\begin{array}{l}
S_{1}=1, S_{2}=5, S_{3}=2, S_{4}=1, S_{5}=6, \\
S_{6}=3, S_{7}=2, S_{8}=5, S_{9}=2, S_{10}=1 .
\end{array}
$$

Among them, the equal ones are
$$
\begin{array}{l}
S_{1}=S_{4}=S_{10}=1, S_{2}=S_{8}=5, S_{3}=S_{7}=S_{9}=2 . \\
S_{4}-S_{1}, S_{10}-S_{4}, S_{8}-S_{2}, S_{7}-S_{3}, S_{9}-S_{3}, S_{9}-S_{7}, S_{10}-S_{1} .
\end{array}
$$

There are 7 groups that can be divided by 11.

Answer: 7


In $\triangle A B C$, $A B=A C=m, B C=n$, $\angle B A C=160^{\circ}$. Then the relationship between $n^{3}-\sqrt{3} m^{3}$ and $3 m^{2} n$ is $n^{3}-\sqrt{3} m^{3}(\quad) 3 m^{2} n$.
(A) greater than
(B) less than
(C) equal to
(D) uncertain

Solution:
6. C.

From $A B=A C$, we get
$$
\angle B=\angle A C B=\frac{180^{\circ}-\angle B A C}{2}=10^{\circ} \text {. }
$$

Construct $\angle B A D=\angle B$ inside $\angle B A C$, with point $D$ on side $B C$, and draw $C E \perp D A$ at point $E$. Then
$$
\begin{array}{l}
A D=B D, \\
\angle D A C=\angle B A C-\angle B A D=150^{\circ} .
\end{array}
$$
$$
\text { Hence } \angle C A E=180^{\circ}-\angle D A C=30^{\circ} \text {. }
$$

In Rt $\triangle A C E$,
$$
\begin{array}{l}
C E=\frac{1}{2} A C=\frac{1}{2} m, \\
A E=A C \cos \angle C A E=\frac{\sqrt{3}}{2} m .
\end{array}
$$

Therefore, $\triangle A B D \backsim \triangle C B A$
$$
\Rightarrow \frac{B D}{B A}=\frac{A B}{C B} \Rightarrow B D=\frac{A B^{2}}{C B}=\frac{m^{2}}{n} \text {. }
$$

In Rt $\triangle C D E$,
$$
\begin{array}{l}
D C^{2}=D E^{2}+C E^{2} \\
\Rightarrow(B C-B D)^{2}=(A D+A E)^{2}+C E^{2} \\
\Rightarrow\left(n-\frac{m^{2}}{n}\right)^{2}=\left(\frac{m^{2}}{n}+\frac{\sqrt{3}}{2} m\right)^{2}+\frac{1}{4} m^{2} \\
\Rightarrow n^{3}-\sqrt{3} m^{3}=3 m^{2} n .
\end{array}
$$

Answer: C


On each kilometer of the highway between the villages of Yolkiino and Palkino, there is a post with a sign. On one side of the sign, it shows how many kilometers are left to Yolkiino, and on the other side, how many kilometers are left to Palkino. Borya noticed that on each post, the sum of all the digits is 13. What is the distance from Yolkiino to Palkino?

Solution:
Let the distance from Yolkiino to Palkiino be $n$ kilometers. Clearly, $n \geq 10$. Moreover, $n \leq 49$ (otherwise, the sum of the digits on the 49th milestone would be greater than 13).

On the tenth milestone from Yolkiino, one side reads 10, and the other side reads $n-10$, which does not exceed 39. The sum of its digits is 12. Therefore, $n-10=39$ (the sum of the digits of smaller numbers does not exceed 11).

Answer: 49


Let $A B C$ be a triangle with orthocenter $H$; suppose that $A B=13, B C=14, C A=15$. Let $G_{A}$ be the centroid of triangle $H B C$, and define $G_{B}, G_{C}$ similarly. Determine the area of triangle $G_{A} G_{B} G_{C}$.

Solution: Answer: $28 / 3$ Let $D, E, F$ be the midpoints of $B C, C A$, and $A B$, respectively. Then $G_{A} G_{B} G_{C}$ is the $D E F$ about $H$ with a ratio of $\frac{2}{3}$, and $D E F$ is the dilation of $A B C$ about $H$ with a ratio of $-\frac{1}{2}$, so $G_{A} G_{B} G_{C}$ is the dilation of $A B C$ about $H$ with ratio $-\frac{1}{3}$. Thus $\left[G_{A} G_{B} G_{C}\right]=\frac{[A B C]}{9}$. By Heron's formula, the area of $A B C$ is $\sqrt{21 \cdot 8 \cdot 7 \cdot 6}=84$, so the area of $G_{A} G_{B} G_{C}$ is $[A B C] / 9=84 / 9=28 / 3$.

Answer: \frac{28}{3}


The common ratio of the geometric sequence $a+\log _{2} 3, a+\log _{4} 3, a+\log _{8} 3$ is

Solution:
$$
\frac{1}{3}
$$
9. [Analysis and Solution] According to the problem, let's assume the common ratio is $q$, so we have $q=\frac{a+\log _{4} 3}{a+\log _{2} 3}=\frac{a+\log _{8} 3}{a+\log _{4} 3}$. According to the properties of ratios, we have
$$
q=\frac{a+\log _{4} 3-\left(a+\log _{8} 3\right)}{a+\log _{2} 3-\left(a+\log _{4} 3\right)}=\frac{\log _{4} 3-\log _{8} 3}{\log _{2} 3-\log _{4} 3}=\frac{\frac{1}{2} \log _{2} 3-\frac{1}{3} \log _{2} 3}{\log _{2} 3-\frac{1}{2} \log _{2} 3}=\frac{1}{3} .
$$

Answer: \frac{1}{3}


In a row, the numbers $\sqrt{7.301}, \sqrt{7.302}, \sqrt{7.303}, \ldots, \sqrt{16.002}, \sqrt{16.003}$ are written (under the square root - consecutive terms of an arithmetic progression with a common difference of 0.001). Find the number of rational numbers among the listed ones.

Solution:
# Answer: 13

Solution. Multiply the numbers by 100, we get $\sqrt{73010}, \sqrt{73020}, \sqrt{73030}, \ldots, \sqrt{160030}$ (in this case, rational numbers will remain rational, and irrational numbers will remain irrational). The square root of a natural number $n$ is a rational number if and only if $n$ is a perfect square. Furthermore, a natural number ending in 0 (i.e., divisible by 2 and 5) can only be a perfect square if it ends in 00 (i.e., divisible by $2^{2}$ and $5^{2}$).

Thus, the required number is the number of perfect squares in the sequence of numbers $731, 732, 733, \ldots$, 1600. Since $27^{2}<731<28^{2}$ and $40^{2}=1600$, the answer is $40-27=13$.

Answer: 13


Solution: Solution. $\int_{0}^{1 / 2} \frac{d x}{\sqrt{1-x^{2}}}=\left.\arcsin x\right|_{0} ^{1 / 2}=\frac{\pi}{6}$.

Answer: \frac{\pi}{6}


For the digits of the decimal number $\overline{a b c d}$, it holds that $a>b>c>d$. These same digits, in some order, are also the digits of the difference $\overline{a b c d}-\overline{d c b a}$. Which is this four-digit number?

Solution:
When the subtraction is written in the usual scheme, in the last two columns (1) the lower digit is larger than the upper one; here, therefore, we exchange one ten of the minuend for units, or one hundred for tens, so the difference in the units and tens place value is:

$$
\begin{array}{ll}
\text { units: } & e=10+d-a=10-(a-d) \\
\text { tens: } & t=10+(c-1)-b=9-(b-c)
\end{array}
$$

However, in the first two columns, the upper digit is larger, so the other digits of the difference are:

$$
\begin{aligned}
& \text { hundreds: } \quad s=(b-1)-c=(b-c)-1, \\
& \text { thousands: } \quad k=a-d
\end{aligned}
$$

( $k$ as the first letter of "kilometer"). We see that

$$
k+e=10 \text { and }
$$

$$
\text { (7) } \quad s+t=8
$$

We can also determine the magnitude relationships for the last two and the first two digits of the difference based on (1):

$$
t=9-b+c \geq 9-(a-1)+(d+1)=e+1>e, \quad k=a-d>a-c>b-c>s
$$

Furthermore, $k, s, t, e$ represent the original digits in some order, so from the above, we conclude that the smallest digit, $d$, is either $e$ or $s$ in the difference, and the digit $a$ is either $k$ or $t$.

