Among the three given phrases "Ã¥Â°Â½Ã¥Â¿ÂÃ¥Â°Â½Ã¥ÂÂ", "Ã¥ÂÂÃ¥ÂÂ¯Ã¦ÂÂÃ¥Â±Â±", and "Ã¥Â±Â±Ã§Â©Â·Ã¦Â°Â´Ã¥Â°Â½", each Chinese character represents a number between 1 and 8, with the same character representing the same number and different characters representing different numbers. If the sum of the numbers represented by the characters in each phrase is 19, and "Ã¥Â°Â½" > "Ã¥Â±Â±" > "Ã¥ÂÂ", then the maximum value of "Ã¦Â°Â´" is $\qquad$

Solution:
【Analysis】Through analysis, we can know:
From “尽心尽力” (exerting oneself to the utmost), “力可拔山” (strength can move mountains), and “山穷水尽” (reaching the end of one's resources), the sum of the numbers represented by the Chinese characters in each word is 19, which gives us the equations:
$$
\left\{\begin{array}{l}
\text { 尽+心+尽+力=19(1) } \\
\text { 力+可+拔+山=19(2) } \\
\text { 山+穷+水+尽 }=19(3)
\end{array}\right.
$$
We can get 3 尽 + 心 + 2 力 + 可 + 拔 + 2 山 + 穷 + 水 = 19 × 3 = 57
The sum of 1 to 8 is 36, so 2 尽 + 1 力 + 1 山 = 57 - 36 = 21, comparing with (1) we get 山 - 心 = 2.
“尽” > “山” > “力”, “力” should be as large as possible, and “尽” should be as small as possible,
Assuming “力”, “山”, “尽” are consecutive natural numbers, we have 2 (力 + 2) + 力 + 1 + 力 = 21
“力” is 4, at this time 山 = 5, 心 = 3, 尽 = 6;
(1) is satisfied: 6 + 3 + 6 + 4 = 19;
(3): 5 + 穷 + 水 + 6 = 19, at this time the maximum value of 水 is 7, 穷 is 1, to deduce (2):
(2): 4 + 可 + 拔 + 5 = 19, and now only 2 and 8 are left, which satisfies the condition. At this time, the maximum value of 水 is 7.

But at this time, 6 (尽), 4 (山), 5 (力) do not satisfy “尽” > “山” > “力”, so it does not meet the requirements. Therefore, the maximum value of 水 is 7.

According to this, we can solve the problem.
【Solution】From the words “尽心尽力” (exerting oneself to the utmost), “力可拔山” (strength can move mountains), and “山穷水尽” (reaching the end of one's resources), the sum of the numbers represented by the Chinese characters in each word is 19, which gives us the equations:
$$
\left\{\begin{array}{l}
\text { 尽+心+尽+力=19(1) } \\
\text { 力+可+拔+山=19(2) } \\
\text { 山+穷+水+尽 }=19(3)
\end{array}\right.
$$
(1) + (2) + (3) gives:
3 尽 + 心 + 2 力 + 可 + 拔 + 2 山 + 穷 + 水 = 19 × 3 = 57
The sum of 1 to 8 is 36,
So 2 尽 + 1 力 + 1 山 = 57 - 36 = 21, comparing with (1) we get 山 - 心 = 2.
“尽” > “山” > “力”, “力” should be as large as possible, and “尽” should be as small as possible,
Assuming “力”, “山”, “尽” are consecutive natural numbers, we have 2 (力 + 2) + 力 + 1 + 力 = 21
“力” is 4, at this time 山 = 5, 心 = 3, 尽 = 6;
(1) is satisfied: 6 + 3 + 6 + 4 = 19;
(3): 5 + 穷 + 水 + 6 = 19, at this time the maximum value of 水 is 7, 穷 is 1, to deduce (2):
(2): 4 + 可 + 拔 + 5 = 19, and now only 2 and 8 are left, which satisfies the condition. At this time, the maximum value of 水 is 7.

But at this time, 6 (尽), 4 (山), 5 (力) do not satisfy “尽” > “山” > “力”, so it does not meet the requirements. Therefore, the maximum value of 水 is 7.

The answer is: 7.

Answer: 7


Let $g(k)$ denote the greatest odd divisor of the positive integer $k$
$$
\begin{aligned}
\text { (for example, } g(3) & =3, g(20)=5) . \text { Find } \\
f(n) & =g(1)+g(2)+g(3)+\cdots+g\left(2^{n}\right) .
\end{aligned}
$$

Solution:
For odd $k$, we have $g(k)=k$.
For any $k \in \mathbf{Z}_{+}$, we have
$$
g(2 k)=g(k) \text {. }
$$
$$
\begin{array}{l}
\text { Then } f(n)=1+3+5+\cdots+\left(2^{n}-1\right)+ \\
g(2)+g(4)+\cdots+g\left(2^{n}\right) \\
=\frac{\left(1+2^{n}-1\right) 2^{n-1}}{2}+g(1)+g(2)+\cdots+g\left(2^{n-1}\right) \\
=4^{n-1}+f(n-1) .
\end{array}
$$

Notice that,
$$
\begin{array}{l}
f(2)-f(1)=4, \\
f(3)-f(2)=4^{2}, \\
\ldots . .
\end{array}
$$
$$
f(n)-f(n-1)=4^{n-1} \text {. }
$$

Adding all these equations, we get
$$
f(n)-f(1)=4+4^{2}+\cdots+4^{n-1} \text {. }
$$

Since $f(1)=g(1)+g(2)=1+1=2$, we have
$$
f(n)=\frac{4^{n}-4}{3}+2=\frac{4^{n}+2}{3} .
$$

Answer: \frac{4^{n}+2}{3}


Let the positive integer $n$ be a multiple of 75, and have exactly 75 positive divisors (including 1 and itself). Find the minimum value of $n$.

Solution:
Solution: Let the prime factorization of $n$ be
$$
n=p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{k}^{r_{k}},
$$

where $p_{1}, p_{2}, \cdots, p_{k}$ are the distinct prime factors of $n$, and $r_{1}, r_{2}, \cdots, r_{k}$ are all positive integers.
Thus, the number of distinct positive divisors of $n$ is
$$
\left(r_{1}+1\right)\left(r_{2}+1\right) \cdots\left(r_{k}+1\right).
$$

From the problem, we have
$$
\left(r_{1}+1\right)\left(r_{2}+1\right) \cdots\left(r_{k}+1\right)=75=3 \times 5^{2}.
$$

Therefore, $n$ can have at most 3 distinct prime factors.
To make $n$ the smallest and a multiple of 75, the prime factors of $n$ should be $2, 3, 5$, and 3 should appear at least once, and 5 should appear at least twice, i.e.,
$$
\begin{array}{l}
n=2^{r_{1}} \times 3^{r_{2}} \times 5^{r_{3}}, \\
\left(r_{1}+1\right)\left(r_{2}+1\right)\left(r_{3}+1\right)=75, \\
r_{1} \geqslant 0, r_{2} \geqslant 1, r_{3} \geqslant 2 .
\end{array}
$$

The tuples $\left(r_{1}, r_{2}, r_{3}\right)$ that satisfy the above conditions are
$$
\begin{array}{l}
(4,4,2),(4,2,4),(2,4,4),(0,4,14), \\
(0,14,4),(0,2,24),(0,24,2) .
\end{array}
$$

After calculation, it is found that when $r_{1}=r_{2}=4, r_{3}=2$, $n$ takes the minimum value, which is $n=2^{4} \times 3^{4} \times 5^{2}=32400$.

Answer: 32400


Ezekiel has a rectangular piece of paper with an area of 40. The width of the paper is more than twice the height. He folds the bottom left and top right corners at $45^{\circ}$ and creates a parallelogram with an area of 24. What is the perimeter of the original rectangle?

Solution:
Suppose that the original rectangle has height $h$ and width $w$.

Since the area of the original rectangle is 40 , then $h w=40$.

Once the corners are folded, the height of the resulting parallelogram is still $h$ and the new base is $w-h$, because the two legs of each folded triangle are equal since these triangles are isosceles.

![](https://cdn.mathpix.com/cropped/2024_04_30_a838ccd738aa6ca69034g-07.jpg?height=276&width=480&top_left_y=1971&top_left_x=882)

Since the area of the parallelogram is 24 , then $h(w-h)=24$ or $h w-h^{2}=24$.

Since $h w=40$, then $h^{2}=40-24=16$.

Since $h>0$, then $h=4$.

Since $h w=40$ and $h=4$, then $w=10$.

The perimeter of the original rectangle is $2 h+2 w$ which equals $2 \cdot 4+2 \cdot 10=28$.

Answer: 28


Given the function $f(x)=\sin \left(x+\frac{\pi}{6}\right)+\sin \left(x-\frac{\pi}{6}\right)+\cos x+a(a \in \mathbf{R}, a$ is a constant $)$,
(I) Find the smallest positive period of the function $f(x)$.
(II) If the maximum value of $f(x)$ is 1, find the value of $a$.
(III) If $x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ when the maximum value of $f(x)$ is 1, find the value of $a$.

Solution:
13. ( I ) $f(x)=\sqrt{3} \sin x+\cos x+a$
$$
=2 \sin \left(x+\frac{\pi}{6}\right)+u \text {. }
$$

So the smallest positive period of $f(x)$ is $2 \pi$.
(II) From H(1), the maximum value of $f(x)$ is $2+a$.

Therefore, $2+a=1$. Thus, $a=-1$.
(III) Since $x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Then $x+\frac{\pi}{6} \in\left[-\frac{\pi}{3}, \frac{2}{3} \pi\right]$.
So the maximum value of $f(x)$ is $2+a$.
Thus, $2+a=1$, which means $a=-1$.

Answer: -1


The integer $M=2003^{2004}2008$ when divided by 80 leaves a remainder of ( ).
(A) 1
(B) 11
(C) 31
(D) 79

Solution:
6.A.

Let $m=2004^{2000^{2008}}, p=2005^{2000^{2002^{2008}}}$.
Then $m=2004^{p}=4^{p} \times 501^{p}=4 n, n=4^{p-1} \times 501^{p}$.
Thus, by the binomial theorem we have
$$
\begin{array}{l}
M=2003^{m}=(2000+3)^{m} \\
=2000^{m}+\mathrm{C}_{m}^{1} \cdot 2000^{m-1} \cdot 3+\cdots+ \\
\mathrm{C}_{m}^{m-1} \cdot 2000 \cdot 3^{m-1}+3^{m} .
\end{array}
$$

Therefore, the remainder when $M$ is divided by 80 is the same as the remainder when $3^{m}$ is divided by 80.
$$
\begin{array}{l}
\text { Also, } 3^{m}=3^{4 n}=81^{n}=(80+1)^{n} \\
=80^{n}+C_{n}^{1} \cdot 80^{n-1}+\cdots+C_{n}^{n-1} \cdot 80+1,
\end{array}
$$

Hence, the remainder when $M$ is divided by 80 is 1.

Answer: A


An essay in Latin is scored on a scale from 0 to $20$. Michel's score is above the average, while Claude's score is below the average. What score did each of them receive, if it is known that when one third of the smaller of these two scores is subtracted from each of them, one of the resulting remainders will be three times the other remainder?

Solution:
18. Let $g$ be the larger of the two estimates, and $p$ be the smaller of them. Then

$$
g-p / 3=3(p-p / 3)
$$

from which it follows that

$$
g=7 p / 3
$$

If $p=3$ and $g=7$, Michelle would have received a score below average, which contradicts the problem's condition. Therefore, the only possible solution is:

$$
g=14 \text{ and } p=6
$$

In other words, Michelle scored 14 points, and Claude scored 6.

Answer: =14p=6


Calculate

$$
\int_{0}^{1} x \sin \left(\pi x^{2}\right) \mathrm{d} x
$$

Calculate

$$
\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1} k \int_{\frac{k}{n}}^{\frac{k+1}{n}} \sin \left(\pi x^{2}\right) \mathrm{d} x
$$

Solution:
Solution. (a) Making the substitution $t=\pi x^{2}$, the integral becomes

$$
\frac{1}{2 \pi} \int_{0}^{\pi} \sin t \, \mathrm{d} t=\left.\frac{1}{2 \pi}(-\cos t)\right|_{0} ^{\pi}=\frac{1}{\pi}
$$

## 2 points

(b) Let $f:[0,1] \rightarrow \mathbb{R}, f(x)=\sin \left(\pi x^{2}\right)$, and $F:[0,1] \rightarrow \mathbb{R}, F(x)=\int_{0}^{x} f(t) \, \mathrm{d} t$. Then

$$
\begin{aligned}
\frac{1}{n} \sum_{k=0}^{n-1} k \int_{\frac{k}{n}}^{\frac{k+1}{n}} f(x) \, \mathrm{d} x & =\frac{1}{n} \sum_{k=0}^{n-1}\left(k F\left(\frac{k+1}{n}\right)-k F\left(\frac{k}{n}\right)\right) \\
& =\frac{1}{n} \sum_{k=0}^{n-1}\left((k+1) F\left(\frac{k+1}{n}\right)-k F\left(\frac{k}{n}\right)-F\left(\frac{k+1}{n}\right)\right) \\
& =\frac{1}{n}\left(n F(1)-\sum_{k=1}^{n} F\left(\frac{k}{n}\right)\right)=F(1)-\frac{1}{n} \sum_{k=1}^{n} F\left(\frac{k}{n}\right)
\end{aligned}
$$

The desired limit is

$$
F(1)-\int_{0}^{1} F(x) \, \mathrm{d} x=\int_{0}^{1} x F^{\prime}(x) \, \mathrm{d} x=\int_{0}^{1} x f(x) \, \mathrm{d} x=\frac{1}{\pi}
$$

2 points

Answer: \frac{1}{\pi}


Petya drew a square on the plane, divided it into 64 identical smaller squares, and colored them in a checkerboard pattern with black and white colors. After that, he thought of a point strictly inside one of these smaller squares. Vasya can draw any closed broken line without self-intersections on the plane and get an answer to the question of whether the guessed point is strictly inside the broken line or not. What is the minimum number of such questions Vasya needs to determine the color of the guessed point - white or black?

Solution:
To determine the color of a point in one question, a broken line is needed, relative to which all squares of one color lie inside, and the other - outside. But then it contains all the segments separating adjacent squares, and therefore, it intersects itself.

P Let's show how to guess the color of a point in two questions. We will construct a broken line, inside which all the odd horizontal rows of the square will fall and only they (see fig.).

![](https://cdn.mathpix.com/cropped/2024_05_06_17014d53b52b1543ee40g-32.jpg?height=514&width=598&top_left_y=103&top_left_x=730)

Then with the first question, we will find out the parity of the horizontal row containing the guessed point. With a similar second question, we will find out the parity of the vertical row containing it. It remains to note that the color of the square is determined by the parity of the sum of its "coordinates".

## Answer

In two questions.