However, the assumption $d=e$ leads to a contradiction: from (2) it follows that $a=10$; therefore, the smallest digit can only be $d=s$.

Knowing this, the assumption $a=k$, i.e., $a>t$, also leads to a contradiction: from (5) it follows that $d=s=0$, from (7) $t=8$, so clearly $a=k=9$ and from (6) $e=1$ follows; the only possible values for the remaining digits are 8 for $b$ and 1 for $c$, but then the condition $b-c=1$ from (4) is not satisfied. Therefore, the largest digit can only be $a=t$, and from (7) $a \leq 8$, and if we do not allow $d=0$ as the leading digit, then $a \leq 7$.

Then $b$ and $c$ are one $k$ and one $e$, and from (6) $b+c=10$, so $b$ and $c$ have the same parity; therefore, their difference is also even: $b-c \geq 2$, so $c \leq 4$ and $b \geq 6$. From this, $a \geq 7$; and if we accept the restriction $d \geq 1$, then only

$$
a=7, \quad d=1, \quad b=6, \quad c=4
$$

is possible. This is also a solution to the problem, and the sought number is 7641.

Starting from $d=0$ would yield $a=8, e=2=c$ and $b=8=a$, so there is no other solution.

Answer: 7641


Positive integers $a_1, a_2, \ldots, a_{101}$ are such that $a_i+1$ is divisible by $a_{i+1}$ for all $1 \le i \le 101$, where $a_{102} = a_1$. What is the largest possible value of $\max(a_1, a_2, \ldots, a_{101})$?

Solution:
1. **Assume the largest value:**
   Let \( a_{101} \) be the largest of the \( a_i \), and denote \( a_{101} = k \). We aim to find the maximum possible value of \( k \).

2. **Claim:**
   We claim that \( a_{101-i} = k - i \) for all \( 0 \leq i \leq 100 \).

3. **Base case:**
   For \( i = 0 \), we have \( a_{101-0} = a_{101} = k \), which is trivially true.

4. **Inductive step:**
   Assume \( a_{101-i} = k - i \) for some \( i \). We need to show that \( a_{101-(i+1)} = k - (i+1) \).

5. **Divisibility condition:**
   Since \( a_{101-i} = k - i \), we have \( k - i \mid a_{101-(i+1)} + 1 \).

6. **Bounding \( a_{101-(i+1)} \):**
   If \( a_{101-(i+1)} + 1 \geq 2(k - i) \), then:
   \[
   k + 1 \geq a_{101-(i+1)} + 1 \geq 2(k - i)
   \]
   Simplifying, we get:
   \[
   k + 1 \geq 2k - 2i \implies k + 1 \geq k + k - 2i \implies 1 \geq k - 2i
   \]
   Since \( k > 201 \) and \( i \leq 100 \), this inequality cannot hold. Therefore, \( a_{101-(i+1)} + 1 < 2(k - i) \).

7. **Conclusion from the bound:**
   Thus, \( a_{101-(i+1)} + 1 = k - i \), which implies \( a_{101-(i+1)} = k - (i+1) \).

8. **Inductive proof:**
   By induction, \( a_{101-i} = k - i \) for all \( 0 \leq i \leq 100 \).

9. **Relating \( a_1 \) and \( a_{101} \):**
   From the claim, \( a_1 = k - 100 \). Since \( a_1 \mid a_{101} + 1 \), we have:
   \[
   k - 100 \mid k + 1
   \]
   This implies:
   \[
   k - 100 \mid 101
   \]

10. **Divisors of 101:**
    The divisors of 101 are 1 and 101. Since \( k - 100 > 101 \), the only possible value is \( k - 100 = 101 \), giving \( k = 201 \).

11. **Verification:**
    If \( k = 201 \), then \( a_1 = 101 \), \( a_2 = 102 \), ..., \( a_{101} = 201 \). Each \( a_i + 1 \) is divisible by \( a_{i+1} \), satisfying the problem's conditions.

The final answer is \( \boxed{201} \).

Answer: 201


In the sequence $\left\{a_{n}\right\}$, $a_{1}=1$, and its first $n$ terms sum $S_{n}$ satisfies the relation
$$
3 t S_{n}-(2 t+3) S_{n-1}=3 t(t>0, n=2,3, \cdots) \text {. }
$$
(1) Prove that the sequence $\left\{a_{n}\right\}$ is a geometric sequence;
(2) Let the common ratio of the sequence $\left\{a_{n}\right\}$ be $f(t)$, and construct the sequence $\left\{b_{n}\right\}$ such that $b_{1}=1, b_{n}=f\left(\frac{1}{b_{n-1}}\right)(n=2,3, \cdots)$, find $b_{n}$;
(3) Find the value of $b_{1} b_{2}-b_{2} b_{3}+b_{3} b_{4}-b_{4} b_{5}+\cdots+b_{2 n-1} b_{2 n}-b_{2 n} b_{2 n+1}$.

Solution:
17. (1) From the given $3 S_{2}-(2 t+3) S_{1}=3 t$,

i.e., $3 t\left(a_{1}+a_{2}\right)-(2 t+3) a_{1}=3 t$.
Given $a_{1}=1$, solving for $a_{2}$ yields $a_{2}=\frac{2 t+3}{3 t}$.
Thus, $\frac{a_{2}}{a_{1}}=\frac{2 t+3}{3 t}$.
For $n \geqslant 2$, we have
$$
\begin{array}{l}
3 t S_{n+1}-(2 t+3) S_{n}=3 t, \\
3 t S_{n}-(2 t+3) S_{n-1}=3 t .
\end{array}
$$
(1) - (2) gives $3 t a_{n+1}-(2 t+3) a_{n}=0$.

Thus, $\frac{a_{n+1}}{a_{n}}=\frac{2 t+3}{3 t}$.
In summary, $\frac{a_{n+1}}{a_{n}}=\frac{2 t+3}{3 t}(n \geqslant 1)$.
Therefore, $\left\{a_{n}\right\}$ is a geometric sequence.
(2) From (1), we know $f(t)=\frac{2 t+3}{3 t}$, then
$$
b_{1}=1, b_{n}=\frac{2 \cdot \frac{1}{b_{n-1}}+3}{3 \cdot \frac{1}{b_{n-1}}}=\frac{2}{3}+b_{n-1} .
$$

Thus, $b_{n}-b_{n-1}=\frac{2}{3}(n=2,3, \cdots)$.
Therefore, $\left\{b_{n}\right\}$ is an arithmetic sequence, and
$$
b_{1}=1, d=b_{n}-b_{n-1}=\frac{2}{3} \text {. }
$$

Hence, $b_{n}=b_{1}+(n-1) d=\frac{2}{3} n+\frac{1}{3}$.
$$
\begin{array}{l}
\text { (3) } b_{1} b_{2}-b_{2} b_{3}+b_{3} b_{4}-b_{4} b_{5}+\cdots+b_{2 n-1} b_{2 n}-b_{2 n} b_{2 n+1} \\
=b_{2}\left(b_{1}-b_{3}\right)+b_{4}\left(b_{3}-b_{5}\right)+\cdots+b_{2 n}\left(b_{2 n-1}-b_{2 n+1}\right) \\
=-\frac{4}{3}\left(b_{2}+b_{4}+\cdots+b_{2 n}\right)=-\frac{4}{3} \cdot \frac{n\left(b_{2}+b_{2 n}\right)}{2} \\
=-\frac{4}{3} \cdot \frac{n\left(\frac{5}{3}+\frac{4 n+1}{3}\right)}{2}=-\frac{8}{9} n^{2}-\frac{4}{3} n .
\end{array}
$$

Answer: -\frac{8}{9} n^{2}-\frac{4}{3} n


On the circle $C:(x-1)^{2}+y^{2}=2$ there are two moving points $A$ and $B$, and they satisfy the condition $\angle AOB=90^{\circ}$ ($O$ being the origin). Find the equation of the trajectory of point $P$ of the parallelogram $OAPB$ with $OA, OB$ as adjacent sides.

Solution:
Three, let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), P(x, y)$. According to the problem, we have
$$
\begin{array}{l}
\left(x_{1}-1\right)^{2}+y_{1}^{2}=2, \\
\left(x_{2}-1\right)^{2}+y_{2}^{2}=2 .
\end{array}
$$

By adding (1) and (2), we get
$$
x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}-2\left(x_{1}+x_{2}\right)=2 .
$$

Since $O A P B$ is a parallelogram, we have
$$
\begin{array}{l}
x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}=x^{2}+y^{2}, \\
x_{1}+x_{2}=x .
\end{array}
$$

Substituting (4) and (5) into (3), we get
$$
(x-1)^{2}+y^{2}=(\sqrt{3})^{2} \text {. }
$$

Therefore, the locus of point $P$ is a circle with center $(1,0)$ and radius $\sqrt{3}$.

Answer: (x-1)^{2}+y^{2}=(\sqrt{3})^{2}


If the function $f(x)=\frac{x}{x^{2}+a}$ has a maximum value of $\frac{\sqrt{2}}{2}$ on $[1,+\infty)$, then $a=$

Solution:
1. $\sqrt{2}-1$.

First, $f(x)=\frac{x}{x^{2}+a}=\frac{1}{x+\frac{a}{x}}(x \geqslant 1)$. When $a \geqslant 1$, the maximum value of the function $f(x)$ is $\frac{1}{2 \sqrt{a}}=\frac{\sqrt{2}}{2}$, solving for $a$ gives $a=\frac{1}{2}<1$, which is a contradiction. When $0<a<1$, $f(x)$ is monotonically decreasing on $[1,+\infty)$, and its maximum value is $\frac{1}{1+a}=\frac{\sqrt{2}}{2}$, yielding $a=\sqrt{2}-1$. Therefore, $a=\sqrt{2}-1$.

Answer: \sqrt{2}-1


Given positive integers $x, y, z$ satisfying $x y z=(22-x)(22-y)(22-z)$, and $x+y+z<44, x^{2}+y^{2}+z^{2}$, the maximum and minimum values are denoted as $M$ and $N$, respectively, then $M+N=$ $\qquad$.

Solution:
Rong 926.
Analysis From the given information, we know that $x, y, z$ are all positive integers less than 22. From the given equation, we have
$$
2 x y z=22^{3}-22^{2}(x+y+z)+22(x y+y z+z x)
$$

Thus, $11 \mid x y z$, so at least one of $x, y, z$ must be equal to 11.
Assume without loss of generality that $x=11$. From the given equation, we get $y+z=22$. Since $y, z$ are both positive integers, we also have
$$
y^{2}+z^{2}=\frac{(y+z)^{2}+(y-z)^{2}}{2}=\frac{22^{2}+(y-z)^{2}}{2}=242+\frac{1}{2}(y-z)^{2}
$$

It is easy to see that when $y=z=11$, $y^{2}+z^{2}$ takes its minimum value of 242, and when $|y-z|=|21-1|=20$, $y^{2}+z^{2}$ takes its maximum value. In this case, $M=11^{2}+21^{2}+1^{2}=563$, so $M+N=563+363=926$.

Answer: 926


In the plane of equilateral $\triangle ABC$, the condition that line $m$ satisfies is: the distances from the 3 vertices of equilateral $\triangle ABC$ to line $m$ take only 2 values, and one of these values is twice the other. The number of such lines $m$ is ( ).
(A) 16
(B) 18
(C) 24
(D) 27

Solution:
5.C.

As shown in Figure 4, there are a total of $8 \times 3=24$ lines.

Answer: C


Given that $f(x)$ is a function defined on
$\mathbf{R}$, $f(1)=1$ and for any $x \in \mathbf{R}$ we have
$$
f(x+5) \geqslant f(x)+5, f(x+1) \leqslant f(x)+1 \text {. }
$$

If $g(x)=f(x)+1-x$, then $g(2002)=$

Solution:
10.1.
$$
\begin{array}{l}
\text { From } g(x)=f(x)+1-x \text { we get } f(x)=g(x)+x-1 . \\
\text { Then } g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+5 \text {, } \\
g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1 . \\
\text { Therefore, } g(x+5) \geqslant g(x), g(x+1) \leqslant g(x) . \\
\therefore g(x) \leqslant g(x+5) \leqslant g(x+4) \leqslant g(x+3) \\
\quad \leqslant g(x+2) \leqslant g(x+1) \leqslant g(x) . \\
\therefore g(x+1)=g(x) .
\end{array}
$$

Thus, $g(x)$ is a periodic function with a period of 1.
Given $g(1)=1$, hence $g(2002)=1$.

Answer: 1


Form the equation of the tangent to the parabola $y=x^{2}-4 x$ at the point with abscissa $x_{0}=1$.

Solution:
Solution. Let's determine the ordinate $y_{0}$ of the point of tangency by substituting the abscissa value $x_{0}=1$ into the equation of the parabola; we have $y_{0}=1-4 \cdot 1=$ $=-3$.

To find the slope of the tangent, we compute the value of the derivative at the point of tangency: $f^{\prime}(x)=2 x-4 ; k=f^{\prime}(1)=-2$.

Now, knowing the point ( $1 ;-3$ ) and the slope $k=y^{\prime}=-2$, we can write the equation of the tangent line:

$$
y+3=-2(x-1), y+3=-2 x+2 \text { or } y=-2 x-1
$$

Answer: -2x-1


A market vendor's basket contains eggs. The first buyer buys half of the eggs and one more. The second buyer buys half of the remaining eggs and one more. The third buyer also gets exactly half of the eggs left in the basket and one more. How many eggs were in the basket?

Solution:
I. solution (with equation): If we denote the number of eggs by $x$, then according to the problem, the first buyer takes $\frac{x}{2}+1$ eggs and leaves $\frac{x}{2}-1$. The second buyer takes $\frac{1}{2}\left(\frac{x}{2}-1\right)+1$ eggs and leaves $\frac{1}{2}\left(\frac{x}{2}-1\right)-1$. The third buyer takes $\frac{1}{2}\left[\frac{1}{2}\left(\frac{x}{2}-1\right)-1\right]^{2}+1$ eggs and leaves $\frac{1}{2}\left[\frac{1}{2}\left(\frac{x}{2}-1\right)-1\right]-1$, which, according to the problem, equals zero. Therefore,

$$
\begin{array}{r}
\frac{1}{2}\left[\frac{1}{2}\left(\frac{x}{2}-1\right)-1\right]-1=0 \\
\frac{x}{4}-\frac{1}{2}-1=2
\end{array}
$$

from which

$$
x=14
$$

Orlik Péter (Bp. V., Eötvös g. I. o. t.)

II. solution (by reasoning): Let's put the eggs back into the basket. The third buyer returns the last 1 egg they took, which was half of the eggs they found, so they return 1 more egg. The second buyer returns the extra egg they took, and since the 3 eggs in the basket represent half of what the second buyer found, the second buyer adds 3 more eggs to make the total 6. The first buyer returns the extra 1 egg they took, and the 7 eggs in the basket represent half of what the first buyer found, so there were 14 eggs in the basket.

Szabó Anna (Jászberény, Tanítóképző II. o. t.)

III. solution: Generally,

the number of eggs before the last purchase is 2,

the number of eggs before the second-to-last purchase is $2(2+1)=2^{2}+2=6$,

the number of eggs before the purchase before that is $2\left(2^{2}+2+1\right)=2^{3}+2^{2}+2=14$, and so on

$\vdots$

for $n$ buyers, the number of eggs before the first purchase is

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

according to the formula for the sum of a geometric series.

Biszterszky Sára (Miskolc, 13. sz. gépip. techn. II. o. t.)

Note: This problem is a typical example (especially when considering more than 3 buyers) that equations are not always the shortest path. Sometimes simple arithmetic can lead to the solution more quickly. (L. Perelman: "Entertaining Algebra" p. 108.)

Answer: 14


In the sequence $\left\{a_{n}\right\}$,
$$
a_{0}=a\left(a \in \mathbf{Z}_{+}\right), a_{n+1}=\frac{a_{n}^{2}}{a_{n}+1} \text {. }
$$

When $0 \leqslant n \leqslant \frac{a}{2}+1$, the greatest integer not exceeding $a_{n}$ is $\qquad$

Solution:
5. $a-n$.