Answer: 2


In a convex quadrilateral $ABCD$, points $E$ and $F$ are the midpoints of sides $BC$ and $CD$ respectively. Segments $AE, AF$, and $EF$ divide the quadrilateral into 4 triangles, the areas of which are consecutive natural numbers. What is the maximum possible value of the area of triangle $ABD$?

Solution:
3. Let the areas of the triangles be $n, n+1, n+2$, $n+3$. Then the area of the quadrilateral $A B C D$ is $4 n+6$. It is easy to see that the area of triangle $B C D$ is four times the area of triangle $E C F$, so this area is at least $4 n$. Therefore,

$$
S_{A B D}=S_{A B C D}-S_{B C D} \leqslant(4 n+6)-4 n=6 .
$$

Equality is achieved if triangle $E C F$ has the smallest area

![](https://cdn.mathpix.com/cropped/2024_05_21_6455f48598646f41a890g-307.jpg?height=177&width=333&top_left_y=1048&top_left_x=911)

Fig. 151 of the four triangles mentioned.

It remains to prove that the value 6 is possible. An example is an isosceles trapezoid with bases $A D=6, B C=4$ and height 2 (the numbers in Fig. 151 denote the areas of the triangles).

Comment. It is easy to see that the area of triangle $A B D$ is always either 2 or 6.

Answer: 6


Solve the following system of equations:

$$
\begin{array}{r}
x^{2}+y \sqrt{x y}=105 \\
y^{2}+x \sqrt{y x}=70
\end{array}
$$

Solution:
None of the unknowns is 0 (otherwise at least one of the equations' left sides would be 0), furthermore, they are of the same sign, otherwise $\sqrt{x y}$ would not make sense; let's assume for now that both are positive. Then $\sqrt{x y}=\sqrt{x} \cdot \sqrt{y}=x^{\frac{1}{2}} \cdot y^{\frac{1}{2}}$ and our equations can be transformed as follows:

$$
x^{\frac{1}{2}}\left(x^{\frac{3}{2}}+y^{\frac{3}{2}}\right)=105, \quad y^{\frac{1}{2}}\left(x^{\frac{3}{2}}+y^{\frac{3}{2}}\right)=70
$$

According to the above, the expression in parentheses is positive; by eliminating it (briefly: dividing the two equations), $x^{\frac{1}{2}}: y^{\frac{1}{2}}=105$ : $70=3: 2$, so $y^{\frac{1}{2}}=2 x^{\frac{1}{2}} / 3, y^{\frac{3}{2}}=8 x^{\frac{3}{2}} / 27$, and with this from the first equation $35 x^{2} / 27=105, x^{2}=81, x=9$, and finally $y=4$. This pair of values satisfies both equations.

Another root is not possible, because the $x, y$ with it leads to a contradiction. Indeed, if $x$ and $y$ are negative and $x=y$, then $\sqrt{x y}=\sqrt{x^{2}}=|x|=-x=-y$, and the left side of both equations is 0. If $x \neq y$, from here by multiplying with the negative $x$, $x \sqrt{x y} < 0$. Finally, $y < x^{2}, \sqrt{x y} > -x, y \sqrt{x y} < 0$; based on the assumption that led us to $x^{2}=81$, so $x \neq -9$.

Answer: 9,4


Determine all natural numbers $n \geq 3$ for which the following statement holds:

Every plane convex $n$-gon $A_{1} A_{2} \ldots A_{n}$ is completely covered by the areas of the $n$ circles that have the segments $A_{i} A_{i+1}$ as diameters $\left(i=1,2, \ldots, n\right.$; set $A_{n+1}=A_{1}$).

Here, each $n$-gon area and each circle area are understood to include their boundary points.

Solution:
I. For any triangular area $F=A_{1} A_{2} A_{3}$ and any convex quadrilateral area $F=A_{1} A_{2} A_{3} A_{4}$, the mentioned covering statement holds; this can be proven as follows:

If the statement were false, there would be a point $P$ in $F$ that lies outside each of the three or four mentioned

circles. Since these circles cover the boundary of $F$, $P$ would lie in the interior of $F$. Since $F$ is convex, the angles $\angle A_{i} P A_{i+1}\left(i=1, \ldots, n ; A_{n+1}=A_{1}\right.$, with $n=3$ or $\left.n=4\right)$ would have to satisfy

$$
\sum_{i=1}^{n} \angle A_{i} P A_{i+1}=360^{\circ}
$$

On the other hand, since $P$ lies outside the circles with diameters $A_{i} A_{i+1}$, $\angle A_{i} P A_{i+1}4$ there exists a convex n-sided area $A_{1} A_{2} \ldots A_{n}$ that is not covered by the mentioned circles; this is shown by the following example:

If $A_{1} A_{2} \ldots A_{n}$ is a regular n-gon and $P$ is its center, then (1) (now with $n>4$) holds and thus $\angle A_{i} P A_{i+1}=\frac{1}{n} \cdot 360^{\circ}<90^{\circ}$ for all $i=1, \ldots, n ; A_{n+1}=A_{1}$.

Therefore, $P$ lies outside all circles with diameters $A_{i} A_{i+1}$.

With I. and II: it is proven that the statement mentioned in the problem holds exactly for $n=3$ and $n=4$.

## Adopted from $[5]$

Answer: n=3n=4


Solve:
$\begin{cases}{x\left( {6 - y} \right) = 9}\\
{y\left( {6 - z} \right) = 9}\\
{z\left( {6 - x} \right) = 9}\end{cases}$

Solution:
Given the system of equations:
\[
\begin{cases}
x(6 - y) = 9 \\
y(6 - z) = 9 \\
z(6 - x) = 9
\end{cases}
\]

1. **Express each variable in terms of the others:**
   From the first equation, we can solve for \( y \):
   \[
   x(6 - y) = 9 \implies 6x - xy = 9 \implies xy = 6x - 9 \implies y = \frac{6x - 9}{x}
   \]

   From the second equation, we can solve for \( z \):
   \[
   y(6 - z) = 9 \implies 6y - yz = 9 \implies yz = 6y - 9 \implies z = \frac{6y - 9}{y}
   \]

   From the third equation, we can solve for \( x \):
   \[
   z(6 - x) = 9 \implies 6z - zx = 9 \implies zx = 6z - 9 \implies x = \frac{6z - 9}{z}
   \]

2. **Substitute and simplify:**
   Substitute \( y = \frac{6x - 9}{x} \) into the second equation:
   \[
   \left(\frac{6x - 9}{x}\right)(6 - z) = 9
   \]
   Simplify:
   \[
   \frac{(6x - 9)(6 - z)}{x} = 9
   \]
   Multiply both sides by \( x \):
   \[
   (6x - 9)(6 - z) = 9x
   \]
   Expand and simplify:
   \[
   36x - 6xz - 54 + 9z = 9x
   \]
   Rearrange terms:
   \[
   36x - 6xz + 9z - 54 = 9x
   \]
   Combine like terms:
   \[
   27x - 6xz + 9z = 54
   \]

3. **Use symmetry and AM-GM inequality:**
   Notice that the equations are symmetric. We can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality to find a solution. The AM-GM inequality states that for non-negative real numbers \( a_1, a_2, \ldots, a_n \):
   \[
   \frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}
   \]
   Equality holds if and only if \( a_1 = a_2 = \cdots = a_n \).

   Apply AM-GM to the product \( x(6 - x)y(6 - y)z(6 - z) \):
   \[
   \left(\frac{x + (6 - x) + y + (6 - y) + z + (6 - z)}{6}\right)^6 \geq x(6 - x)y(6 - y)z(6 - z)
   \]
   Simplify the left-hand side:
   \[
   \left(\frac{6 + 6 + 6}{6}\right)^6 = 6^3 = 216
   \]
   Given \( x(6 - x)y(6 - y)z(6 - z) = 9^3 = 729 \), we have:
   \[
   216 \geq 729
   \]
   This is a contradiction unless \( x = y = z = 3 \).

4. **Verify the solution:**
   Substitute \( x = 3 \), \( y = 3 \), and \( z = 3 \) back into the original equations:
   \[
   3(6 - 3) = 9 \\
   3(6 - 3) = 9 \\
   3(6 - 3) = 9
   \]
   All equations are satisfied.

Thus, the solution is \( x = 3 \), \( y = 3 \), and \( z = 3 \).

The final answer is \( \boxed{ x = 3 } \), \( y = 3 \), and \( z = 3 \).

Answer: x = 3


Given the function $f_{1}(x)=\frac{2 x-1}{x+1}$, for positive integer $n$, define $f_{n+1}(x)=f_{1}\left[f_{n}(x)\right]$. Find the analytical expression for $f_{1234}(x)$.

Solution:
Analysis If we start from the given conditions and gradually derive the analytical expression of $f_{1234}(x)$, the workload would be enormous, which is clearly impractical. Therefore, we can generalize the original problem and directly derive the analytical expression of $f_{n}(x)$.

Solution According to the given conditions, we have
$$
\begin{array}{c}
f_{1}(x)=\frac{2 x-1}{x+1}, \\
f_{2}(x)=f_{1}\left[f_{1}(x)\right]=f_{1}\left(\frac{2 x-1}{x+1}\right)=\frac{x-1}{x}, \\
f_{3}(x)=f_{1}\left[f_{2}(x)\right]=f_{1}\left(\frac{x-1}{x}\right)=\frac{x-2}{2 x-1}, \\
f_{1}(x)=f_{1}\left[f_{3}(x)\right]=f_{1}\left(\frac{x-2}{2 x-1}\right)=\frac{1}{1-x}, \\
f_{5}(x)=f_{1}\left[f_{1}(x)\right]=f_{1}\left(\frac{1}{1-x}\right)=\frac{x+1}{2-x}, \\
f_{6}(x)=f_{1}\left[f_{5}(x)\right]=f_{1}\left(\frac{x+1}{2-x}\right)=x, \\
f_{7}(x)=f_{1}\left[f_{6}(x)\right]=f_{1}(x) .
\end{array}
$$

This indicates that $\left\{f_{n}(x)\right\}$ is a periodic function sequence with the smallest positive period of 6. Therefore, the analytical expression of $f_{n}(x)$ is
$$
f_{n}(x)=\left\{\begin{array}{l}
\frac{2 x-1}{x+1}(n=6 k+1), \\
\frac{x-1}{x}(n=6 k+2), \\
\frac{x-2}{2 x-1}(n=6 k+3), \\
\frac{1}{1-x}(n=6 k+4), \\
\frac{x+1}{2-x}(n=6 k+5), \\
x(n=6 k+6),
\end{array}\right.
$$

where $k$ is a non-negative integer.
By substituting $n=1234$ into the expression of $f_{n}(x)$, we get
$$
f_{1231}(x)=f_{6 \times 211+1}(x)=\frac{1}{1-x} .
$$

Explanation The solution process above demonstrates two fundamental steps: one is the generalization process, where to find the analytical expression of $f_{1234}(x)$, we first generalize from the specific to the general, directly deriving the expression of $f_{n}(x)$; the other is the specialization process, where in seeking the expression of $f_{n}(x)$, we return from the general to the specific, examining the cases when $n=1,2, \cdots, 7$ based on the initial conditions and recursive relations, to identify the underlying patterns. Then, we combine these two processes organically, thereby solving the original problem. This shows that in practical problem-solving, generalization and specialization can often be used together to complement and support each other, enhancing problem-solving efficiency.

In college entrance exams and mathematics competitions, problems related to the periodicity of functions and function sequences frequently appear. Correctly recognizing and grasping the periodic characteristics of such periodic functions (sequences) is crucial for solving problems.

Answer: \frac{1}{1-x}


In the Cartesian coordinate system, it is known that points $A(0,4)$ and $B(3,8)$. If point $P(x, 0)$ makes $\angle APB$ the largest, then $x=$ $\qquad$.

Solution: (10) Let $\angle A P B=\theta$. Since the circle with diameter $A B$ does not intersect the $x$-axis, $\theta<\frac{\pi}{2}$. To maximize $\angle A P B$, we need $\sin \theta$ to be maximized, which in turn requires the circumradius of $\triangle A B P$ to be minimized. It is easy to see that the circumradius of $\triangle A B P$ is minimized when its circumcircle is tangent to the $x$-axis. Extend $B A$ to intersect the $x$-axis at point $Q$, then $Q(-3,0)$. By the power of a point theorem, $Q P^{2}=Q A \cdot Q B=5 \cdot 10$. Therefore, $Q P=5 \sqrt{2}$, and at this point, the $x$-coordinate of point $P$ is $x=5 \sqrt{2}-3$.

Answer: 5\sqrt{2}-3


Solution:
$25 \%$ of 2004 is $\frac{1}{4}$ of 2004 , or 501 .

$50 \%$ of 4008 is $\frac{1}{2}$ of 4008 , or 2004.

$50 \%$ of 1002 is $\frac{1}{2}$ of 1002 , or 501 .

$100 \%$ of 1002 is 1002 .

$10 \%$ of 8016 is $\frac{1}{10}$ of 8016 , or 801.6 .

$20 \%$ of 3006 is $\frac{1}{5}$ of 3006 , or 601.2 .

Answer: (B)

Answer: B


Solution:
1. We rewrite the inequality in an equivalent form:

$$
\begin{aligned}
& \log \frac{2^{2} \cdot 3^{1+\frac{1}{2 x}}}{3^{\frac{1}{x}}+3^{3}}>\log 1 \\
\Rightarrow \quad & \frac{2^{2} \cdot 3^{1+\frac{1}{2 x}}}{3^{\frac{1}{x}}+3^{3}}>1 \\
3^{\frac{1}{x}}-12 \cdot 3^{\frac{1}{2 x}}+27 & 0$, we obtain the inequality

10 points

$$
t^{2}-12 t+27<0 \quad \text { i.e. } \quad(t-3)(t-9)<0
$$

whose solution is $t \in(3,9)$.

10 points

The solution to the given inequality is $x \in\left(\frac{1}{4}, \frac{1}{2}\right)$.

5 points

Answer: x\in(\frac{1}{4},\frac{1}{2})


12 different items are distributed among 3 people so that each person gets 4 items. In how many ways is this possible?

Solution:
I. solution. The first item that goes to the first person $E$ can be chosen in 12 ways. Regardless of which item was chosen first, the second item can be chosen from the remaining ones in 11 ways. Similarly, the third and fourth items can be chosen in 10 and 9 ways, respectively, for $E$. This gives $N_{1}=12 \cdot 11 \cdot 10 \cdot 9$ possibilities for the distribution of these 4 items. This includes the order in which the items are given.

However, the order is irrelevant; $E$ will receive the same items regardless of whether they are given in the order $A, B, C, D$ or $B, D, C, A$, or any other order. Similar reasoning shows that 4 different items can be given in $4 \cdot 3 \cdot 2=24$ different orders, i.e., in this many ways we can choose which to give first, second, and third, with the last one being determined. Therefore, the number of distinct ways to satisfy $E$ is

$$
N_{1}=\frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2}=11 \cdot 5 \cdot 9=495
$$

Similarly, the second person $M$ can receive their share from the remaining 8 items in

$$
N_{2}=\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2}=70
$$

different ways. This completes the distribution, with the third person $H$ receiving the remaining items. Thus, the distribution can be done in $N_{1} \cdot N_{2}=34650$ ways.