For any positive integer $n$, it is known by recursion that $a_{n}>0$.
By $a_{n}-a_{n+1}=\frac{a_{n}}{a_{n}+1}=1-\frac{1}{1+a_{n}} \in(0,1)$, we know the sequence $\left\{a_{n}\right\}$ is decreasing, and for any $n \in \mathbf{Z}_{+}$,
$a_{n}=a_{0}+\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)>a-n$.
Thus, $a_{n-1}>a-(n-1) \ (n \geqslant 2)$.
Therefore, when $n=1$,
$\sum_{i=1}^{n} \frac{1}{1+a_{i-1}}=\frac{1}{1+a_{0}}<1$.
When $2 \leqslant n \leqslant \frac{a}{2}+1$, by the sequence $\left\{a_{n}\right\}$ being decreasing,
$\sum_{i=1}^{n} \frac{1}{1+a_{i-1}}<\frac{n}{1+a_{n-1}}<\frac{n}{a-n+2} \leqslant 1$.
Hence, $a-n<a_{n}=a-n+\sum_{i=1}^{n} \frac{1}{1+a_{i-1}}<a-n+1$.
Therefore, the greatest integer not exceeding $a_{n}$ is $a-n$. In particular, when $n=0$, it also holds.

Answer: a-n


If the two numbers of each book are different, then the same number may correspond to two different books: one book is numbered as such according to the new number, and another book is numbered as such according to the old number. This can easily cause confusion. Now, consider placing these books in separate piles so that in each pile, one number corresponds to only one book. Then, how many piles are needed to achieve this?

Our answer is: Just 3 piles are enough.

Solution: The 1st book can be placed in any pile at will. Suppose that the first $\mathrm{k}-1$ books have been successfully sorted into piles, we now consider the placement of the $\mathrm{k}$-th book: the old number in this book may match the new number of one of the $\mathrm{k}-1$ books already sorted, in which case this book cannot be placed in that pile; the new number of this book may also match the old number of one of the books already sorted, in which case this book cannot be placed in that pile either; apart from these, no other duplicate numbers will occur, so at most, it cannot be placed in two piles, meaning there is always one pile available for it. Therefore, the $\mathrm{k}$-th book can be successfully placed. By mathematical induction: these $\mathrm{n}$ books can be divided into three piles, ensuring that in each pile, one number corresponds to one book.

Answer: 3


Find the largest natural number $n$ such that the value of the sum

$$
\lfloor\sqrt{1}\rfloor+\lfloor\sqrt{2}\rfloor+\lfloor\sqrt{3}\rfloor+\ldots+\lfloor\sqrt{n}\rfloor
$$

is a prime number. The notation $\lfloor x\rfloor$ denotes the greatest integer not greater than $x$.

Solution:
SOLUTION. Consider the infinite sequence $\left(a_{n}\right)_{n=1}^{\infty}$ of natural numbers $a_{n}=\lfloor\sqrt{n}\rfloor$. This sequence is clearly non-decreasing, and since

$$
k=\sqrt{k^{2}}6$ the value of the fraction

$$
\frac{(k-1) k(4 k+1)}{6}
$$

remains a natural number even after canceling out six by a natural number divisible by some prime factor $p$ of the number $k$, which naturally divides another addend in the derived expression for the sum $s_{n}$. The prime number $p$ will thus be a divisor of the number $s_{n}$, with $p \leqq k<s_{n}$, so the number $s_{n}$ cannot be a prime. Therefore, the solution to the given problem only needs to be sought among such natural numbers $n$ for which $k=\lfloor\sqrt{n}\rfloor \leqq 6$, hence only numbers $n<(6+1)^{2}=49$ remain in play. For $n=48$, substituting into the derived formula for $s_{n}$ we calculate $s_{48}=203$, which is a number divisible by seven. For $n=47$, we get $s_{47}=197$, which is a prime number.

Answer. The sought number is $n=47$.

SUPPLEMENTARY AND ADDITIONAL PROBLEMS:

N1. Prove that $1+3+5+\ldots+(2 n-1)=n^{2}$.

N2. Prove that $1^{2}+2^{2}+\ldots+n^{2}=\frac{1}{6} n(n+1)(2 n+1)$.

N3. Determine how many integer solutions the equation

![](https://cdn.mathpix.com/cropped/2024_04_17_bd202734538ab0b74df6g-07.jpg?height=113&width=943&top_left_y=846&top_left_x=682)

$[38-\mathrm{B}-\mathrm{II}-4]$

D1. Prove that for every natural number $n \geqq 2$ the following holds:

$$
\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor+\ldots+\lfloor\sqrt[n]{n}\rfloor=\left\lfloor\log _{2} n\right\rfloor+\left\lfloor\log _{3} n\right\rfloor+\ldots+\left\lfloor\log _{n} n\right\rfloor
$$

[Fill an $n \times n$ table with numbers such that the cell in the $a$-th row and $b$-th column contains the number $a^{b}$, and count the number of cells where the number does not exceed $n$ in two ways: by columns and by rows.]

Answer: 47


The area of an equilateral triangle inscribed in a circle is 81 cm$^{2}$. Find the radius of the circle.

Solution:
Solution: $S_{\Delta}=81 c M^{2}, S_{\Delta}=\frac{1}{2} a \cdot h=\frac{1}{2} a \cdot a \cos 30^{\circ}=\frac{a^{2} \sqrt{3}}{4}, r=$ ?

$S_{\triangle}=3 S_{\triangle A O C}=3 \cdot \frac{1}{2} a \cdot r \cos 60^{\circ}=\frac{3}{4} a r=\frac{3}{2} r \cdot r \cdot \sin 120^{\circ}=\frac{3 r^{2} \sqrt{3}}{4}$,

$\frac{3 r^{2} \sqrt{3}}{4}=81, r^{2}=\frac{81 \cdot 4}{3 \sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}=36 \sqrt{3}, r=6 \sqrt[4]{3}$

![](https://cdn.mathpix.com/cropped/2024_05_06_8eeeb8d6ace591217257g-07.jpg?height=319&width=302&top_left_y=326&top_left_x=1531)

Answer: $r=6 \sqrt[4]{3}$.

Answer: 6\sqrt[4]{3}


The integers from $1$ through $9$ inclusive, are placed in the squares of a $3 \times 3$ grid. Each square contains a different integer. The product of the integers in the first and second rows are $60$ and $96$ respectively. Find the sum of the integers in the third row.

Solution:
1. We are given a \(3 \times 3\) grid with integers from 1 to 9, each appearing exactly once. The product of the integers in the first row is 60, and the product of the integers in the second row is 96. We need to find the sum of the integers in the third row.

2. First, calculate the product of all integers from 1 to 9:
   \[
   9! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 = 362880
   \]

3. The product of the integers in the third row can be found by dividing the total product by the product of the integers in the first and second rows:
   \[
   \text{Product of third row} = \frac{9!}{60 \times 96}
   \]

4. Calculate \(60 \times 96\):
   \[
   60 \times 96 = 5760
   \]

5. Now, divide \(362880\) by \(5760\):
   \[
   \frac{362880}{5760} = 63
   \]

6. Factorize 63 to find the integers in the third row:
   \[
   63 = 3^2 \times 7
   \]

7. Since each integer in the grid is distinct and must be from 1 to 9, the factors 3, 3, and 7 must correspond to the integers 1, 7, and 9 (since 3 appears twice, it must be 1 and 9).

8. Therefore, the integers in the third row are 1, 7, and 9 in some order.

9. The sum of the integers in the third row is:
   \[
   1 + 7 + 9 = 17
   \]

Conclusion:
\[
\boxed{17}
\]

Answer: 17


In a convex pentagon $A B C D E$, a point $M$ is taken on side $A E$, and a point $N$ is taken on side $D E$. Segments $C M$ and $B N$ intersect at point $P$. Find the area of pentagon $A B C D E$ if it is known that quadrilaterals $A B P M$ and $D C P N$ are parallelograms with areas 5 and 8, respectively, and the area of triangle $B C P$ is 10.

| param1 | param2 | param3 | Answer |
| :---: | :---: | :---: | :---: |
| 5 | 8 | 10 | 25 |
| 6 | 10 | 3 | 29 |
| 7 | 9 | 7 | 27.5 |
| 8 | 11 | 20 | 41.2 |
| 9 | 12 | 4 | 38.5 |

## Condition

In a convex pentagon $A B C D E$, a point $M$ is taken on side $A E$, and a point $N$ is taken on side $D E$. Segments $C M$ and $B N$ intersect at point $P$. Find the area of pentagon $A B C D E$ if it is known that quadrilaterals $A B P M$ and $D C P N$ are parallelograms with areas 5 and 8, respectively, and the area of triangle $B C P$ is 10.