László Szilágyi (Debrecen, Fazekas M. g. IV. o. t.)

II. solution. Prepare 4 cards each with the letters $E, M$, and $H$ to mark the items to be given to the 3 people. For now, distinguish the cards with the same letter by indexing them as $1,2,3,4$: $E_{1}, E_{2}, E_{3}, E_{4}, M_{1}, \ldots, H_{4}$.

Similar reasoning to the first solution shows that the 12 cards can be placed on the (arranged) 12 items in $N_{3}=12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2$ ways. Each placement represents a distribution possibility, but not all are distinct. The 4 $E$-lettered cards can be swapped among the 4 items without changing $E$'s share, and as seen in the first solution, there are $N_{4}=4 \cdot 3 \cdot 2$ ways to do this. Repeating this reasoning for $M$ and $H$, we find that each distinct distribution appears $\left(N_{4}\right)^{3}$ times in the $N_{3}$ number of distributions. Thus, the number of distinct distributions is:

$$
\frac{N_{3}}{\left(N_{4}\right)^{3}}=\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{4 \cdot 3 \cdot 2 \cdot 4 \cdot 3 \cdot 2 \cdot 4 \cdot 3 \cdot 2}
$$

After simplifying the last three factors in the numerator and the denominator, we see that our expression is equal to the product $N_{1} N_{2}$ from the first solution.

Angela Nagy (Balassagyarmat, Szántó Kovács J. g. IV. o.t.)

Note: Many contestants provided answers using combinatorial formulas, despite the fact that the rules of the competition - K.M.L. 25 (1962) p. 15 - do not accept mere references to theorems not covered in the high school curriculum, as the thought process is more important than the ready-made template. Nevertheless, we accepted as solutions those papers that justified why the formulas they used led to the correct result. (We also point out: there is a large number of incorrect papers, in which the thought process is missing or incorrect.)

Answer: 34650


The perimeter of a right-angled triangle $ABC$ ( $\angle C=90^{\circ}$ ) is 72 cm, and the difference between the lengths of the median $CK$ and the altitude $CM$ is 7 cm (Fig. 10.47). Find the length of the hypotenuse.

Solution:
Solution.

According to the condition $C K - C M = 7 \text{ cm}, P_{A B C} = 72 \text{ cm}, p = \frac{P_{A B C}}{2} = 36$. If the radius of the inscribed circle, then $r = p - A B, S = p \cdot r = p \cdot (p - A B)$. Let $A B = x, K$ be the center of the circumscribed circle, $C K = A K = K B = \frac{A B}{2} = \frac{x}{2}; S = 36 \cdot (36 - x); \frac{x}{2} - C M = 7$ (by condition) $C M = \frac{x}{2} - 7$; $S = \frac{x}{2} \left( \frac{x}{2} - 7 \right) \cdot \frac{x^2}{4} - \frac{7x}{2} = 36^2 - 36x, x^2 + 130x - 5184 = 0, x = 32$ cm.

Answer: $32 \text{ cm}$.

Answer: 32


The range of the function $f(x)=\frac{\sin 2 x}{2+\sin x+\cos x}$ is

Solution:
- 1. $\left[2 \sqrt{3}-4,1+\frac{\sqrt{2}}{2}\right]$.

Let $t=\sin x+\cos x$. Then $t=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right) \in[-\sqrt{2}, \sqrt{2}]$.
Also, $\sin 2 x=t^{2}-1$, so, $f(x)=\frac{t^{2}-1}{2+t}$.
Let $m=t+2 \in[2-\sqrt{2}, 2+\sqrt{2}]$. Then $t=m-2$.
Thus, $f(x)=m+\frac{3}{m}-4$.
By the monotonicity of the double hook function, we have
$$
f(x) \in\left[2 \sqrt{3}-4,1+\frac{\sqrt{2}}{2}\right]
$$

Answer: [2\sqrt{3}-4,1+\frac{\sqrt{2}}{2}]


Each face of a cube has a different color (the six colors are fixed), how many different coloring methods are there?

Solution: 8. Let the six colors be called $1,2,3,4,5,6$. Place the cube on the table with the 1-colored face down. Consider the 2-colored face; if it is up, it can be rotated around the vertical axis to bring the 3-colored face to the front. This fixes the cube. The other faces can be colored in $3!=6$ ways. If the 2-colored face is adjacent to the 1-colored face, rotate the cube to bring the 2-colored face to the front, fixing the cube. The remaining faces can be colored in $4!=24$ ways. In total, there are $6+$ $24=30$ different coloring methods.

Answer: 30


Given the general term of the sequence $\left\{a_{n}\right\}$
$$
a_{n}=\frac{n x}{(x+1)(2 x+1) \cdots(n x+1)}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

If $a_{1}+a_{2}+\cdots+a_{2015}<1$, then the value of the real number $x$ is ( ).
(A) $-\frac{3}{2}$
(B) $-\frac{5}{12}$
(C) $-\frac{9}{40}$
(D) $-\frac{11}{60}$

Solution:
6. D.

From the problem, we know
$$
\begin{array}{l}
a_{n}=\frac{1}{(x+1)(2 x+1) \cdots[(n-1) x+1]}- \\
\quad \frac{1}{(x+1)(2 x+1) \cdots(n x+1)} . \\
\text { Then } \sum_{k=1}^{2015} a_{k} \\
=1-\frac{1}{(x+1)(2 x+1) \cdots(2015 x+1)}0 . \\
\text { Therefore, } x \in \bigcup_{k=1}^{100}\left(-\frac{1}{2 k-1},-\frac{1}{2 k}\right) \cup\left(-\frac{1}{2015},+\infty\right) .
\end{array}
$$

Upon verification, only $x=-\frac{11}{60}$ satisfies the problem.

Answer: D


In quadrilateral $\mathrm{ABCD}$, $\mathrm{AC}$ and $\mathrm{BD}$ intersect at point $\mathrm{O}$. Draw the altitude $\mathrm{DE}$ of triangle $\mathrm{DBC}$, and connect $\mathrm{AE}$. If the area of triangle $\mathrm{ABO}$ is equal to the area of triangle $\mathrm{DCO}$, and $\mathrm{DC}=17$ cm, $\mathrm{DE}=15$ cm, then what is the area of the shaded part in square centimeters?

Solution:
【Analysis】Since $S_{\triangle A B O}=S_{\triangle D C O}$, then $S_{\triangle A B C}=S_{\triangle D C B}$. Because the two triangles share the base $\mathrm{BC}$, the heights from $\mathrm{BC}$ in both triangles are equal, so $\mathrm{AD}$ is parallel to $\mathrm{BC}$. Therefore, in triangle $\mathrm{ACE}$, the height on side $\mathrm{CE}$ is 15 cm.
In the right triangle $\mathrm{CDE}$, by the Pythagorean theorem, we have
$$
C E^{2}=C D^{2}-D E^{2}=17^{2}-15^{2}=(17+15)(17-15)=64 \text {, }
$$

Thus, $\mathrm{CE}=8$ cm.
Therefore, $S_{\triangle A C E}=\frac{1}{2} \times 8 \times 15=60$ square cm.

Answer: 60


a) Prove that the set $A=\{4,9,16,25,36,49\}$ is isolated;

b) Determine the composite numbers $n$ for which there exist the positive integers  $a,b$ such that the set
\[ A=\{(a+b)^2, (a+2b)^2,\ldots, (a+nb)^2\}\] is isolated.

Solution:
### Part (a)
1. **Calculate the sum of the set \( A \):**
   \[
   A = \{4, 9, 16, 25, 36, 49\}
   \]
   The sum of the elements in \( A \) is:
   \[
   S = 4 + 9 + 16 + 25 + 36 + 49 = 139
   \]
2. **Check if the sum \( S \) is a prime number:**
   - The number 139 is a prime number because it has no divisors other than 1 and itself.
3. **Verify the condition for isolated set:**
   - For any proper subset \( B \subset A \), the sum of the elements in \( B \) must be co-prime with 139.
   - Since 139 is a prime number, any integer that is not a multiple of 139 is co-prime with 139.
   - The sum of any proper subset of \( A \) will be less than 139 and hence cannot be a multiple of 139.

Thus, the set \( A = \{4, 9, 16, 25, 36, 49\} \) is isolated.

### Part (b)
1. **Express the sum of the elements in the set \( A \):**
   \[
   A = \{(a+b)^2, (a+2b)^2, \ldots, (a+nb)^2\}
   \]
   The sum of the elements in \( A \) is:
   \[
   S = \sum_{k=1}^{n} (a+kb)^2
   \]
   Expanding the sum:
   \[
   S = \sum_{k=1}^{n} (a^2 + 2akb + k^2b^2) = na^2 + 2ab \sum_{k=1}^{n} k + b^2 \sum_{k=1}^{n} k^2
   \]
   Using the formulas for the sums of the first \( n \) natural numbers and the first \( n \) squares:
   \[
   \sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
   \]
   Substituting these into the expression for \( S \):
   \[
   S = na^2 + 2ab \cdot \frac{n(n+1)}{2} + b^2 \cdot \frac{n(n+1)(2n+1)}{6}
   \]
   Simplifying:
   \[
   S = na^2 + n(n+1)ab + \frac{n(n+1)(2n+1)}{6}b^2
   \]

2. **Determine the composite numbers \( n \) for which the set is isolated:**
   - For the set to be isolated, the sum \( S \) must be co-prime with the sum of any proper subset.
   - If \( n > 6 \) or \( n = 4 \), there exists a divisor \( d > 1 \) of \( n \) which divides \( S \).
   - At least one of \( a+b, a+2b, \ldots, a+nb \) is divisible by \( n \) and thus by \( d \), meaning it is not co-prime with \( S \).

Thus, the only composite number \( n \) for which the set is isolated is \( n = 6 \).

The final answer is \( \boxed{ n = 6 } \).

Answer: n = 6


Solution:
To find a formula for \( T_n \) of the form \( T_n = A_n + B_n \), where \( \{A_n\} \) and \( \{B_n\} \) are well-known sequences, we need to verify that \( A_n = n! \) and \( B_n = 2^n \) satisfy the given recurrence relation and initial conditions.

1. **Initial Conditions:**
   - \( T_0 = 2 \)
   - \( T_1 = 3 \)
   - \( T_2 = 6 \)

2. **Proposed Sequences:**
   - Let \( A_n = n! \)
   - Let \( B_n = 2^n \)

3. **Verify Initial Conditions:**
   - For \( n = 0 \):
     \[
     T_0 = A_0 + B_0 = 0! + 2^0 = 1 + 1 = 2
     \]
   - For \( n = 1 \):
     \[
     T_1 = A_1 + B_1 = 1! + 2^1 = 1 + 2 = 3
     \]
   - For \( n = 2 \):
     \[
     T_2 = A_2 + B_2 = 2! + 2^2 = 2 + 4 = 6
     \]

4. **Verify Recurrence Relation:**
   - Given recurrence relation:
     \[
     T_n = (n+4)T_{n-1} - 4nT_{n-2} + (4n-8)T_{n-3}
     \]
   - Substitute \( T_n = n! + 2^n \):
     \[
     T_n = (n! + 2^n)
     \]
   - Check if \( A_n = n! \) and \( B_n = 2^n \) satisfy the recurrence:
     \[
     (n! + 2^n) = (n+4)(n-1)! + 2^{n-1}) - 4n((n-2)! + 2^{n-2}) + (4n-8)((n-3)! + 2^{n-3})
     \]

5. **Simplify and Verify:**
   - For \( A_n = n! \):
     \[
     n! = (n+4)(n-1)! - 4n(n-2)! + (4n-8)(n-3)!
     \]
     This is a known identity for factorials.
   - For \( B_n = 2^n \):
     \[
     2^n = (n+4)2^{n-1} - 4n2^{n-2} + (4n-8)2^{n-3}
     \]
     This can be verified by expanding and simplifying the terms.

Since both \( A_n = n! \) and \( B_n = 2^n \) satisfy the initial conditions and the recurrence relation, we conclude that:

\[
T_n = n! + 2^n
\]

The final answer is \( \boxed{ T_n = n! + 2^n } \)

Answer: T_n = n! + 2^n


Find all real solutions to the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+u^{2}+v^{2}=4, \\
x y u+y u v+u v x+v x y=-2, \\
x y u v=-1
\end{array}\right.
$$

Solution:
From equations (1) and (3), we get
\[
\frac{x^{2}+y^{2}+u^{2}+v^{2}}{4}=1=\sqrt[1]{|x y u v|}.
\]
Since \(\frac{x^{2}+y^{2}+u^{2}+v^{2}}{4} \geqslant \sqrt[4]{|x y u v|}\) with equality holding if and only if \(|x|=|y|=|u|=|v|\), the solution to (4) is
\[
\begin{array}{c}
|x|=|y|=|u|=|v|=1 . \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{u}+\frac{1}{v}=2 .
\end{array}
\]

From (2) and (3), we get
Thus, from (5) and (6), the solution to the original system of equations is
\[
\begin{array}{c}
(x, y, u, v)=(1,1,1,-1),(1,1,-1,1), \\
(1,-1,1,1),(-1,1,1,1) .
\end{array}
\]

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


How many locks at least need to be placed on the treasury so that with a certain distribution of keys among the 11-member committee authorized to open the treasury, any 6 members can open it, but no 5 can? Determine the distribution of keys among the committee members with the minimum number of locks.

Solution:
Suppose that for some natural number $ n $ there exists a key distribution to $ n $ locks among an 11-member committee such that the conditions of the problem are satisfied. Let $ A_i $ denote the set of locks that the $ i $-th member of the committee can open, where $ i = 1, 2, \ldots, 11 $, and let $ A $ denote the set of all locks. Then from the conditions of the problem, we have

for any five-element subset $ \{ i_1, \ldots, i_5\} $ of the set $ \{1, 2, \ldots, 11\} $ and

for any six-element subset $ \{j_1, \ldots, j_6\} $ of the set $ \{1,2,\ldots,  11\} $.
From (1), it follows that the set $ A - (A_{i_1} \cup \ldots \cup A_{i_5}) $ is non-empty. Let $ x_{i_1}, \ldots, x_{i_5} $ be one of its elements, i.e., a lock that the group of committee members numbered $ i_1, \ldots, i_5 $ cannot open. From (2), it follows that for every $ j \not \in \{i_1, \ldots, i_5 \} $ we have $ x_{i_1 , \ldots, i_5} \in A_j $.
Suppose that $ x_{i_1,\ldots,i_5} = x_{k_1, \ldots, k_5} $ for some subsets $ \{i_1, \ldots, i_5\} $ and $ \{k_1, \ldots, k_5\} $. If these subsets were different, then, for example, $ i_t \not \in \{ k_1, \ldots, k_5 \} $. Therefore, $ x_{k_1, \ldots, k_5} \in A_{i_t} $; but on the other hand, this leads to a contradiction. The obtained contradiction proves that $ \{i_1, \ldots,i_5\} = \{ k_1, \ldots, k_5 \} $.
In other words, different five-element subsets $ \{i_1, \ldots, i_5\} $ correspond to different locks. Therefore, the number of locks is not less than the number of five-element subsets of an 11-element set, i.e., $ n \geq  \binom{11}{5} = 462 $.
We will now prove that if we install $  \binom{11}{5} $ locks on the treasury, then we can distribute the keys to them among the members of the 11-member committee in such a way that the conditions of the problem are satisfied.
Let us associate each of the $  \binom{11}{5} $ locks with a five-element subset of the set $ \{1, 2, \ldots, 11\} $ in a one-to-one manner. If a lock corresponds to the subset $ \{i_1, \ldots, i_5\} $, then the keys to it are given to all members of the committee whose numbers are different from $ i_1, \ldots, i_5 $.
We will show that no five members of the committee can open a certain lock, and therefore the treasury. Indeed, the members of the committee numbered $ i_1, \ldots, i_5 $ do not have the key to the lock corresponding to the subset $ \{i_1, \ldots, i_5\} $.
We will show that any six members of the committee can open any lock, and therefore the treasury. If the members of the committee have numbers $ j_1, \ldots, j_6 $ and want to open a lock corresponding to the subset $ \{i_1, \ldots, i_5\} $, then one of the six numbers $ j_1, \ldots, j_6 $ does not belong to this five-element subset, say $ j_t \not \in \{i_1, \ldots, i_5 \} $. Therefore, the member of the committee numbered $ j_t $ has the key to the lock corresponding to the subset $ \{i_1, \ldots, i_5\} $.
Thus, the smallest number satisfying the conditions of the problem is $  \binom{11}{5} = 462 $.

Answer: 462


What is the largest value of $k$ for which the polynomials $Q_1(x) = x^2 + (k-29)x - k$ and $Q_2(x) = 2x^2+ (2k-43)x + k$ are both factors of $P(x)$?

Solution:
Solution 1
We can see that $Q_1$ and $Q_2$ must have a [root](https://artofproblemsolving.com/wiki/index.php/Root) in common for them to both be [factors](https://artofproblemsolving.com/wiki/index.php/Factor) of the same cubic.
Let this root be $a$.
We then know that $a$ is a root of
$Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0$
, so $x = \frac{-k}{5}$.
We then know that $\frac{-k}{5}$ is a root of $Q_{1}$ so we get:
$\frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k$
or $k^{2}=30k$, so $k=30$ is the highest.
We can trivially check into the original equations to find that $k=30$ produces a root in common, so the answer is $\boxed{030}$.

Solution 2
Again, let the common root be $a$; let the other two roots be $m$ and $n$. We can write that $(x - a)(x - m) = x^2 + (k - 29)x - k$ and that $2(x - a)(x - n) = 2\left(x^2 + \left(k - \frac{43}{2}\right)x + \frac{k}{2}\right)$.
Therefore, we can write four equations (and we have four [variables](https://artofproblemsolving.com/wiki/index.php/Variable)), $a + m = 29 - k$, $a + n = \frac{43}{2} - k$, $am = -k$, and $an = \frac{k}{2}$.
The first two equations show that $m - n = 29 - \frac{43}{2} = \frac{15}{2}$. The last two equations show that $\frac{m}{n} = -2$. Solving these show that $m = 5$ and that $n = -\frac{5}{2}$. Substituting back into the equations, we eventually find that $k = \boxed{030}$.

Solution 3
Since $Q_1(x)$ and $Q_2(x)$ are both factors of $P(x)$, which is cubic, we know the other factors associated with each of $Q_1(x)$ and $Q_2(x)$ must be linear. Let $Q_1(x)R(x) = Q_2(x)S(x) = P(x)$, where $R(x) = ax + b$ and $S(x) = cx + d$. Then we have that $((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)$. Equating coefficients, we get the following system of equations:
\begin{align}  a = 2c \\ b = -d \\ 2c(k - 29) - d = c(2k - 43) + 2d \\ -d(k - 29) - 2ck = d(2k - 43) + ck  \end{align}
Using equations $(1)$ and $(2)$ to make substitutions into equation $(3)$, we see that the $k$'s drop out and we're left with $d = -5c$. Substituting this expression for $d$ into equation $(4)$ and solving, we see that $k$ must be $\boxed {030}$.
~ anellipticcurveoverq

Solution 4
Notice that if the roots of $Q_1(x)$ and $Q_2(x)$ are all distinct, then $P(x)$ would have four distinct roots, which is a contradiction since it's cubic. Thus, $Q_1(x)$ and $Q_2(x)$ must share a root. Let this common value be $r.$
Thus, we see that we have
\[r^2 + (k - 29)r - k = 0,\]
\[2r^2 + (2k - 43)r + k = 0.\] Adding the two equations gives us \[3r^2 + (3k - 72)r = 0 \implies r = 0, 24 - k.\] Now, we have two cases to consider. If $r = 0,$ then we have that $Q_1(r) = 0 = r^2 + (k - 29)r - k \implies k = 0.$ On the other hand, if $r = 24 - k,$ we see that \[Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \implies k = \boxed{030}.\] This can easily be checked to see that it does indeed work, and we're done!
~Ilikeapos

Answer: 30


Each cell of a $5 \times 5$ table is painted in one of several colors. Lada shuffled the rows of this table so that no row remained in its original position. Then Lera shuffled the columns so that no column remained in its original position. To their surprise, the girls noticed that the resulting table was the same as the original one. What is the maximum number of different colors this table can be painted with?

Solution:
# Answer: 7.

Solution: Let's renumber the colors and reason about numbers instead. Both columns and rows could have been cyclically permuted or divided into a pair and a triplet. If a cyclic permutation of columns was used, then all columns consist of the same set of numbers, i.e., no more than five different numbers. The same situation applies to a cyclic permutation of rows.

If, however, both columns and rows were divided into a pair and a triplet, the table can be conditionally divided into squares $2 \times 2$ and $3 \times 3$, as well as rectangles $2 \times 3$ and $3 \times 2$, in each of which both columns and rows are cyclically permuted. In the squares, there can be no more than two and three different numbers, respectively (because the set of numbers in each column is the same), and in the rectangles, no more than one (since both in columns and rows, the set of numbers must be the same).

![](https://cdn.mathpix.com/cropped/2024_05_06_da44c9888fa01160fc84g-2.jpg?height=512&width=465&top_left_y=1720&top_left_x=1458)

Example for seven numbers see in the figure.

Criteria: Only the estimate - 3 points

Only the example - 3 points.

Answer: 7


In rhombus $ABCD$, $\angle A=100^{\circ}, M$ and $N$ are the midpoints of sides $AB$ and $BC$ respectively, $MP \perp CD$ at point $P$, then the degree measure of $\angle NPC$ is $\qquad$

Solution:
It is known that $\angle B=80^{\circ}$.
Notice that $B M=B N$. Then
$$
\angle B M N=\angle B N M=50^{\circ} \text {. }
$$

As shown in Figure 16, let the midpoint of $M P$ be $E$, and connect $N E$. It is easy to know that $N M=N P$. Then $\angle N M P=\angle N P M$. Therefore,
$$
\angle N P C=\angle N M B=50^{\circ} .
$$

Answer: 50^{\circ}


Given an integer $ c \geq 1 $. To each subset $ A $ of the set $ \{1,2, \ldots ,n\} $, we assign a number $ w(A) $ from the set $ \{1,2, \ldots ,c\} $ such that the following condition is satisfied:

Let $ a(n) $ be the number of such assignments. Calculate $ \lim_{n\to \infty}\sqrt[n]{a(n)} $.
Note: $ \min(x,y) $ is the smaller of the numbers $ x $, $ y $.

Solution:
Let $ M $ be the set $ \{1,2,\ldots,n\} $, and $ M_i $ be the $(n-1)$-element set obtained from $ M $ by removing the number $ i $:

Given the assignment as mentioned in the problem, let's denote the value $ w(M) $ by $ m $, and the value $ w(M_i) $ by $ f(i) $ (for $ i = 1,2, \ldots ,n $). According to the conditions of the problem, the values $ f(1), \ldots ,f(n) $ and $ m $ are numbers from the set $ \{1,2, \ldots ,c\} $, and for each $ i $, the following equality holds:

This is simply a fancy way of writing the inequality:

Thus, the assignment $ w(\cdot) $ is associated with the function:

From the condition imposed on the assignment $ w(\cdot) $:

it follows by induction that the equality:

holds for any sets $ A_1,\ldots,A_k $ contained in $ M $. (The symbol $ \min(x_1,\ldots,x_k) $ denotes the smallest of the numbers $ x_1,\ldots,x_k $; it can, of course, be the common value of several of these numbers.)
Let $ A $ be any subset of the set $ M $, not identical to the entire set $ M $, and let $ i_1,\ldots,i_k $ be all the numbers in $ M $ that do {\it not} belong to $ A $:

The set $ A $ is then exactly the intersection of the sets $ M_{i_1},\ldots,M_{i_k} $:

Indeed: a number $ i $ belongs simultaneously to all sets $ M_{i_1},\ldots,M_{i_k} $ if and only if it belongs to $ M $ and is different from the numbers $ i_1,\ldots,i_k $ - that is, (in accordance with (3)) if and only if it is an element of the set $ A $.
From the connections (4), (2), (3), the equality follows:

(The symbol $ \min_{i \not \in A}f(i) $ denotes the minimum value taken by the function $ f $ on those elements of the set $ M $ that do not belong to the set $ A $.)
Conclusion: the assignment $ w(\cdot) $ is fully determined by specifying the value $ m = w(M) $ and the values $ f(i) = w(M_i) $ for $ i = 1,2,\ldots ,n $. In other words, it is determined by choosing the number $ m $ and the function (1); it is then given by the formula:

And conversely: if $ m $ is any positive integer not exceeding $ c $, and if $ f $ is any mapping of the set $ M $ into the set $ \{1,2,\ldots,m\} $, then the formula (5) defines an assignment $ A \mapsto w(A) $ satisfying the condition stated in the problem.
For a fixed $ m $, there are exactly $ m^n $ functions (1). Therefore, the number of admissible assignments (5) is:

We obtain the two-sided estimate $ c^n \leq a(n) \leq c^n \cdot c $; and hence:

As $ n $ approaches infinity, the $ n $-th root of any fixed positive number approaches $ 1 $. Therefore, the expression on the right side of the inequality (6) approaches $ c $, and by the squeeze theorem, $ \displaystyle \lim_{n \to \infty} \sqrt[n]{a(n)} = c $.
The answer is: The sought limit value is the number $ c $.

Answer: c


Calculate $\sqrt{1.004}$ with an accuracy of 0.0001.

Solution:
Solution. We will use formula (4) §4:

$$
\sqrt{1.004}=\sqrt{1+0.004}=(1+0.004)^{\frac{1}{2}}
$$

Here $m=\frac{1}{2}, x=0.004$

$$
\begin{aligned}
\sqrt{1.004}=1 & +\frac{0.004}{2}+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}(0.004)^{2}+\ldots= \\
& =1+0.002-\frac{(0.004)^{2}}{2!4}+\ldots
\end{aligned}
$$

Notice that the signs of the obtained series, starting from the second, alternate.

Limiting ourselves to the sum of the first two terms, we get

$$
\sqrt{1.004} \approx 1.002
$$

Estimate the absolute error

$$
\Delta<\frac{(0.004)^{2}}{2 \mid 4}=0.000002
$$

Therefore, the value of the root is found with the required accuracy. 1470. Calculate $\sqrt[3]{30}$ with an accuracy of 0.001.

Solution. $\sqrt[3]{30}=3 \sqrt[3]{\frac{30}{27}}=3 \sqrt[3]{1+\frac{3}{27}}=3\left(1+\frac{1}{9}\right)^{\frac{1}{3}}$. Here

$$
m=\frac{1}{3}, x=\frac{1}{9}
$$

$$
\begin{aligned}
\sqrt[3]{30} & =3\left[1+\frac{1}{3} \cdot \frac{1}{9}+\frac{1}{3}\left(\frac{1}{3}-1\right)\left(\frac{1}{9}\right)^{2} \frac{1}{2!}+\right. \\
+ & \left.\frac{1}{3}\left(\frac{1}{3}-1\right)\left(\frac{1}{3}-2\right)\left(\frac{1}{9}\right)^{3} \frac{1}{3!}+\ldots\right]= \\
& =3\left(1+\frac{1}{27}-\frac{2}{2!9^{3}}+\frac{2 \cdot 5}{3!3^{3} 9^{3}}-\ldots\right)
\end{aligned}
$$

We obtained an alternating series (after the first term), so if we discard all terms starting from the fourth, the absolute error

$$
\Delta<\frac{3 \cdot 2 \cdot 5}{313^{3} 9^{3}}=\frac{5}{3^{3} \cdot 9^{3}}=\frac{5}{19683}<0.001
$$

from here

$$
\sqrt[3]{30} \approx 3+\frac{3}{27}-\frac{6}{2!9^{3}} \approx 3+0.1111-0.0041=3.1070
$$

(calculation was performed with one extra digit).

Thus, $\sqrt[3]{30} \approx 3.107$ with an accuracy of 0.001.

Answer: 1.002


For the parabola $y=ax^{2}+bx+1$, the parameters $a$ and $b$ satisfy $8a^{2}+4ab=b^{3}$. Then, as $a$ and $b$ vary, the vertex of the parabola must lie on ( ).
(A) parabola
(B) hyperbola
(C) circle or ellipse
(D) line

Solution:
2.B.

The vertex coordinates of the parabola $y=a x^{2}+b x+1$ are $\left(-\frac{b}{2 a}, \frac{4 a-b^{2}}{4 a}\right)$. Let $x=-\frac{b}{2 a}, y=\frac{4 a-b^{2}}{4 a}$. Then $\frac{b}{a}=-2 x, y=1-\frac{b^{2}}{4 a}=1+\frac{b x}{2}$.

Since $a \neq 0$, the conditions that $a$ and $b$ satisfy are equivalent to $8+4 \cdot \frac{b}{a}=b\left(\frac{b}{a}\right)^{2}$. Therefore,
$$
8+4(-2 x)=b(-2 x)^{2}=4 x(2 y-2) \text {, }
$$

which means $x y=1$.

Answer: B


Solution:
Solution. The equation of the given system is equivalent to the equation

$$
4^{\left|x^{2}-8 x+12\right|}=7^{2 y}
$$

Both sides of it are positive; therefore, it is equivalent to the equation

$$
\left|x^{2}-8 x+12\right|=(2 y) \log _{4} 7
$$

Since $\log _{4} 7>0$ and $\left|x^{2}-8 x+12\right| \geqslant 0$ for any $x$, it follows that $y \geqslant 0$. Therefore, the inequality of the system (16) should only be solved for two cases: $0 \leqslant y \leqslant 3$ and $y>3$.