## Answer.

25

Solution:
# Solution

Let the areas of parallelograms ABP M and DCPN, triangle BCP, and quadrilateral

![](https://cdn.mathpix.com/cropped/2024_05_06_a839f63dd79b7140018dg-12.jpg?height=63&width=1559&top_left_y=1279&top_left_x=288)
$A E \sqcup B N N C D$. Similarly, $D E \sqcup C M \sqcup\{B$. Then M PNE is also a parallelogram, and thus $S_{4}: S_{1}=N P: P B$ (parallelograms ABPM and M PNE have equal heights). Similarly, triangle $B C P$ and parallelogram $D C P N$ have equal heights, so $S_{3}: S_{2}=$ $\frac{1}{2} B P: P N$. Multiplying the written equalities, we get: $\frac{S_{4}}{S_{1}} \cdot \frac{S_{3}}{S_{2}}=\frac{1}{2} \cdot$ Therefore, $S=S_{1}+S_{2}+$ $S_{3}+S_{4}=S_{1}+S_{2}+S_{3}+\frac{S_{1} S_{2}}{2 S_{3}}=25$

Answer: 25


Let the sets be
$$
\begin{array}{l}
A=\left\{x \left\lvert\, x+\frac{5-x}{x-2}=2 \sqrt{x+1}\right., x \in \mathbf{R}\right\}, \\
B=\{x \mid x>2, x \in \mathbf{R}\} .
\end{array}
$$

Then the elements of the set $A \cap B$ are

Solution:
2. $\frac{5+\sqrt{13}}{2}$.

Notice that, when $x>2$,
$$
x+\frac{5-x}{x-2}=x-2+\frac{x+1}{x-2} \geqslant 2 \sqrt{x+1} \text {. }
$$

By the equality, we have $x-2=\frac{x+1}{x-2}$.
Solving this gives $x=\frac{5+\sqrt{13}}{2}$.

Answer: \frac{5+\sqrt{13}}{2}


Solution:
$$
\begin{array}{c}
\text { Solve the original expression }=\frac{(\sqrt{2}+\sqrt{3})^{2}-(\sqrt{5})^{2}}{\sqrt{2}+\sqrt{3}+\sqrt{5}} \\
=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{3}-\sqrt{5})}{\sqrt{2}+\sqrt{3}+\sqrt{5}} . \\
=\sqrt{2}+\sqrt{3}-\sqrt{5} .
\end{array}
$$

This method is very effective, but as a math problem, this alone cannot achieve the desired effect.

Solving math problems using the idea of combining numbers and shapes often gives people a fresh and extraordinary feeling.

Construct $\triangle A B C$ such that $a=\sqrt{2}, b=\sqrt{3}, c=$ $\sqrt{5}$,
let $p=$
$$
\frac{a+b+c}{2} \text {. }
$$

It is easy to see that $\triangle A B C$ is a right triangle.

Construct $\odot O$ inscribed in the right triangle $\triangle A B C$, with radius $r$, and note that $r=\frac{a+b-c}{2}=$
$$
\begin{array}{l}
\sqrt{2}+\frac{\sqrt{3}-v^{\prime}}{2} . \\
\because S_{\triangle A B C}=\frac{a b}{2}=\frac{\sqrt{2} \times \sqrt{3}}{2}=\frac{\sqrt{6}}{2}, \\
\therefore 2 \sqrt{6}=4 S_{\triangle A B C} . \\
\text { Also, } S_{\triangle A B C}=r p, \\
\therefore \text { the original expression }=\frac{4 S_{\triangle A B C}}{2 p}=\frac{4 r p}{2 p}=2 r \\
=\sqrt{2}+\sqrt{3}-\sqrt{5} .
\end{array}
$$

From this, it can be seen that combining numbers and shapes is a powerful tool for broadening thinking, a 'higher level of thinking'.

Answer: \sqrt{2}+\sqrt{3}-\sqrt{5}


Arrange all proper fractions into a sequence $\left\{a_{n}\right\}$ :
$$
\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \cdots
$$

The sorting method is: from left to right, first arrange the denominators in ascending order, and for fractions with the same denominator, arrange them in ascending order of the numerators. Then $a_{2017}=$ $\qquad$ .

Solution:
7. $\frac{1}{65}$.

Divide the sequence into segments by the denominator, there are $k$ fractions with the denominator $k+1$, noticing that,
$$
\frac{63 \times 64}{2}=2016, \frac{64 \times 65}{2}=2080,
$$

we know that 2017 belongs to the 64th segment, so $a_{2017}$ should be the first number with the denominator 65, which is $\frac{1}{65}$.

Answer: \frac{1}{65}


In an $m \times n$ grid, each square is either filled or not filled. For each square, its value is defined as $0$ if it is filled and is defined as the number of neighbouring filled cells if it is not filled. Here, two squares are neighbouring if they share a common vertex or side. Let $f(m,n)$ be the largest total value of squares in the grid. Determine the minimal real constant $C$ such that $$\frac{f(m,n)}{mn} \le C$$holds for any positive integers $m,n$

Solution:
1. **Understanding the Grid and Values:**
   - Each square in the grid can have up to 8 neighbors (if it is not on the edge or corner).
   - The value of a filled square is \( 0 \).
   - The value of an unfilled square is the number of its neighboring filled squares.

2. **Maximizing the Total Value \( f(m,n) \):**
   - To maximize the total value, we need to maximize the number of unfilled squares that have the maximum possible number of filled neighbors.
   - Consider a checkerboard pattern where filled and unfilled squares alternate. This pattern ensures that each unfilled square has the maximum number of filled neighbors.

3. **Checkerboard Pattern Analysis:**
   - In a checkerboard pattern, each unfilled square (white square) is surrounded by filled squares (black squares).
   - For an interior unfilled square, it has 8 neighbors, but only 4 of them are filled (since it is a checkerboard pattern).
   - For edge and corner unfilled squares, the number of filled neighbors will be less.

4. **Calculating the Total Value \( f(m,n) \):**
   - Let’s denote the number of unfilled squares as \( U \) and the number of filled squares as \( F \).
   - In a checkerboard pattern, approximately half of the squares are filled and half are unfilled.
   - Each unfilled square has approximately 4 filled neighbors (on average, considering edge effects).

5. **Bounding \( f(m,n) \):**
   - The total value \( f(m,n) \) is approximately \( 4U \) since each unfilled square contributes 4 to the total value.
   - Since \( U \approx \frac{mn}{2} \), we have:
     \[
     f(m,n) \approx 4 \cdot \frac{mn}{2} = 2mn
     \]

6. **Finding the Constant \( C \):**
   - We need to find \( C \) such that:
     \[
     \frac{f(m,n)}{mn} \le C
     \]
   - From the above calculation, we have:
     \[
     \frac{f(m,n)}{mn} \approx \frac{2mn}{mn} = 2
     \]
   - Therefore, \( C \ge 2 \).

7. **Verifying the Bound:**
   - To ensure that \( C = 2 \) is the minimal constant, we need to check if there are configurations where \( \frac{f(m,n)}{mn} \) exceeds 2.
   - Given the checkerboard pattern provides a tight bound, and no other configuration can provide a higher average value per square, \( C = 2 \) is indeed the minimal constant.

\(\blacksquare\)

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

Answer: C = 2


A circle touches the extensions of two sides $AB$ and $AD$ of square $ABCD$, and the point of tangency cuts off a segment of length 2 cm from vertex $A$. Two tangents are drawn from point $C$ to this circle. Find the side of the square if the angle between the tangents is $30^{\circ}$, and it is known that $\sin 15^{\circ}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$.

Solution:
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-37.jpg?height=442&width=563&top_left_y=1498&top_left_x=815)

Fig. 1

The segment cut off from vertex $A$ by the point of tangency of the circle is equal to the radius of this circle. If radii of the circle are drawn to the points of tangency, a square with a side of 2 cm is formed. Then $O A=2 \sqrt{2}$, $O C=\frac{O K}{\sin 15^{\circ}}=\frac{4 \sqrt{2}}{(\sqrt{3}-1)}$, then the diagonal of the square $A B C D \quad A C=\frac{4 \sqrt{2}}{\sqrt{3}-1}-2 \sqrt{2}$, and the side of the square $A B=\frac{A C}{\sqrt{2}}=\frac{4}{\sqrt{3}-1}-2=2 \sqrt{3}$.

Answer. $2 \sqrt{3}$.

Answer: 2\sqrt{3}


Let positive real numbers $a, b, c$ satisfy $\frac{2}{a}+\frac{1}{b}=\frac{\sqrt{3}}{c}$. Then the minimum value of $\frac{2 a^{2}+b^{2}}{c^{2}}$ is $\qquad$ .