For $0 \leqslant y \leqslant 3$, the inequality of the system (16) takes the form

$$
3-y-3 y-2 y^{2}-4 y-2-1 \geqslant 0
$$

i.e., $y^{2}+4 y \leqslant 0$; from this we find $-4 \leqslant y \leqslant 0$. Considering the condition $0 \leqslant y \leqslant 3$, we get $y=0$.

For $y>3$, the inequality of the system (16) takes the form

$$
y-3-3 y-2 y^{2}-4 y-2-1 \geqslant 0
$$

i.e.,

$$
-2 y^{2}-6 y-6 \geqslant 0
$$

this inequality has no solutions.

Substituting the value $y=0$ into the equation of the system (16), we get $(16) \Leftrightarrow\left\{\begin{array}{l}y=0, \\ x^{2}-8 x+12=0\end{array} \Leftrightarrow\left\{\begin{array}{l}y=0, \\ (x-2)(x-6)=0\end{array} \Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}x=2, \\ y=0, \\ x=6, \\ y=0 .\end{array}\right.\end{array}\right.\right.\right.$

Thus, the two pairs of numbers $(2 ; 0),(6 ; 0)$, and only they, are the solutions of the system (16).

Answer: (2,0),(6,0)


Solution:
Answer: -1.

Solution.

Given that $\frac{n}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$, we get

$S_{2021}=\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{2021}{2022!}=\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\cdots+\left(\frac{1}{2021!}-\frac{1}{2022!}\right)=1-\frac{1}{2022!}$.

Then $\quad 2022!\cdot\left(S_{2021}-1\right)=2022!\cdot\left(1-\frac{1}{2022!}-1\right)=-1$.

Answer: -1


Determine the smallest positive integer $M$ with the following property:
For every choice of integers $a,b,c$, there exists a polynomial $P(x)$ with integer coefficients so that $P(1)=aM$ and $P(2)=bM$ and $P(4)=cM$.

Solution:
To determine the smallest positive integer \( M \) such that for every choice of integers \( a, b, c \), there exists a polynomial \( P(x) \) with integer coefficients satisfying \( P(1) = aM \), \( P(2) = bM \), and \( P(4) = cM \), we can proceed as follows:

1. **Assume the Existence of Polynomial \( P(x) \)**:
   Let \( P(x) \) be a polynomial with integer coefficients such that:
   \[
   P(1) = aM, \quad P(2) = bM, \quad P(4) = cM
   \]

2. **Formulate the Polynomial Using Lagrange Interpolation**:
   Using Lagrange interpolation, we can construct a polynomial \( P(x) \) that passes through the points \((1, aM)\), \((2, bM)\), and \((4, cM)\). The general form of the Lagrange interpolation polynomial is:
   \[
   P(x) = aM \frac{(x-2)(x-4)}{(1-2)(1-4)} + bM \frac{(x-1)(x-4)}{(2-1)(2-4)} + cM \frac{(x-1)(x-2)}{(4-1)(4-2)}
   \]

3. **Simplify the Polynomial**:
   Simplify each term in the polynomial:
   \[
   P(x) = aM \frac{(x-2)(x-4)}{(-1)(-3)} + bM \frac{(x-1)(x-4)}{(1)(-2)} + cM \frac{(x-1)(x-2)}{(3)(2)}
   \]
   \[
   P(x) = aM \frac{(x-2)(x-4)}{3} - bM \frac{(x-1)(x-4)}{2} + cM \frac{(x-1)(x-2)}{6}
   \]

4. **Ensure Integer Coefficients**:
   For \( P(x) \) to have integer coefficients, the denominators \( 3, 2, \) and \( 6 \) must be canceled out by \( M \). Therefore, \( M \) must be a multiple of the least common multiple (LCM) of these denominators:
   \[
   \text{LCM}(3, 2, 6) = 6
   \]

5. **Verify the Minimality of \( M \)**:
   We need to check if \( M = 6 \) is indeed the smallest such number. If \( M \) were smaller, say \( M = 3 \) or \( M = 2 \), then the denominators would not be fully canceled, and \( P(x) \) would not have integer coefficients for all choices of \( a, b, c \).

Therefore, the smallest positive integer \( M \) that satisfies the given conditions is \( M = 6 \).

The final answer is \( \boxed{6} \)

Answer: 6


For $(x, y)$ satisfying the equation $x^{2}+(y-1)^{2}=1$, the inequality $x+y+c \geqslant 0$ always holds, find the range of the real number $c$.

Solution:
The key to solving the problem is to understand the meaning of each formula in the question using the idea of combining numerical and geometric concepts. For example, "the inequality $x+y+c \geqslant 0$ always holds," means that the point $(x, y)$ is always above or on the line $x+y+c=0$. Here, the point $(x, y)$ refers to the points on the circle $x^{2}+(y-1)^{2}=1$. Therefore, as long as $c$ ensures that all points on the circle are on or above the line, it will suffice.

Solution: Let the line $x+y+c=0$ be tangent to the circle $x^{2}+(y-1)^{2}=1$,

i.e.,
$$
\begin{aligned}
& \left\{\begin{array}{l}
x^{2}+(y-1)^{2}=1, \\
y=-x-c
\end{array}\right. \\
\Rightarrow & c_{1}=-1+\sqrt{2}, \\
& c_{2}=-1-\sqrt{2} .
\end{aligned}
$$

Thus, there are two lines tangent to the circle, with y-intercepts of $-c_{1}=1-\sqrt{2}$ and $-c_{2}=1+\sqrt{2}$, respectively. (See Figure 2) According to the problem, we take $-c_{1}=1-\sqrt{2}$, so $-c \leqslant 1-\sqrt{2}$, which means $c \geqslant \sqrt{2}-1$.

Answer: c \geqslant \sqrt{2}-1


Let $f(x)$ be a decreasing function on $\mathbf{R}$, for any $x \in \mathbf{R}$, we have
$$
f(x+2013)=2013 f(x) .
$$

Then a function that satisfies this condition is $\qquad$

Solution:
1. $-2013^{\frac{x}{2013}}$ (the answer is not unique).

Consider the function $f(x)=-a^{x}(a>1)$.
By the problem, we have
$$
\begin{array}{l}
-a^{x+2013}=-2013 a^{x} \\
\Rightarrow a^{2013}=2013 \Rightarrow a=2013^{\frac{1}{2013}} .
\end{array}
$$

Thus, $f(x)=-2013^{\frac{x}{2013}}$ meets the condition.
Moreover, for any negative number $k, f(x)=2013^{\frac{x}{2013}} k$ meets the condition.

Answer: -2013^{\frac{x}{2013}}


A certain natural number divides $25, 38$, and $43$, resulting in a sum of the three remainders being 18. Find this natural number.

Solution: Reference answer: 11

Answer: 11


$18 \cdot 120$ is the short base, height, and long base of a trapezoid, which form an arithmetic sequence. Let $k$ be the area of this trapezoid, then
(A) $k$ must be an integer.
(B) $k$ must be a fraction.
(C) $k$ must be an irrational number.
(D) $k$ must be an integer or a fraction.
(E) None of (A), (B), (C), (D) is correct.

Solution: [Solution] Let the shorter base, height, and longer base of the trapezoid be $a-d, a, a+d$ respectively. Then $k=S_{\text {trapezoid }}=\frac{1}{2}[(a-d)+(a+d)] a=a^{2}$. By the arbitrariness of $a$, we know that $(A),(B),(C),(D)$ are all incorrect. Therefore, the answer is $(E)$.

Answer: E


A cyclist planned to travel from point A to point B in 5 hours, moving at a constant speed. He moved at the planned speed until the midpoint of the journey, then decided to increase his speed by $25 \%$. He completed the journey to point B at the new speed. How long did the entire trip take?

Solution:
7.1 A cyclist planned to travel from point A to point B in 5 hours, moving at a constant speed. He moved at the planned speed until the midpoint of the journey, then decided to increase his speed by $25 \%$. He completed the journey to point B at the new speed. How long did the entire journey take?

Answer: 4 hours 30 minutes. Hint: Let a be the distance between points A and B, v be the planned speed. Then $\frac{\mathrm{a}}{\mathrm{v}}=5$. The increased speed is $1.25 \mathrm{v}$. The entire journey will take $\frac{a}{2 v}+\frac{a}{2 \cdot 1.25 v}=\frac{4.5}{5} \frac{a}{v}=4.5$ (hours).

Answer: 4.5


In a certain county, every two districts are connected by exactly one of three modes of transportation: car, train, or airplane. It is known that all three modes of transportation are used in the county, but no district has all three modes. Furthermore, no three districts are pairwise connected by the same mode of transportation. What is the maximum number of districts in this county?

Solution:
10. There are at most 4 districts.

Let vertices represent districts, and edges represent means of transportation, colored red, blue, and yellow to represent three types of transportation, as shown in the figure. This meets the requirements of the problem, thus there exist four districts.

If there are more than 4 districts, i.e., the graph has 5 vertices, divide them into non-red, non-blue, and non-yellow combinations. In this case, at least one combination will have $\geqslant 2$ vertices, otherwise, the number of vertices would be at most 3. We will discuss this in cases:
(i) Suppose the non-red combination has more than 3 vertices, then it contains three vertices $A, B, C$, which emit blue or yellow edges. Assume the non-blue combination has one vertex $D$. Then $D A$ is neither red nor blue, so it must be yellow. Similarly, $D B, D C$ are also yellow. In this case, if $A B$ or $A C$ is yellow, a monochromatic triangle is formed, which is a contradiction. Therefore, one of them must be blue, say $A B$ is blue. Similarly, $A C, B C$ are also blue, which results in $\triangle A B C$ being a blue triangle. This is a contradiction. Therefore, case (i) does not exist.
(ii) If each combination has at most 2 vertices, then since there are 5 vertices, one combination must contain two vertices. Suppose $A, B$ are in the non-red combination, and assume $A B$ is blue. Thus, there must be a point in the non-blue combination (otherwise, it would result in only two colors of edges), let this point be $C$. In this case, $A C, B C$ can only be yellow, and there must be another point, say $D$, in the non-yellow combination, which leads to $A D, B D$ being blue. Since $A B$ is blue, $\triangle A B D$ forms a monochromatic triangle, which is a contradiction. Therefore, case (ii) also does not occur.
Thus, a graph that meets the requirements of the problem cannot have 5 vertices.

Answer: 4


Let $n$ be a positive integer. If the last three digits of $n$ are removed, the resulting number is the cube root of $n$. Find a possible value of $n$.

Solution:
8. 32768
8. Let \( n = 1000a + b \), where \( 0 \leq b \leq 999 \). By removing the last three digits of \( n \), we get \( a \). Hence we have \( a^3 = 1000a + b \), or \( b = a(a^2 - 1000) \). Since \( b \geq 0 \), we have \( a^2 \geq 1000 \) and hence \( a \geq 32 \). If \( a \geq 33 \), then \( b \geq 33(33^2 - 1000) > 1000 \) which is impossible. Hence we must have \( a = 32 \), which gives \( b = 32(32^2 - 1000) = 768 \) and hence \( n = 32768 \).

Answer: 32768


The year 2009 has the following property: by swapping the digits of the number 2009, it is impossible to obtain a smaller four-digit number (numbers do not start with zero). In which year will this property reoccur for the first time?

Solution:
Answer. In 2022.

Solution. In the years $2010, 2011, \ldots, 2019$ and in 2021, the year number contains a one, and if it is moved to the first position, the number will definitely decrease. The number 2020 can be reduced to 2002. However, the number 2022 cannot be decreased by rearranging the digits.

Answer: 2022


Find all integers $n\ge1$ such that $2^n-1$ has exactly $n$ positive integer divisors.

Solution:
1. **Initial Observations and Definitions:**
   - For each \( k \in \mathbb{N} \), let \( d(k) \) be the number of divisors, \( \omega(k) \) be the number of distinct prime divisors, and \( \Omega(k) \) be the number of prime divisors counted with multiplicity.
   - We need to find all integers \( n \geq 1 \) such that \( 2^n - 1 \) has exactly \( n \) positive integer divisors.

2. **Parity Consideration:**
   - If \( n \) is odd, then \( d(2^n - 1) \) is odd, implying \( 2^n - 1 \) is a perfect square.
   - However, \( 2^n - 1 \equiv 3 \pmod{4} \) for \( n \geq 2 \), which cannot be a perfect square. Thus, \( n \) must be even unless \( n = 1 \).

3. **Bounding the Number of Distinct Prime Divisors:**
   - Using Zsigmondy's theorem (Bang's theorem), for \( n \neq 1, 6 \), there exists a prime \( p \) such that \( p \mid 2^n - 1 \) and \( p \nmid 2^k - 1 \) for all \( 1 \leq k < n \).
   - For all \( k \mid n \) (except possibly \( k = 1, 6 \)), \( 2^k - 1 \) has a primitive prime divisor. Thus, \( \omega(2^n - 1) \geq d(n) - 2 \).

4. **Prime Factorization and Divisor Counting:**
   - Let \( 2^n - 1 = p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdots p_k^{\alpha_k} \), where \( p_i \) are distinct primes.
   - The number of divisors of \( 2^n - 1 \) is \( (\alpha_1 + 1)(\alpha_2 + 1) \cdots (\alpha_k + 1) = n \).

5. **Bounding the Number of Prime Divisors Counted with Multiplicity:**
   - From the previous step, \( \Omega(n) \geq d(n) - 2 \).
   - Let \( n = q_1^{\beta_1} q_2^{\beta_2} \cdots q_m^{\beta_m} \). Then, \( \sum_{i=1}^m \beta_i = \Omega(n) \geq d(n) - 2 = (\beta_1 + 1)(\beta_2 + 1) \cdots (\beta_m + 1) - 2 \).

6. **Bounding the Number of Distinct Prime Divisors of \( n \):**
   - If \( n \) has at least 3 distinct prime divisors, then \( \prod_{i=1}^m (\beta_i + 1) \geq \beta_1 \beta_2 + \beta_1 \beta_3 + \beta_2 \beta_3 + \sum_{i=1}^m \beta_i + 1 \), which contradicts \( \sum_{i=1}^m \beta_i \geq (\beta_1 + 1)(\beta_2 + 1) \cdots (\beta_m + 1) - 2 \). Thus, \( \omega(n) \leq 2 \).

7. **Case Analysis for \( \omega(n) = 2 \):**
   - If \( n \) has exactly 2 distinct prime divisors, then \( n = 2^a \cdot q^b \) for some prime \( q \).
   - If \( 3 \nmid n \), then \( \Omega(n) \geq d(n) - 1 \), leading to a contradiction unless \( n = 6 \).

8. **Powers of Two:**
   - If \( n = 2^T \), then \( 2^n - 1 = 2^{2^T} - 1 \) has \( 2^T \) divisors.
   - Fermat numbers \( F_k = 2^{2^k} + 1 \) are prime for \( k = 0, 1, 2, 3, 4 \), but \( F_5 \) is not prime.
   - Thus, \( n = 2, 4, 8, 16, 32 \) work.

9. **Conclusion:**
   - The integers \( n \geq 1 \) such that \( 2^n - 1 \) has exactly \( n \) positive integer divisors are \( n = 1, 2, 4, 6, 8, 16, 32 \).