Solution:
4. 9 .

From the given, we have $\frac{2 c}{a}+\frac{c}{b}=\sqrt{3}$.
By the Cauchy-Schwarz inequality and the AM-GM inequality, we get
$$
\begin{array}{l}
\frac{2 a^{2}+b^{2}}{c^{2}} \\
=\frac{1}{3}(2+1)\left[2\left(\frac{a}{c}\right)^{2}+\left(\frac{b}{c}\right)^{2}\right] \\
\geqslant \frac{1}{3}\left(\frac{2 a}{c}+\frac{b}{c}\right)^{2} \\
=\frac{1}{9}\left[\left(\frac{2 a}{c}+\frac{b}{c}\right)\left(\frac{2 c}{a}+\frac{c}{b}\right)\right]^{2} \\
=\frac{1}{9}\left(4+\frac{2 a}{b}+\frac{2 b}{a}+1\right)^{2} \\
\geqslant \frac{1}{9}(4+2 \times 2+1)^{2}=9,
\end{array}
$$

Equality holds if and only if $a=b=\sqrt{3} c$. Therefore, the minimum value of $\frac{2 a^{2}+b^{2}}{c^{2}}$ is 9.

Answer: 9


Find all natural numbers $n, k$ such that
$$ 2^n - 5^k = 7. $$

Solution:
To find all natural numbers \( n \) and \( k \) such that \( 2^n - 5^k = 7 \), we can proceed as follows:

1. **Initial Check for Small Values of \( n \):**
   - For \( n = 1 \):
     \[
     2^1 - 5^k = 2 - 5^k = 7 \implies 5^k = -5 \quad \text{(no solution since \( 5^k \) is positive)}
     \]
   - For \( n = 2 \):
     \[
     2^2 - 5^k = 4 - 5^k = 7 \implies 5^k = -3 \quad \text{(no solution since \( 5^k \) is positive)}
     \]

2. **Consider the Equation Modulo 8:**
   - We take the equation \( 2^n - 5^k = 7 \) modulo 8:
     \[
     2^n \equiv 7 + 5^k \pmod{8}
     \]
   - Since \( 2^n \mod 8 \) cycles through 2, 4, 8, 16, etc., we have:
     \[
     2^n \equiv 0, 2, 4 \pmod{8}
     \]
   - We need \( 5^k \equiv 1 \pmod{8} \) for \( 2^n \equiv 7 + 5^k \pmod{8} \) to hold. This implies \( 5^k \equiv 1 \pmod{8} \).

3. **Check \( 5^k \mod 8 \):**
   - \( 5^1 \equiv 5 \pmod{8} \)
   - \( 5^2 \equiv 25 \equiv 1 \pmod{8} \)
   - Therefore, \( k \) must be even for \( 5^k \equiv 1 \pmod{8} \).

4. **Rewrite the Equation:**
   - Since \( k \) is even, let \( k = 2m \). The equation becomes:
     \[
     2^n - 5^{2m} = 7
     \]
   - This can be rewritten as:
     \[
     2^n - 7 = 5^{2m}
     \]
   - Let \( x = 5^m \). Then the equation becomes:
     \[
     2^n - 7 = x^2
     \]
   - This is a well-known equation in number theory.

5. **Known Solutions to \( 2^n - 7 = x^2 \):**
   - The known solutions to \( 2^n - 7 = x^2 \) are:
     \[
     (x, n) = (1, 3), (3, 4), (5, 5), (11, 7), (181, 15)
     \]
   - We need to find the corresponding \( k \) for each \( n \):
     - For \( n = 3 \):
       \[
       2^3 - 7 = 1 \implies x = 1 \implies 5^m = 1 \implies m = 0 \implies k = 2m = 0 \quad \text{(not a natural number)}
       \]
     - For \( n = 4 \):
       \[
       2^4 - 7 = 9 \implies x = 3 \implies 5^m = 3 \quad \text{(no solution since \( 5^m \) is a power of 5)}
       \]
     - For \( n = 5 \):
       \[
       2^5 - 7 = 25 \implies x = 5 \implies 5^m = 5 \implies m = 1 \implies k = 2m = 2
       \]
     - For \( n = 7 \):
       \[
       2^7 - 7 = 121 \implies x = 11 \implies 5^m = 11 \quad \text{(no solution since \( 5^m \) is a power of 5)}
       \]
     - For \( n = 15 \):
       \[
       2^{15} - 7 = 32761 \implies x = 181 \implies 5^m = 181 \quad \text{(no solution since \( 5^m \) is a power of 5)}
       \]

6. **Conclusion:**
   - The only solution in natural numbers is \( (n, k) = (5, 2) \).

The final answer is \( \boxed{ (n, k) = (5, 2) } \).

Answer: (n, k) = (5, 2)


For any pair of numbers, a certain operation Â«*Â» is defined, satisfying the following properties: $a *(b * c)=(a * b) \cdot c$ and $a * a=1$, where the operation Â«$\cdot$Â» is the multiplication operation. Find the root $x$ of the equation: $\quad x * 2=2018$.

Solution:
Answer: 4036.

## Solution:

Given the condition of the problem, we have $x * 1=x *(x * x)=(x * x) \cdot x=1 \cdot x=x$. Then

1) $(x * 2) \cdot 2=2018 \cdot 2=4036$,
2) $(x * 2) \cdot 2=x *(2 * 2)=x \cdot 1=x$.

Therefore, $x=4036$.

Answer: 4036


Around a circle with radius 3, an isosceles trapezoid $ABCD$ ($BC \| AD$) is described, the area of which is 48. The circle touches the sides $AB$ and $CD$ at points $K$ and $L$. Find $KL$.

Solution:
Let $B H$ be the height of the trapezoid, $O P$ be the perpendicular dropped from the center $O$ of the circle to the chord $K L$, and $K$ be the point of tangency of the circle with the lateral side $A B$. Then triangles $O P K$ and $A H B$ are similar.

## Solution

Let the circle with radius $r=3$ touch the lateral sides $A B$ and $C D$ of the trapezoid $A B C D$ at points $K$ and $L$ respectively; $B H=2 r$ is the height of the trapezoid. By symmetry, $K L \| B C$. Since the trapezoid $A B C D$ is circumscribed, $B C + A D = A B + C D = 2 A B$.

According to the problem, $1 / 2 (B C + A D) \cdot 6 = 48$, from which $B C + A D = 16$. Therefore, $A B = 8$.

The perpendicular $O P$, dropped from the center $O$ of the circle, bisects the chord $K L$. From the similarity of the right triangles $O P K$ and $A H B$, it follows that

$P K : O K = B H : A B$. Hence, $K L = 2 P K = 2 B H \cdot O K / A B = 9 / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-26.jpg?height=520&width=1066&top_left_y=743&top_left_x=496)

## Answer

4.5.

Answer: 4.5


The function $f(x)$ defined on $\mathbf{R}$ satisfies $\left\{\begin{array}{l}f(x+2016) \leqslant f(x)+2016, \\ f(x+2017) \geqslant f(x)+2017,\end{array}\right.$ and $f(1)=2$. Let $a_{n}=f(n)\left(n \in \mathbf{N}^{*}\right)$, then $a_{2018}=$ $\qquad$

Solution:
$$
f(x)+2017 \leqslant f(x+2017)=f(x+1+2016) \leqslant f(x+1)+2016 \Rightarrow f(x+1) \geqslant f(x)+1 \text {, }
$$

Then $f(x)+2016 \geqslant f(x+2016)=f(x+2015+1) \geqslant f(x+2015)+1=f(x+2014+1)+1 \geqslant f(2014)+2$ $\geqslant \cdots \geqslant f(x+1)+2015 \Rightarrow f(x+1) \leqslant f(x)+1$, thus $f(x+1)=f(x)+1$.
Therefore, $a_{2018}=f(2018)=2019$.

Answer: 2019


As shown in Figure $1, \odot O_{1}$ is externally tangent to $\odot O_{2}$ at point $P$. From point $A$ on $\odot O_{1}$, a tangent line $A B$ is drawn to $\odot O_{2}$, with $B$ as the point of tangency. Line $A P$ is extended to intersect $\odot O_{2}$ at point $C$. Given that the radii of $\odot O_{1}$ and $\odot O_{2}$ are $2$ and $1$ respectively, then
$$
\frac{A C}{A B}=
$$

Solution:
5. $\frac{\sqrt{6}}{2}$.