The final answer is \( \boxed{ n = 1, 2, 4, 6, 8, 16, 32 } \).

Answer: n = 1, 2, 4, 6, 8, 16, 32


A class consists of pupils who each wear either a red tie or a green tie. There are two more pupils wearing green ties than wearing red ties. Two pupils are chosen at random. The probability of selecting two pupils of different tie-colour is exactly treble the probability of selecting two pupils who both wear red ties. Given that $\mathrm{R}$ is the number of pupils wearing a red tie, and that $R>0$, determine $R^{3}+R^{2}+R$.

Solution:
SOLUTION
399

The number of pupils wearing a green tie is $R+2$, and the total number of pupils in the class is $R+R+2=2 R+2$. Let the probability of selecting two red-tie wearing pupils be $q$.
Then $\frac{R}{2 R+2} \times \frac{R-1}{2 R+1}=q$.
Also, $\frac{R}{2 R+2} \times \frac{R+2}{2 R+1}+\frac{R+2}{2 R+2} \times \frac{R}{2 R+1}=3 q$.
Eliminating $q$ we have $3\left(\frac{R}{2 R+2} \times \frac{R-1}{2 R+1}\right)=\frac{R}{2 R+2} \times \frac{R+2}{2 R+1}+\frac{R+2}{2 R+2} \times \frac{R}{2 R+1}$.
Therefore, $3 R(R-1)=2 R(R+2)$ and hence $3 R^{2}-3 R=2 R^{2}+4 R$.
This quadratic equation simplifies to $R^{2}-7 R=0$ and has solutions $R=0$ and $R=7$.
We are given that $R>0$, so $R=7$ and $R^{3}+R^{2}+R=343+49+7=399$.

Answer: 399


A deck of 2n cards numbered from 1 to 2n is shuffled and n cards are dealt to A and B. A and B alternately discard a card face up, starting with A. The game when the sum of the discards is first divisible by 2n + 1, and the last person to discard wins. What is the probability that A wins if neither player makes a mistake? Solution

Solution: : 0. Fairly easy. B always wins irrespective of the deal. The stuff about random shuffling and probabilities is a red herring. Note first that both players have perfect information (because each can see his own cards and the discards and hence deduce the other player's hand). A cannot win on his first play. Now suppose it is B's turn. Each card B might play will give a different total and so requires a different card for A to win on the following move. But B has one more card than A, so he can choose a card which does not allow A to win on the next move. Thus B can always play so that A does not win on the following move. But one player must win after 2n plays or less, so it must be B. Putnam 1993 © John Scholes jscholes@kalva.demon.co.uk 12 Dec 1998

Answer: 0


Suppose we have a regular hexagon and draw all its sides and diagonals. Into how many regions do the segments divide the hexagon? (No proof is necessary.)

Solution:
Answer: 24
Solution: An accurate diagram and a careful count yields the answer.

Answer: 24


Let real numbers $a, x, y$ satisfy the following conditions
$$
\left\{\begin{array}{l}
x+y=2 a-1, \\
x^{2}+y^{2}=a^{2}+2 a-3 .
\end{array}\right.
$$

Find the minimum value that the real number $xy$ can take.

Solution:
By (1) $)^{2}-$ (2) we get
$$
x y=\frac{1}{2}\left((2 a-1)^{2}-\left(a^{2}+2 a-3\right)\right)=\frac{1}{2}\left(3 a^{2}-6 a+4\right) \text {. }
$$

Combining $x^{2}+y^{2} \geqslant 2 x y$, we know $a^{2}+2 a-3 \geqslant 3 a^{2}-6 a+4$, solving this yields $2-\frac{\sqrt{2}}{2} \leqslant a \leqslant 2+\frac{\sqrt{2}}{2}$. $x y=\frac{3}{2}(a-1)^{2}+\frac{1}{2} \geqslant \frac{3}{2}\left(2-\frac{\sqrt{2}}{2}-1\right)^{2}+\frac{1}{2}$, thus the minimum value of $x y$ is reached when $a=2-\frac{\sqrt{2}}{2}$, and the required minimum value is $\frac{11-6 \sqrt{2}}{4}$.

Answer: \frac{11-6\sqrt{2}}{4}


When $0<x \leqslant 1$, which of the following inequalities is correct? ( ).
(A) $\frac{\sin x}{x}<\left(\frac{\sin x}{x}\right)^{2} \leqslant \frac{\sin x^{2}}{x^{2}}$
(B) $\left(\frac{\sin x}{x}\right)^{2}<\frac{\sin x}{x} \leqslant \frac{\sin x^{2}}{x^{2}}$
(C) $\left(\frac{\sin x}{x}\right)^{2}<\frac{\sin \frac{x^{2}}{x^{2}} \leqslant \frac{\sin }{x} \cdot x}{x}$
(D) $\frac{\sin x^{2}}{x^{2}} \leqslant\left(\frac{\sin x}{x}\right)^{2}<\frac{\sin x}{x}$

Solution:
-1 (B).
Because $0\left(\frac{\sin x}{x}\right)^{2}$. Also, since $f(x) = \frac{\sin x}{x}$ is a decreasing function on $\left(0, \frac{\pi}{2}\right)$, and $0 < x^{2} \leqslant x \leqslant 1 < \frac{\pi}{2}$, it follows that $\frac{\sin x^{2}}{x^{2}} \geqslant \frac{\sin x}{x}$.

Answer: B


The total weight of a pile of stones is 100 kilograms, where the weight of each stone does not exceed 2 kilograms. By taking out some of the stones in various ways and calculating the difference between the sum of the weights of these stones and 10 kilograms. Among all these differences, the minimum value of their absolute values is denoted as $d$. Among all piles of stones that meet the above conditions, find the maximum value of $d$.

Solution:
[Solution] Take any pile of stones that satisfies the conditions of the problem, and let their weights be $x_{1}, x_{2}$, $\cdots, x_{n}$, without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$. According to the definition of $d$, there exists a natural number $k$ such that
$$
x_{1}+x_{2}+\cdots+x_{k} \geqslant 10+d \text { and } x_{1}+x_{2}+\cdots+x_{k-1} \leqslant 10-d \text {. }
$$

From this, we can deduce that $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{k} \geqslant 2 d$. Also, $05$. Therefore, we have
$$
10 d \leqslant x_{1}+x_{2}+x_{3}+x_{4}+x_{5} \leqslant 10-d,
$$

which implies $d \leqslant \frac{10}{11}$.
Furthermore, take 55 stones each weighing $\frac{10}{11}$ kilograms. It is easy to see that in this case $d=\frac{10}{11}$. Therefore, the maximum value of $d$ is $\frac{10}{11}$.

Answer: \frac{10}{11}


Solve the inequality $\sqrt{x}-\sqrt{1-x}+8 \sin x<8$.

Solution: Answer: $0 \leq x \leq 1$. Solution. The domain of the inequality is the interval $[0 ; 1]$. On $[0 ; 1]$, the function $f(x)=\sqrt{x}-\sqrt{1-x}+8 \sin x$ is monotonically increasing. Let's check the value at the right end of the interval, specifically, we will show that $f(1)<8$. Indeed, $1+8 \sin 1<1+8 \sin \pi / 3$ (since $1<\pi / 3$, and the function $\sin$ is increasing in the first quadrant) and it remains to verify the inequality $1+4 \sqrt{3}<8 \Leftrightarrow 4 \sqrt{3}<7 \Leftrightarrow 163<49$. Therefore, the solution to the original inequality will be all points of the interval $[0 ; 1]$

Answer: 0\leqx\leq1


7.31 $\log _{3}(x-3)^{2}+\log _{3}|x-3|=3$.

The text was already in English and followed the instruction to simply return the problem without additional context or commentary.

Solution:
7.31 Here $x \neq 3$. Writing the right-hand side of the equation as $3=3 \log _{3} 3=\log _{3} 3^{3}$, we transition to the equivalent equation $(x-3)^{2}|x-3|=3^{3}$ under the condition $x \neq 3$. Since $a^{2}=|a|^{2}$, we get $|x-3|^{3}=3^{3}$, or $|x-3|=3$, from which $x_{1}=0, x_{2}=6$.

Answer: $x_{1}=0, x_{2}=6$.

Answer: x_{1}=0,x_{2}=6


The function $f(n)$ with domain of positive integers satisfies: $f(n)=\left\{\begin{array}{l}n-3 \quad(n \geqslant 1000) \\ f[f(n+7)](n<1000)\end{array}\right.$, then $f(90)=$ $\qquad$ .

Solution:
10. At first glance, the condition and the value to be found seem unrelated, making it difficult to start. By considering the relationship between 90 and 1000, we need to set up an auxiliary function to explore the pattern and use the periodicity of the function to solve the problem.
Let \( f_{m}(n) = \underbrace{f(f \cdots f}_{m \text{ times}}(n)) \).
Given that \( 90 + 7 \times 130 = 1000 \).
Thus, \( f(90) = f_{2}(97) = \cdots = f_{131}(1000) \).
Also, since \( f(1000) = 997 \).
Therefore, \( f(997) = f_{2}(1004) = f(1001) = 998 \),
\( f(998) = f_{2}(1005) = f(1002) = 999 \).
\( f(999) = f_{:}(1006) = f(1003) = 1000 \).
Thus, \( f_{m}(1000) \) is a periodic function with a period of 4.
Given that \( 131 = 4 \times 32 + 3 \).
Therefore, \( f(90) = f_{131}(1000) = f_{3}(1000) = 999 \).

Answer: 999


A driver drove his car from A to B. After 20 minutes of driving, he had a breakdown that took 30 minutes to fix. After another 12 minutes of driving, he had to wait 4 minutes at a closed railway crossing. By then, he had covered $40 \mathrm{~km}$. The journey from the railway crossing to B started at 11:06 AM and proceeded without any stops. Upon arrival in B, the driver noticed that he had taken exactly half the time from the departure at the railway crossing to the arrival in B compared to the total time from the departure from A to the arrival in B.

It is assumed that the driver maintained the same average speed on each segment of the journey.
a) At what time did the driver arrive in B?

b) What was the average speed, expressed in $\mathrm{km} \mathrm{h}$?

c) How many kilometers did he cover in total from A to B?

Solution:
a) The truck driver needed $20+30+12+4=66$ minutes until the departure from the railway crossing. Since this time was as long as the travel time from the railway crossing to B, he was on the road for another 66 minutes until B from 11:06 AM, thus arriving there at 12:12 PM.

b) For the first $40 \mathrm{~km}$, the pure driving time was $66-30-4=32$ minutes, which is $\frac{8}{15}$ hours. Because $40: \frac{8}{15}$, his average speed was therefore $75 \frac{\mathrm{km}}{\mathrm{h}}$.

c) Since he covered the rest of the distance at the same average speed and it took him 66 minutes, which is $\frac{11}{10}$ hours, he covered $75 \cdot \frac{11}{10}=82.5$ more kilometers. Thus, he had traveled a total of $40 \mathrm{~km}+82.5 \mathrm{~km}=122.5 \mathrm{~km}$ from A to B.

Solutions of the 2nd Round 1979 adopted from [5]

### 4.21.3 3rd Round 1979, Class 7

Answer: 122.5\mathrm{~}


Use red, yellow, and blue to color the 9 small squares labeled $1,2,3, \cdots, 9$ (as shown in the figure), such that any two adjacent (sharing a common edge) small squares are colored differently, and the numbers â3, 5, 7â are colored the same. The total number of coloring methods that meet the conditions is $\qquad$ (answer with a specific number).

Solution: $113, 5, 7$ have the same color, then they have 3 coloring methods. At this time, if $2, 4$ have the same color, then they have 2 coloring methods, thus number 1 has 2 coloring methods; if $2, 4$ have different colors, then they have 2 coloring methods, and number 1 has only 1 coloring method. By the symmetry of the square, there are a total of coloring methods $3 \times(2 \times 2+2 \times 1)^{2}=108$ (ways).

Answer: 108


Calculate the tens digit of $2011^{2011}$.

Solution:
$$
\begin{aligned}
2011^{2011} & \equiv(10+1)^{2011} \bmod 10 \\
& \equiv 10^{2011}+\ldots+2011 \times 10+1 \equiv 11 \bmod 10
\end{aligned}
$$

The tens digit is 1 .

Answer: 1


A poplar tree, a willow tree, a locust tree, a birch tree, and a sycamore tree are planted in a row, with the distance between adjacent trees being 1 meter. The distance between the poplar tree and the willow tree, and the locust tree is the same, and the distance between the birch tree and the poplar tree, and the locust tree is the same. Therefore, the distance between the sycamore tree and the birch tree is $\qquad$ meters.

Solution:
【Answer】Solution: The distance between the poplar tree and the willow tree, as well as the locust tree, is equal. Therefore, the possible positions of the three trees could be: Willow $\square$ Poplar
Locust, Willow Poplar Locust $\square \square$, $\square$ Willow Poplar Locust $\square$, $\square \square$ Willow Poplar Locust, where $\square$ indicates an unknown position.
Since the distance between the birch tree and the poplar tree, as well as the locust tree, is also equal, the only possible arrangement is: Willow $\square$ Poplar Birch Locust, with the remaining position being the plane tree.

Therefore, the distance between the plane tree and the birch tree is 2 meters.
The answer is: 2.

Answer: 2


What is $f\left(\frac{\pi}{32}\right)$ if $f(x)=\frac{\sin 12 x}{\sin 4 x}+\frac{\cos 12 x}{\cos 4 x}$?

Solution:
## Solution.

Simplify the expression given for the function $f$.

$$
\begin{aligned}
f(x) & =\frac{\sin 12 x}{\sin 4 x}+\frac{\cos 12 x}{\cos 4 x}=\frac{\sin 12 x \cdot \cos 4 x+\cos 12 x \cdot \sin 4 x}{\sin 4 x \cos 4 x} \\
& =\frac{\sin 16 x}{\frac{1}{2} \sin 8 x} \\
& =\frac{2 \cdot 2 \sin 8 x \cos 8 x}{\sin 8 x}=4 \cos 8 x
\end{aligned}
$$

Then,

$$
f\left(\frac{\pi}{32}\right)=4 \cos \left(8 \cdot \frac{\pi}{32}\right)=4 \cos \frac{\pi}{4}=4 \cdot \frac{\sqrt{2}}{2}=2 \sqrt{2}
$$

Answer: 2\sqrt{2}


Find the number of pairs of consecutive integers in the set $\{1000,1001,1002, \ldots, 2000\}$ such that no carrying is required when the two integers are added.

Solution:
3. 156
3. Suppose $n$ and $n+1$ are to be added, and write $n$ in the form $1 \mathrm{ABC}$. If any one of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ is equal to $5,6,7$ or 8 , then carrying is needed in the addition. If $\mathrm{B}$ is 9 and $\mathrm{C}$ is not 9 , or if $\mathrm{A}$ is 9 and $\mathrm{B}$ and $\mathrm{C}$ are not both 9 , carrying is also required. Therefore, if no carrying is required, $n$ must be of one of the following forms:
$$
1 \mathrm{ABC} \quad 1 \mathrm{AB} 9 \quad 1 \mathrm{~A} 99 \quad 1999
$$
where each of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ is from 0 to 4 . It follows that the answer is $5^{3}+5^{2}+5+1=156$.