Connect $O_{1} O_{2}$. Then $O_{1} O_{2}$ passes through the point of tangency $P$.
Connect $O_{1} A$, $O_{1} P$, $O_{2} P$, $O_{2} C$.
Then isosceles $\triangle O_{1} A P \backsim$ isosceles $\triangle O_{2} P C$
$$
\Rightarrow \frac{A P}{P C}=\frac{O_{1} P}{O_{2} P}=2 \text {. }
$$

Assume $A P=2 t$. Then $P C=t$.
By the secant-tangent theorem,
$$
\begin{array}{l}
A B^{2}=A P \cdot A C=6 t^{2} \Rightarrow A B=\sqrt{6} t . \\
\text { Therefore, } \frac{A C}{A B}=\frac{3 t}{\sqrt{6} t}=\frac{\sqrt{6}}{2} .
\end{array}
$$

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


Given that there are seven points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, P_{6}, P_{7}$ (not necessarily equally spaced) on a line, and let $P$ be any point selected on the line, and let $s$ be the sum of the lengths of the segments $P P_{1}, P P_{2}, \cdots, P P_{7}$. The position of $P$ when and only when $s$ is minimized is
(A) the midpoint of $P_{1} P_{7}$.
(B) the midpoint of $P_{2} P_{6}$.
(C) the midpoint of $P_{3} P_{5}$.
(D) $P_{4}$.
(E) $P_{1}$.

Solution:
[Solution] First, consider the case with two points $P_{1}, P_{2}$. Here, $P$ can be any point on $P_{1} P_{2}$. Notice that
$$
s=P_{1} P+P P_{2}=P_{1} P_{2} .
$$

Next, consider the case with three points $P_{1}, P_{2}, P_{3}$. Clearly, to minimize $s$, we should choose $P$ to be $P_{2}$, in which case
$$
s=P_{1} P_{2}+P_{2} P_{3}=P_{1} P_{3} .
$$

Similarly, for any $n$ points on a line:
If $n=2 k$, then $P$ should lie on $P_{k} P_{k+1}$;
If $n=2 k+1$, then $P$ should coincide with $P_{k+1}$.
Thus, when $n=7$, $P$ should be chosen at $P_{4}$.
Therefore, the answer is $(D)$.
Note: This problem can be translated into an algebraic problem as follows:
If $x_{1}<x_{2}<\cdots<x_{n}$, find the minimum value of $f(x)=\sum_{i=1}^{n}\left|x-x_{i}\right|$.

Answer: D


A natural number is written on the board. Nikolai noticed that he can append a digit to the right of it in two ways so that the resulting number is divisible by 9. In how many ways can he append a digit to the right of the given number so that the resulting number is divisible by 3?

Solution:
Answer: 4 ways.

Solution. Note that the difference between the two numbers "noticed" by Nikolai is less than 10, but is divisible by 9. Therefore, this difference is 9. This is only possible if the appended digits are 0 and 9. Then it is easy to see that for divisibility by 3, in addition to these two digits, the digits 3 and 6 can also be appended. In total, there are 4 ways.

## Criteria

4 6. A complete and justified solution is provided.

In the absence of a correct solution, the highest applicable criterion from those listed below is used

2 6. It is proven that Nikolai could only append 0 or 9 at the beginning.

Answer: 4


Determine the largest real number $z$, such that
$$x+y+z=5, \quad xy+yz+zx=3,$$

and $x, y$ are also real numbers.

Solution:
[Solution] From $x+y+z=5$, we get
$$(x+y)^{2}=(5-z)^{2},$$

From $x y+y z+z x=3$, we get
$$x y=3-z(5-z) .$$

Thus, we have
$$\begin{aligned}
(x-y)^{2} & =(x+y)^{2}-4 x y=(5-z)^{2}-4[3-z(5-z)] \\
& =-3 z^{2}+10 z+13=(13-3 z)(1+z),
\end{aligned}$$

Therefore,
$$\begin{array}{l}
(13-3 z)(1+z) \geqslant 0 \\
-1 \leqslant z \leqslant \frac{13}{3}
\end{array}$$

When $x=y=\frac{1}{3}$, $z=\frac{13}{3}$. Therefore, the maximum value of $z$ is $\frac{13}{3}$.

Answer: \frac{13}{3}


At what angle of inclination of the lateral sides is the cross-sectional area of the channel the largest?

Solution:
Solution. Let the smaller base of the trapezoid be denoted by $a$, the angle of inclination of the lateral sides by $\alpha$, and the area of the section by $S$ (Fig. 124). According to the condition, $|A B|=|A D|=|B C|=a$.

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-261.jpg?height=215&width=356&top_left_y=1598&top_left_x=122)

Fig. 124 From triangle $B C K$, we find $|B K| = a \cos \alpha$. Then $|C D| = a + 2a \cos \alpha$. The height of the trapezoid is $|C K| = a \sin \alpha$. Let's determine the area of the trapezoid:

$$
\begin{gathered}
S = \frac{|A B| + |D C|}{2} |C K| = \frac{a + a + 2a \cos \alpha}{2} a \sin \alpha = \\
= a^2 (1 + \cos \alpha) \sin \alpha.
\end{gathered}
$$

To find the maximum value of the area $S = S(\alpha)$, we need to find the critical points within the interval $(0, \pi / 2)$, compute the values of the function $S(\alpha)$ at these points and at the endpoints of the interval $(0, \pi / 2)$, and then select the largest value from the obtained results.

Let's find the derivative:

$$
\begin{aligned}
& S'(\alpha) = a^2 \left( (1 + \cos \alpha)' \sin \alpha + (1 + \cos \alpha) (\sin \alpha)' \right) = \\
= & a^2 \left( -\sin^2 \alpha + \cos \alpha + \cos^2 \alpha \right) = a^2 \left( 2 \cos^2 \alpha + \cos \alpha - 1 \right)
\end{aligned}
$$

Set the derivative to zero and solve the resulting equation:

$$
2 \cos^2 \alpha + \cos \alpha - 1 = 0 ; \quad \cos \alpha_1 = 1 / 2 ; \quad \cos \alpha_2 = -1
$$

from which $\alpha_1 = \pm \pi / 3 + 2 \pi n, \alpha_2 = \pi + 2 \pi n$. Only the critical point $\alpha_1 = \pi / 3$ belongs to the interval $(0, \pi / 2)$.

Let's find $S(\alpha_1) = S\left(\frac{\pi}{3}\right) = a^2 \left(1 + \cos \frac{\pi}{3}\right) \sin \frac{\pi}{3} = \frac{3 \sqrt{3}}{4} a^2, S(0) = 0$, $S\left(\frac{\pi}{2}\right) = a^2$. Comparing the obtained results, we conclude that the function $S(\alpha)$ attains its maximum value at $\alpha = \pi / 3$.

Thus, the area of the channel section is the largest when the angle of inclination of the lateral sides is $60^{\circ}$.

Answer: 60


A line passing through the origin is tangent to the circle $x^{2}+y^{2}+4 x+3=0$. If the point of tangency is in the third quadrant, then the equation of the line is $\qquad$ .

Solution: 7. $y=\frac{\sqrt{3}}{3} x$ Hint: $(x+2)^{2}+y^{2}=1$, center $(-2,0)$, radius is 1, by plane geometry, the angle of inclination of the required tangent is $30^{\circ}$.

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


Mr. AlgebriÄ decided to encourage his two sons to solve a problem, so he said that they would determine their own pocket money for going out in the following way. From the set $\left\{5,5^{2}, 5^{3}, \ldots, 5^{33}\right\}$, they need to choose two different numbers, $a$ and $b$, such that $\log _{a} b$ is an integer. Each of them will receive as many kuna as the number of such pairs of numbers they write. How much kuna could one of them earn at most?

Solution:
## Solution.

Let $a=5^{x}, b=5^{y}$, where $x$ and $y$ are numbers from the set $\{1,2,3, \ldots, 32,33\}$. Then,

$$
\log _{a} b=\log _{5^{x}} 5^{y}=\frac{y}{x}
$$

This fraction will be an integer if and only if $y$ is a multiple of $x$, greater than $x$, and less than or equal to 33. We have the following cases:

$$
\begin{aligned}
& x=1 \Rightarrow y \in\{2,3, \ldots, 33\} \text { which is a total of } 32 \text { pairs of the form }(1, y) . \\
& x=2 \Rightarrow y \in\{4,6,8, \ldots, 32\} \text { which is a total of }\left\lfloor\frac{33}{2}\right\rfloor-1=15 \text { pairs of the form }(2, y) . \\
& x=3 \Rightarrow y \in\{6,9,12 \ldots, 33\} \text { which is a total of }\left\lfloor\frac{33}{3}\right\rfloor-1=10 \text { pairs of the form }(3, y) . \\
& x=4 \Rightarrow y \in\{8,12,16 \ldots, 32\} \text { which is a total of }\left\lfloor\frac{33}{4}\right\rfloor-1=7 \text { pairs of the form }(4, y) . \\
& x=5 \Rightarrow y \in\{10,15,20,25,30\} \text { which is a total of }\left\lfloor\frac{33}{5}\right\rfloor-1=5 \text { pairs of the form }(5, y) . \\
& x=6 \Rightarrow y \in\{12,18,24,30\} \text { which is a total of }\left\lfloor\frac{33}{6}\right\rfloor-1=4 \text { pairs of the form }(6, y) . \\
& x=7 \Rightarrow y \in\{14,21,28\} \text { which is a total of }\left\lfloor\frac{33}{7}\right\rfloor-1=3 \text { pairs of the form }(7, y) \\
& x=8 \Rightarrow y \in\{16,24,32\} \text { which is a total of }\left\lfloor\frac{33}{8}\right\rfloor-1=3 \text { pairs of the form }(8, y) .
\end{aligned}
$$

Multiples of the numbers $9,10,11$ less than 33 are each 2, and multiples of the numbers $12,13,14,15,16$ are each 1.

The total number of pairs is

$$
32+15+10+7+5+4+3+3+3 \cdot 2+5 \cdot 1=90
$$

We conclude that one of the sons can earn a maximum of 90 kn.

Answer: 90


990 divided by 11 leaves a remainder of ( ).
(A) 1
(B) 3
(C) 5
(D) 7

Solution:
A

Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.

Answer: A


Diagonals of a convex quadrilateral $ABCD$ intersect at point $E$, $AB=AD$, $CA$ is the bisector of angle $C$,

$\angle BAD=140^{\circ}, \angle BEA=110^{\circ}$.

Find the angle $CDB$.

Solution:
Continue sides $B C$ and $A D$ until they intersect at point $F$ and prove that triangle $C D F$ is isosceles.

## Solution

The angles at the base $B D$ of the isosceles triangle $B A D$ are each $20^{\circ}$. Therefore, $\angle C A D = \angle A E B - \angle A D E = 90^{\circ}$.

Extend sides $B C$ and $A D$ until they intersect at point $F$. Since the bisector $C A$ of triangle $C D F$ is also its altitude, triangle $C D F$ is isosceles. Therefore, $F A = A D = A B$.

Since the median $A B$ of triangle $B F D$ is half the length of side $D F$, $\angle D B F = 90^{\circ}$. Therefore, $\angle C D F = \angle B F D = 90^{\circ} - \angle B D F = 70^{\circ}$.

Thus, $\angle C D B = \angle C D F - \angle B D A = 50^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-33.jpg?height=574&width=440&top_left_y=342&top_left_x=815)

Answer

$50^{\circ}$.

Answer: 50


Given $h(0)=2$, for $n \geqslant 1$, $h(n)>0$, and
$$
h^{2}(n)-2 \cdot h^{2}(n-1)=1 \text {, }
$$

Find the general term formula for the sequence $\{h(n)\}_{n \geqslant 0}$.

Solution: 17. $h(n)=\sqrt{5 \cdot 2^{n}-1}$.

Answer: (n)=\sqrt{5\cdot2^{n}-1}


Given that $\left\{a_{n}\right\}$ is a geometric sequence, and $a_{1} a_{2017}=1$, if $f(x)=\frac{2}{1+x^{2}}$, then $f\left(a_{1}\right)+f\left(a_{2}\right)+f\left(a_{3}\right)+\cdots+f\left(a_{2017}\right)=$

Solution:
$$
f\left(a_{1}\right)+f\left(a_{2017}\right)=\frac{2}{1+a_{1}^{2}}+\frac{2}{1+a_{2017}^{2}}=\frac{4+2\left(a_{1}^{2}+a_{2017}^{2}\right)}{a_{1}^{2} a_{2017}^{2}+a_{1}^{2}+a_{2017}^{2}+1}=2 \text{, }
$$

Therefore, $f\left(a_{1}\right)+f\left(a_{2}\right)+f\left(a_{3}\right)+\cdots+f\left(a_{2017}\right)=2017$.

Answer: 2017


Find all natural numbers $n$ such that $n$ has as many digits as it has distinct prime divisors, and the sum of the distinct prime divisors is equal to the sum of the powers of the same divisors.

Solution:
Solution. Let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$. From the condition of the problem, we have

$$
p_{1}+p_{2}+\ldots+p_{k}=\alpha_{1}+\alpha_{2}+\ldots+\alpha_{k}
$$

We consider the number of digits of the number $n$. If it has 4 digits, then it has 4 different prime divisors. Then $n \geq 2^{14} \cdot 3 \cdot 5 \cdot 7>10^{4}$, which is not possible. If $n$ has $k>4$ digits, then

$$
\begin{aligned}
n & \geq 2^{2+3+5+7+p_{5}+\ldots+p_{k}-(k-1)} \cdot 3 \cdot 5 \cdot 7 \cdot p_{5} \cdot \ldots \cdot p_{k} \\
& =2^{14} \cdot 3 \cdot 5 \cdot 7 \cdot 2^{p_{5}+\ldots+p_{k}-(k-4)} p_{5} \cdot \ldots \cdot p_{k} \\
& >10^{4} \cdot 10^{k-4}=10^{k}
\end{aligned}
$$

which is again not possible. Therefore, $n$ is at most a three-digit number.

Let $n$ be a three-digit number. Then $n=p_{1}^{\alpha_{1}}{p_{2}}^{\alpha_{2}}{p_{3}}^{\alpha_{3}}$. If $5 \mid n$, then

$$
n \geq 2^{8} \cdot 3 \cdot 5>10^{3}
$$

We conclude that the prime factors of $n$ are less than or equal to 3. However, the only prime numbers less than or equal to 3 are 2 and 3, and there are 3 prime numbers in the factorization of $n$, which is a contradiction.

Let $n$ be a two-digit number. Then $n=p_{1}^{\alpha_{1}}{p_{2}}^{\alpha_{2}}$. If $5 \mid n$, then

$$
n \geq 2^{6} \cdot 5>10^{2}
$$

Thus, $n=2^{\alpha_{1}} 3^{\alpha_{2}}$, where $\alpha_{1}+\alpha_{2}=5$. By direct verification, we find that

$$
n=2^{4} \cdot 3=48, n=2^{3} \cdot 3^{2}=72
$$

are solutions.

Let $n$ be a one-digit number. Then only $n=2^{2}$ satisfies the condition of the problem.

Answer: 48,72,4


On a shelf, there are three books. The first has 90, the second 110, and the third 150 pages. The covers of the books are of equal thickness, and each of them is 2 mm thick. How many millimeters thick are the books together if it is known that 10 pages have a thickness of $1 \mathrm{~mm}$?

Solution:
Solution. Each book has two covers, so the total number of covers is six, and their total thickness is $12 \mathrm{~mm}$. The total number of pages is

$$
90+110+150=350
$$

So the total thickness of all the pages is $350: 10=35 \mathrm{~mm}$. Finally, the total thickness of all three books together is $12+35=47 \mathrm{~mm}$.

Answer: 47\mathrm{~}


For some positive integers $m$ and $n, 2^{m}-2^{n}=1792$. Determine the value of $m^{2}+n^{2}$.

Solution:
Since $2^{11}=2048$ and $2^{8}=256$, then $2^{11}-2^{8}=2048-256=1792$.

Therefore, $m=11$ and $n=8$, which gives $m^{2}+n^{2}=11^{2}+8^{2}=121+64=185$.

Alternatively, we can factor the left side of the equation $2^{m}-2^{n}=1792$ to obtain $2^{n}\left(2^{m-n}-1\right)=$ 1792 .

Since $m>n$, then $2^{m-n}-1$ is an integer.

Now, $1792=2^{8} \cdot 7$.

Since $2^{n}\left(2^{m-n}-1\right)=2^{8} \cdot 7$, then $2^{n}=2^{8}$ (which gives $n=8$ ) and $2^{m-n}-1=7$ (which gives $m-n=3$ and so $m=11$ ).

ANSWER: 185

Answer: 185