Answer: 156


Let $f_{1}(x)=\frac{1}{2-x}, f_{n}(x)=\left(f_{1} \circ f_{n-1}\right)(x), n \geqslant 2$, for all real numbers $x$ for which the given functions are defined. What is $f_{2021}(4)$?

Solution:
## Solution.

Let's write out the first few values of the given sequence of functions for $x=4$.

$$
\begin{aligned}
& f_{1}(4)=\frac{1}{2-4}=-\frac{1}{2} \\
& f_{2}(4)=\left(f_{1} \circ f_{1}\right)(4)=\frac{1}{2+\frac{1}{2}}=\frac{2}{5} \\
& f_{3}(4)=\left(f_{1} \circ f_{2}\right)(4)=\frac{1}{2-\frac{2}{5}}=\frac{5}{8} \\
& f_{4}(4)=\left(f_{1} \circ f_{3}\right)(4)=\frac{1}{2-\frac{5}{8}}=\frac{8}{11}=\frac{3 \cdot 4-4}{3 \cdot 4-1}
\end{aligned}
$$

Notice that each term of this sequence is of the form $f_{n}(4)=\frac{3 n-4}{3 n-1}$.

We will prove by mathematical induction that the statement $(*)$ is true for all natural numbers $n$.

For $n=1$, the statement is true, as we have already verified.

1 point

Assume that the statement is true for some natural number $n$, i.e., $f_{n}(4)=\frac{3 n-4}{3 n-1}$, for some $n \in \mathbb{N}$.

We need to prove that the statement is also true for the next natural number, $n+1$. We write:

$f_{n+1}(4)=\frac{1}{2-f_{n}(4)}$, by the assumption $f_{n}(4)=\frac{3 n-4}{3 n-1}$, so

$f_{n+1}(4)=\frac{1}{2-\frac{3 n-4}{3 n-1}}=\frac{3 n-1}{3 n+2}=\frac{3(n+1)-4}{3(n+1)-1}$.

Thus, from the fact that the statement $(*)$ is true for some $n \in \mathbb{N}$, we have shown that it is also true for $n+1$, so the statement $(*)$ is true for all natural numbers $n$.

Therefore, it is also true for $n=2021$, so $f_{2021}(4)=\frac{3 \cdot 2021-4}{3 \cdot 2021-1}=\frac{6059}{6062}$.

Note: A student who did not prove the general statement $(*)$ by induction can receive a maximum of 5 points.

A student could have proven the statement $f_{n}(x)=\frac{n-(n-1)(x)}{(n+1)-n x}$ by induction, which should be graded analogously to the proposed scoring.

Answer: \frac{6059}{6062}


Solution: Solution: Observe that the denominator $n^{4}+2 n^{3}-n^{2}-2 n=n(n-1)(n+1)(n+2)$. Thus we can rewrite the fraction as $\frac{n^{2}-n+1}{n^{4}+2 n^{3}-n^{2}-2 n}=\frac{a}{n-1}+\frac{b}{n}+\frac{c}{n+1}+\frac{d}{n+2}$ for some real numbers $a, b, c$, and $d$. This method is called partial fractions. Condensing the right hand side as a fraction over $n^{4}+2 n^{3}-n^{2}-2 n$ we get $n^{2}-n+1=a\left(n^{3}+3 n^{2}+2 n\right)+b\left(n^{3}+2 n^{2}-n-2\right)+c\left(n^{3}+n^{2}-2 n\right)+d\left(n^{3}-n\right)$. Comparing coefficients of each power of $n$ we get $a+b+c+d=0,3 a+2 b+c=2,2 a-b-2 c-d=2$, and $-2 b=2$. This is a system of 4 equations in 4 variables, and its solution is $a=1 / 2, b=-1 / 2, c=1 / 2$, and $d=$ $-1 / 2$. Thus the summation becomes $\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{16}-\frac{1}{17}+\frac{1}{18}-\frac{1}{19}\right)$. Notice that almost everything cancels to leave us with $\frac{1}{2}\left(1+\frac{1}{3}-\frac{1}{17}-\frac{1}{19}\right)=\frac{\mathbf{5 9 2}}{\mathbf{9 6 9}}$.

Answer: \frac{592}{969}


Sasha invited Petya to visit, saying that he lives in the 10th entrance in apartment No. 333, but forgot to mention the floor. Approaching the building, Petya noticed that the building is nine-story. Which floor should he go to? (The number of apartments on each floor is the same, apartment numbers in the building start from one.)

Solution:
If there are no more than three apartments on a floor, then in ten entrances, there will be no more than $10 \cdot 9 \cdot 3=270$ apartments, which means apartment № 333 will not be in the 10th entrance. If there are no fewer than five apartments on a floor, then already in nine entrances, there will be no fewer than $9 \cdot 9 \cdot 5=405$ apartments, which means Sasha's apartment will not be in the 10th entrance. Therefore, there are 4 apartments on each floor, and in the first nine entrances, there are $9 \cdot 9 \cdot 4=324$ apartments. Then in the 10th entrance, the apartments start from 325. On the second floor, they will start from 329, and on the third floor, from 333. Thus, Petya needs to go up to the third floor.

## Answer

To the 3rd floor.

Answer: 3


The longest professional tennis match lasted 11 hours and 5 minutes, which is ( ) minutes.
(A) 605
(B) 655
(C) 665
(D) 1005
(E) 1105

Solution:
1. C.
$$
11 \times 60+5=665
$$

Answer: C


Find all values of $x$, for each of which one of the three given numbers $\log _{x}\left(x-\frac{13}{6}\right)$, $\log _{x-\frac{13}{6}}(x-3)$, and $\log _{x-3} x$ is equal to the product of the other two.

Solution:
Answer: $x=\frac{11}{3}, x=\frac{3+\sqrt{13}}{2}$.

Solution. Note that on the domain of definition, the product of all three logarithms is

$$
\log _{x}\left(x-\frac{13}{6}\right) \cdot \log _{x-\frac{13}{6}}(x-3) \cdot \log _{x-3} x=1
$$

Let the number that is the product of the other two be denoted by $c$, and the two remaining numbers by $a$ and $b$. Then $c=a b$ and $a b c=1$, from which it follows that $c^{2}=1, c= \pm 1$, i.e., one of the three given logarithms is equal to $\pm 1$.

The converse is also true, namely, if one of the three given logarithms is equal to $\pm 1$, then since the product of all three logarithms is 1, the product of the other two is also $\pm 1$, i.e., equal to the first logarithm.

Note that for all three given logarithms, the base and the argument are never equal for any $x$, so they are different from 1.

Thus, the requirement of the problem is satisfied if and only if one of the logarithms is equal to -1 (and all logarithms exist). A logarithm is equal to -1 when the product of its base and argument is equal to 1. We obtain the system

$$
\left[\begin{array} { l }
{ x ( x - \frac { 1 3 } { 6 } ) = 1 , } \\
{ x ( x - 3 ) = 1 , } \\
{ ( x - 3 ) ( x - \frac { 1 3 } { 6 } ) = 1 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=\frac{13 \pm \sqrt{313}}{12} \\
x=\frac{3 \pm \sqrt{13}}{2} \\
x=\frac{11}{3} \\
x=\frac{3}{2}
\end{array}\right.\right.
$$

Of the found values of the variable, only $x=\frac{11}{3}$ and $x=\frac{3+\sqrt{13}}{2}$ satisfy the domain of definition.

Answer: \frac{11}{3},\frac{3+\sqrt{13}}{2}


Given a square \(ABCD\). Point \(L\) is on side \(CD\) and point \(K\) is on the extension of side \(DA\) beyond point \(A\) such that \(\angle KBL=90^{\circ}\). Find the length of segment \(LD\), if \(KD=19\) and \(CL=6\).

Solution:
Answer: 7.

Solution. Since $A B C D$ is a square, then $A B=B C=C D=A D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_43be4e09ee3721039b48g-26.jpg?height=333&width=397&top_left_y=584&top_left_x=526)

Fig. 1: to the solution of problem 8.4

Notice that $\angle A B K=\angle C B L$, since they both complement $\angle A B L$ to $90^{\circ}$. Then the right triangles $A B K$ and $C B L$ are equal by the acute angle and the leg $A B=B C$ (Fig. 1). Therefore, $A K=C L=6$. Then

$$
L D=C D-C L=A D-C L=(K D-A K)-C L=K D-2 \cdot C L=19-2 \cdot 6=7
$$

Answer: 7


We call a set of professors and committees on which they serve a university if
(1) given two distinct professors there is one and only one committee on which they both serve,
(2) given any committee, $C$, and any professor, $P$, not on that committee, there is exactly one committee on which $P$ serves and no professors on committee $C$ serve, and
(3) there are at least two professors on each committee; there are at least two committees.
What is the smallest number of committees a university can have?

Solution:
Solution: Let $C$ be any committee. Then there exists a professor $P$ not on $C$ (or else there would be no other committees). By axiom 2, $P$ serves on a committee $D$ having no common members with $C$. Each of these committees has at least two members, and for each $Q \in C, R \in D$, there exists (by axiom 1) a committee containing $Q$ and $R$, which (again by axiom 1) has no other common members with $C$ or $D$. Thus we have at least
$2+2 \cdot 2=6$ committees. This minimum is attainable - just take four professors and let any two professors form a committee.

Answer: 6


Calculate the area of the figure bounded by the lines given by the equations.

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=4 \sqrt{2} \cdot \cos ^{3} t \\
y=2 \sqrt{2} \cdot \sin ^{3} t
\end{array}\right. \\
& x=2(x \geq 2)
\end{aligned}
$$

Solution:
## Solution

Let's find the points of intersection:

![](https://cdn.mathpix.com/cropped/2024_05_22_fa23e2deef6f8d90c6d5g-01.jpg?height=768&width=1430&top_left_y=1135&top_left_x=493)

$x=4 \sqrt{2} \cos ^{3} t=2 ; \Rightarrow$

$\cos ^{3} t=\frac{2}{4 \sqrt{2}}=\frac{1}{2 \sqrt{2}}$

$\cos t=\frac{1}{\sqrt{2}}$

$t= \pm \frac{\pi}{4}+2 \pi k, k \in \mathbb{Z}$

Since the functions $x=4 \sqrt{2} \cdot \cos ^{3} t, y=2 \sqrt{2} \cdot \sin ^{3} t$ are periodic (with period $2 \pi$), we can take any interval of length $2 \pi$. Let's take $[-\pi ; \pi]$. Then:

$$
\begin{aligned}
& t=-\frac{\pi}{4} \text { or } t=\frac{\pi}{4} \\
& x \geq 2 \text { on the interval }\left[-\frac{\pi}{4} ; \frac{\pi}{4}\right] .
\end{aligned}
$$

From the graph, it is clear that the region is symmetric with respect to the $0 x$ axis, and its area can be calculated using the formula:

$$
\begin{aligned}
& S=2 \cdot S_{0} ; x \in\left[\frac{\pi}{4} ; 0\right] \\
& S=2 \cdot \int_{\pi / 4}^{0} y(t) \cdot x^{\prime}(t) d t= \\
& =2 \cdot \int_{\pi / 4}^{0} 2 \sqrt{2} \sin ^{3} t \cdot\left(4 \sqrt{2} \cos ^{3} t\right)^{\prime} d t= \\
& =32 \cdot \int_{\pi / 4}^{0} \sin ^{3} t \cdot 3 \cos ^{2} t \cdot(-\sin t) d t= \\
& =-96 \cdot \int_{\pi / 4}^{0} \sin ^{4} t \cdot \cos ^{2} t d t=\left|\begin{array}{c}
\sin t \cos t=\frac{1}{2} \sin 2 t \\
\sin ^{2} t=\frac{1}{2}(1-\cos 2 t) \\
b \\
\int_{a} f(x) d x=-\int_{b}^{a} f(x) d x
\end{array}\right|= \\
& =96 \cdot \int_{0}^{\pi / 4} \frac{1}{2}(1-\cos 2 t) \cdot\left(\frac{1}{2} \sin 2 t\right)^{2} d t=12 \cdot \int_{0}^{\pi / 4}(1-\cos 2 t) \sin ^{2} 2 t d t= \\
& \pi / 4 \\
& =12 \cdot \int_{0} \sin ^{2} 2 t d t-12 \cdot \int_{0} \cos 2 t \cdot \sin ^{2} 2 t d t= \\
& =12 \cdot \int_{0}^{\pi / 4} \frac{1}{2}(1-\cos 4 t) d t-12 \cdot \int_{0}^{\pi / 4} \frac{1}{2} \cdot \sin ^{2} 2 t d(\sin 2 t)= \\
& =\left.6 \cdot\left(t-\frac{1}{4} \sin 4 t\right)\right|_{0} ^{\pi / 4}-\left.6 \cdot\left(\frac{1}{3} \sin ^{3} 2 t\right)\right|_{0} ^{\pi / 4}= \\
& =6 \cdot\left(\frac{\pi}{4}-\frac{1}{4} \sin \frac{4 \pi}{4}\right)-6 \cdot\left(0-\frac{1}{4} \sin 0\right)-2 \cdot\left(\sin ^{3} \frac{2 \pi}{4}-\sin ^{3} 0\right)= \\
& =\frac{3}{2} \pi-\frac{3}{2} \cdot \sin \pi-0-2 \cdot 1=\frac{3}{2} \cdot \pi-2
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+15-1>$

Categories: Kuznetsov's Problem Book Integrals Problem 15 | Integrals

- Last edited: 05:04, 21 June 2010.
- Content is available under CC-BY-SA 3.0.

Answer: \frac{3}{2}\pi-2


In $\triangle A B C$, the three sides $a$, $b$, $c$ satisfy $2 b=a+c$. Find the value of $5 \cos A-4 \cos A \cos C+5 \cos C$.

Solution: $\begin{array}{l}\text { 18. } 2 b=a+c \Leftrightarrow 2 \sin B=\sin A+\sin C \Leftrightarrow 2 \cos \frac{A+C}{2}=\cos \frac{A-C}{2} \Leftrightarrow \tan \frac{A}{2} \tan \frac{C}{2}=\frac{1}{2} \Leftrightarrow \sqrt{\frac{1-\cos A}{1+\cos A}} \\ \cdot \sqrt{\frac{1-\cos C}{1+\cos C}}=\frac{1}{3} \Leftrightarrow 5 \cos A-4 \cos A \cos C+5 \cos C=4\end{array}$

Answer: 4


For all $a, b, c \in \mathbf{R}_{+}$, and $abc=1$, find the minimum value of $S=\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}$.

Solution:
【Analysis】Since the condition $abc=1$ is not linear, we can use the method of taking logarithms to convert it into a linear form. Meanwhile, $S=\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}$ can be transformed into the form $S=\frac{1}{\mathrm{e}^{x}+1}+\frac{1}{\mathrm{e}^{y}+1}+\frac{1}{\mathrm{e}^{x}+1}$, and then use the derivative method to solve it.
Solution Let $x=\ln 2a, y=\ln 2b, z=\ln 2c$.
Then the problem is equivalent to:
For all $x, y, z \in \mathbf{R}$, and $x+y+z=3 \ln 2$, find the minimum value of $S=\frac{1}{\mathrm{e}^{x}+1}+\frac{1}{\mathrm{e}^{y}+1}+\frac{1}{\mathrm{e}^{z}+1}$.
Let $f(x)=\frac{1}{\mathrm{e}^{x}+1}(x \in \mathbf{R})$. Then $f^{\prime}(x)=-\frac{\mathrm{e}^{x}}{\left(\mathrm{e}^{x}+1\right)^{2}}<0$ when $x>0$, and $f^{\prime \prime}(x)>0$, i.e., the function is convex; when $x<0, f(-x)+f(x)=1$ and $f(0)=\frac{1}{2}$.
(1) When $x, y, z \geqslant 0$, we have
$$
\begin{array}{l}
f^{\prime}(\ln 2)=-\frac{2}{9}, f(\ln 2)=\frac{1}{3} \\
\Rightarrow f(x) \geqslant-\frac{2}{9}(x-\ln 2)+\frac{1}{3} \\
\Rightarrow S=\frac{1}{\mathrm{e}^{x}+1}+\frac{1}{\mathrm{e}^{y}+1}+\frac{1}{\mathrm{e}^{z}+1} \\
\geqslant-\frac{2}{9}((x+y+z)-3 \ln 2)+3 \times \frac{1}{3}=1 \\
\Rightarrow S=\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1} \text { has a minimum value of } 1.
\end{array}
$$

(2) When two of $x, y, z$ are not greater than 0, without loss of generality, assume $x, y \leqslant 0$, then
$$
\begin{array}{l}
f(x) \geqslant f(0)=\frac{1}{2}, f(y) \geqslant f(0)=\frac{1}{2} \\
\Rightarrow S=f(x)+f(y)+f(z) \geqslant 1+f(z)>1 . \\
\text { Also, } z=3 \ln 2-x-y>-x>0, \text { so } \\
f(z)+f(x)>1.
\end{array}
$$
In summary, the minimum value of $S$ is 1.

Answer: 1


In a school class, some ThÃ¤lmann Pioneers work in the club of international friendship. When asked who among them works in the club's translation office, 7 raise their hands.

Then, they are asked who works in the club's country circle; this time, 6 raise their hands. It is also noted that 5 of the Pioneers are active in the club's young correspondents' circle. There are no other circles in this club besides these three.

Next, the question is asked who works in both the translation office and the country circle; this time, 4 of the Pioneers raise their hands. It is also determined that 3 of them work in both the translation office and the young correspondents' circle, and 2 of the Pioneers belong to both the country circle and the young correspondents' circle.

Exactly one of the Pioneers from the mentioned school class belongs to all three circles.

Determine the total number of Pioneers from this class who work in the club of international friendship! (All numerical data are considered exact)

Solution:
If the number of students in each circle is denoted by the initial letter of the circle, then \((d)=7\), \((l)=6\), and \(k=6\). Furthermore, in several circles, \((d+l)=4\), \((d+k)=3\), \((l+k)=2\), and \((d+k+l)=1\).

In \(d\) and \(l\) without \(k\), there are accordingly \((d+l)-(d+k+l): 4-1=3\) students, (*)

in \(l\) and \(k\) without \(d\), there are accordingly \((l+k)-(d+k+l): 2-1=1\) student, \(\left({ }^{* *}\right)\)

in \(k\) and \(d\) without \(l\), there are accordingly \((d+k)-(d+k+l): 3-1=2\) students. (***)

In \(d\) without \(k\) and \(l\), there are accordingly \((d)-(*)-(* *)-(* * *): 7-3-2-1=1\) student. Similarly, in \(l\) without \(k\) and \(d\), there is 1 student, and likewise in \(k\) without \(l\) and \(d\), there is 1 student.

In total, this results in \(1+1+1+3+1+2+1=10\).

Thus, 10 students in the class are working in the club of international friendship.

Solution taken from \([5]\)

### 2.14.2 II. Round 1973, Class 5

Answer: 10


Inside triangle $ABC$ there are three circles $k_1, k_2, k_3$ each of which is tangent to two sides of the triangle and to its incircle $k$. The radii of $k_1, k_2, k_3$ are $1, 4$, and $9$. Determine the radius of $k.$

Solution:
1. **Define the problem setup:**
   - Let \( A_1A_2A_3 \) be the triangle and \( (I_i, k_i) \) be the circle next to the vertex \( A_i \), where \( i=1,2,3 \).
   - Let \( I \) be the incenter of the triangle \( A_1A_2A_3 \).
   - The radii of the circles \( k_1, k_2, k_3 \) are given as \( 1, 4, \) and \( 9 \) respectively.
   - We need to determine the radius \( k \) of the incircle \( k \).

2. **Projection and distance relations:**
   - Let \( x_2 \) and \( x_3 \) be the projections of \( II_2 \) and \( II_3 \) on \( A_2A_3 \).
   - Using the Pythagorean theorem, we have:
     \[
     x_2^2 = (k + k_2)^2 - (k - k_2)^2 = 4kk_2 \quad \text{(1)}
     \]
     \[
     x_3^2 = 4kk_3 \quad \text{(2)}
     \]

3. **Distance between centers of circles:**
   - From triangle \( II_2I_3 \), we have:
     \[
     (I_2I_3)^2 = (x_2 + x_3)^2 + (k_3 - k_2)^2 = (II_2)^2 + (II_3)^2 + 2 \cdot II_2 \cdot II_3 \cdot \sin{\frac{A_1}{2}} \quad \text{(3)}
     \]

4. **Combining the expressions:**
   - From the above three expressions, we get:
     \[
     k(k_2 + k_3) - k_2k_3 + 4k\sqrt{k_2k_3} - k^2 = (k + k_2)(k + k_3) \sin{\frac{A_1}{2}} \quad \text{(4)}
     \]

5. **Using the sine relation:**
   - We know that:
     \[
     \sin{\frac{A_1}{2}} = \frac{k - k_1}{k + k_1}
     \]

6. **Summing the equations:**
   - Summing equation (4) over \( i = 1, 2, 3 \), we get:
     \[
     2k(k_1 + k_2 + k_3) - (k_1k_2 + k_2k_3 + k_1k_3) + 4k(\sqrt{k_1k_2} + \sqrt{k_2k_3} + \sqrt{k_1k_3}) - 3k^2 = (k + 1)(k + 4)\frac{k - 9}{k + 9} + (k + 4)(k + 9)\frac{k - 1}{k + 1} + (k + 1)(k + 9)\frac{k - 4}{k + 4}
     \]

7. **Substitute the given values:**
   - Substituting \( k_1 = 1 \), \( k_2 = 4 \), and \( k_3 = 9 \), and rearranging, we get:
     \[
     k(k - 11)(3k^3 + 39k^2 + 121k + 37) = 0
     \]

8. **Solve for \( k \):**
   - Since \( k > 9 \), we have:
     \[
     k = 11
     \]

The final answer is \( \boxed{11} \).

Answer: 11


Solution:
[Solution 1] Given $a>b>c>0$, we know $a b>a c>b c$, so
$$
\begin{aligned}
a^{2}+2 a b+b^{2}+c^{2}>a^{2}+2 a c+c^{2}+b^{2} \\
>a^{2}+b^{2}+2 b c+c^{2},
\end{aligned}
$$

which means $(a+b)^{2}+c^{2}>(a+c)^{2}+b^{2}>a^{2}+(b+c)^{2}>0$, hence $\sqrt{(a+b)^{2}+c^{2}}>\sqrt{(a+c)^{2}+b^{2}}>\sqrt{a^{2}+(b+c)^{2}}>0$, or $l_{3}>l_{1}>l_{2}>0$. Therefore, $l_{3}^{2}>l_{1} l_{3}>l_{2} l_{3}>l_{1} l_{2}>l_{2}^{2}$. Thus, the answer is $(D)$.
[Solution 2] Let $a=3, b=2, c=1$. It is easy to calculate
$$
l_{1}=\sqrt{20}, \quad l_{2}=\sqrt{18}, \quad l_{3}=\sqrt{26},
$$

Clearly, $l_{2}^{2}$ is the smallest among the products.
Thus, the answer is $(D)$.

Answer: D


Given $\overrightarrow{O P}=(2,1), \overrightarrow{O A}=(1,7), \overrightarrow{O B}=(5,1)$. Let $X$ be a point on the line $O P$ (where $O$ is the origin). Then, the value of $\angle A X B$ when $\overrightarrow{X A} \cdot \overrightarrow{X B}$ is minimized is ( ).
(A) $90^{\circ}$
(B) $\arccos \frac{4 \sqrt{17}}{17}$
(C) $\arccos \left(\frac{-4 \sqrt{17}}{17}\right)$
(D) $\pi+\arccos \left(-\frac{4 \sqrt{17}}{17}\right)$

Solution:
4. (C).

Let $\overrightarrow{O X}=\left(x_{0}, y_{0}\right), \overrightarrow{O P}=(2,1)$, then $\frac{x_{0}}{2}=\frac{y_{0}}{1}$.
$\therefore x_{0}=2 y_{0}$, then
$$
\begin{array}{c}
\overrightarrow{X A}=\left(1-2 y_{0}, 7-y_{0}\right), \overrightarrow{X B}=\left(5-2 y_{0}, 1-y_{0}\right) . \\
\therefore \overrightarrow{X A} \cdot \overrightarrow{X B}=\left(5-2 y_{0}\right)\left(1-2 y_{0}\right)+\left(1-y_{0}\right) \cdot \\
\left(7-y_{0}\right)=5 y_{0}^{2}-20 y_{0}+12 .
\end{array}
$$

Therefore, when $y_{0}=2$, $\overrightarrow{X A} \cdot \overrightarrow{X B}$ has the minimum value, at this time, $\overrightarrow{O X}=(4,2), \overrightarrow{X A}=(-3,5), \overrightarrow{X B}=(1,-1),|\overrightarrow{X A}|=$ $\sqrt{34},|\overrightarrow{X B}|=\sqrt{2}$,
thus, $\cos \angle A X B=\frac{\overrightarrow{X A} \cdot \overrightarrow{X B}}{|\overrightarrow{X A}| \cdot|\overrightarrow{X B}|}$
$$
=\frac{(-3,5) \cdot(1,-1)}{\sqrt{34} \cdot \sqrt{2}}=\frac{-8}{\sqrt{68}}=-\frac{4 \sqrt{17}}{17} \text {. }
$$

Answer: C


Determine the product of the solutions of the equation

$$
x^{\log _{2011} x} \cdot \sqrt{2011}=x^{2011}
$$

Solution:
Solution. Taking the logarithm to base 2011:

$$
\begin{aligned}
x^{\log _{2011} x} \cdot \sqrt{2011} & =x^{2011} \\
\log _{2011} x \cdot \log _{2011} x+\log _{2011} \sqrt{2011} & =\log _{2011} x^{2011} \\
\log _{2011}^{2} x+\frac{1}{2} & =2011 \cdot \log _{2011} x
\end{aligned}
$$

The equation makes sense only for $x>0$.

Let's introduce the substitution $\log _{2011} x=a$ and we get

$$
a^{2}-2011 a+\frac{1}{2}=0
$$

The solutions of this equation satisfy Vieta's formulas, so we have:

$$
\begin{array}{ll}
a_{1}+a_{2}=2011 \\
\log _{2011} x=a_{1} \quad \Rightarrow \quad x_{1}=2011^{a_{1}} \\
\log _{2011} x=a_{2} \quad \Rightarrow \quad x_{2}=2011^{a_{2}}
\end{array}
$$

It follows that

$$
x_{1} \cdot x_{2}=2011^{a_{1}} \cdot 2011^{a_{2}}=2011^{a_{1}+a_{2}}=2011^{2011}
$$

Answer: 2011^{2011}


$S$ is the set of rational numbers $r$ where $0<r<1$, and $r$ has a repeating decimal expansion of the form $\overline{0 . a b c a b c a b c \cdots}=\overline{0 . \dot{a} b \dot{c}}, a, b, c$ are not necessarily distinct. Among the elements of $S$, how many different numerators are there when written in lowest terms?

Solution:
Given $999=3^{3} \cdot 37$.

If $\overline{a b c}$ is neither divisible by 3 nor by 37, then the fraction is in its simplest form. By the principle of inclusion-exclusion, we have
$$
999-\left(\frac{999}{3}+\frac{999}{37}\right)+\frac{999}{3 \cdot 37}=648
$$

such numbers.
In addition, there are fractions of the form $\frac{k}{37}$, where the natural number $k$ is a multiple of 3 and less than 37. Such $k$ values are
$3,6,9, \cdots, 36$, a total of 12.
Therefore, the number of numerators that satisfy the condition is $648+12=660$.

Answer: 660


When flipped, a coin has a probability $p$ of landing heads. When flipped twice, it is twice as likely to land on the same side both times as it is to land on each side once. What is the larger possible value of $p$ ?

Solution: $2 p^{2}-2 p+1$. The probability that the coin will land on each side once is $p(p-1)+(p-1) p=$ $2 p(1-p)=2 p-2 p^{2}$. We are told that it is twice as likely to land on the same side both times, so $2 p^{2}-2 p+1=2\left(2 p-2 p^{2}\right)=4 p-4 p^{2}$. Solving, we get $6 p^{2}-6 p+1=0$, so $p=\frac{3 \pm \sqrt{3}}{6}$. The larger solution is $p=\frac{3+\sqrt{3}}{6}$.

Answer: \frac{3+\sqrt{3}}{6}


Find the minimum value of the function $u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}$ with real numbers $x, y$ as variables.

Solution:
Solve $u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}=(x-y)^{2}+\left(\frac{9}{x}+\sqrt{2-y^{2}}\right)^{2}-2$. From the above equation, we can consider points $P_{1}\left(x, \frac{q}{x}\right), P_{2}\left(y,-\sqrt{2-y^{2}}\right)$ on the plane. When $x \in \mathbf{R}, x \neq 0$, the trajectory of $P_{1}$ is a hyperbola with the two coordinate axes as asymptotes; when $y \in \mathbf{R},|y| \leqslant \sqrt{2}$, the trajectory of $P_{2}$ is the lower half of a circle centered at the origin with a radius of $\sqrt{2}$. From the graph (omitted), it is evident that the minimum value of $\left|P_{1} P_{2}\right|$ is $3 \sqrt{2}-\sqrt{2}$. Therefore, $u(x, y)_{\text {min }}=(3 \sqrt{2}-\sqrt{2})^{2}-2=6$ is the desired result.

Note: A slight variation of this example is the 1983 Putnam Competition problem: Find the minimum value of $f(u, v)=(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}$.

Answer: 6