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## Task A-2.5.
Ivica made a large cube with a side length of $n$ from $n^3$ unit cubes and then painted some of the six faces of the large cube, while leaving others unpainted. When he disassembled the large cube, he found that exactly 1000 unit cubes had no painted faces. Show that this is indeed possible and determi... | ## Solution.
After painting some sides of a cube composed of $n^{3}$ smaller cubes, the number of unpainted smaller cubes is certainly less than $n^{3}$ and greater than $(n-2)^{3}$.
Thus, $(n-2)^{3}<1000<n^{3}$, so it must be that $(n-1)^{3}=10^{3}$, i.e., $n=11$.
Now let's count the unpainted smaller cubes in all ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.2.
Exactly three interior angles of a convex polygon are obtuse. What is the maximum number of sides that this polygon can have? | ## First solution.
The sum of the exterior angles of a convex polygon is $360^{\circ}$.
Three exterior angles of the observed polygon are acute (opposite to the obtuse interior angles), while all other exterior angles must be obtuse or right.
2 points
At most three exterior angles can be obtuse or right, otherwise ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.3.
Solve the equation in the set of real numbers
$$
4^{2 x+\sqrt{-1+x^{2}}}-5 \cdot 2^{2 x-1+\sqrt{-1+x^{2}}}=6
$$ | ## Solution.
We solve the problem by substitution $2 x+\sqrt{-1+x^{2}}=y$ under the condition $x^{2}-1 \geq 0$.
(Or by substitution $2^{2 x+\sqrt{-1+x^{2}}}=t$.)
Then we have $2 \cdot\left(2^{y}\right)^{2}-5 \cdot 2^{y}-12=0$.
The solution is only $2^{y}=4$ since $2^{y}>0 \forall y \in \mathbb{R}$. Then $y=2$.
It ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.5.
A bus left from place $A$ to place $B$. 50 minutes later, a car left from place $A$ and arrived at place $B$ 10 minutes before the bus. If they had left simultaneously, one from place $A$ and the other from place $B$ (one heading towards the other), they would have met after one hour and 12 minutes. If ... | ## Solution.
The bus spent $x$ hours on the journey between locations $A$ and $B$, while the car spent $x-1$ hours, so it must be that $x>1$.
The speed of the bus is $\frac{s}{x}$ (where $s$ is the distance between locations $A$ and $B$).
The speed of the car is $\frac{s}{x-1}$.
Starting from location $A$, the bus ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.1.
Determine the value of the real parameter $p$ so that the solutions of the equation
$$
(p-3) x^{2}+\left(p^{2}+1\right) x-11 p+18=0
$$
are the lengths of the legs of a right triangle with a hypotenuse of length $\sqrt{17}$. | ## Solution.
Let $x_{1}$ and $x_{2}$ be the solutions of the given equation. The condition given in the problem then states $x_{1}^{2}+x_{2}^{2}=17$. According to Viète's formulas, we have
$$
x_{1}+x_{2}=-\frac{p^{2}+1}{p-3} \quad \text { and } \quad x_{1} x_{2}=\frac{-11 p+18}{p-3}
$$
so
$$
x_{1}^{2}+x_{2}^{2}=\le... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.5.
On a board of dimensions $8 \times 8$, kings and rooks are placed such that no figure is under attack. A king attacks the adjacent fields (eight of them, unless the king is on the edge of the board), and a rook attacks all the fields in the row and column it is in. What is the maximum number of figures ... | ## Solution.
One rook covers its own square and attacks 7 additional squares in its row and column.
The second rook covers its own square and attacks 6 additional squares in its row and column (not counting those already attacked by the first rook).
Continuing this way, the sixth rook covers its own square and 2 add... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.4.
Determine all triples $(x, y, z)$ of real numbers for which
$$
\begin{aligned}
& \left(x^{2}+1\right) y=z^{2}+1 \\
& \left(y^{2}+1\right) z=x^{2}+1 \\
& \left(z^{2}+1\right) x=y^{2}+1
\end{aligned}
$$ | ## First solution.
If we multiply all three given equations, we see that
$$
\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right) y z x=\left(z^{2}+1\right)\left(x^{2}+1\right)\left(y^{2}+1\right)
$$
Notice that for all real numbers $t$, $t^{2}+1 \geqslant 1>0$, so from the above equation, we conclude that $x... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. For which real numbers $m$ does the equation $3 x+9=m(m-x)$ have a unique solution?
(8) | Solution. By rearranging we get $(m+3) x=m^{2}-9$.
For $m \neq-3$, $m+3 \neq 0$ so there is a unique solution $x=m-3$.
(for a complete solution 8 points)
(for the solution $x=m-3$ (always), 0 points) | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. If $a^{2}+b^{2}+c^{2}=a b+b c+c a$, what is $(a+2 b-3 c)^{2009}$? | Solution. If we multiply the given equation by 2 and move everything to the left side, we get
$$
2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a=0
$$
If we rearrange the equation, we get
$$
a^{2}+b^{2}-2 a b+b^{2}+c^{2}-2 b c+c^{2}+a^{2}-2 c a=0
$$
or
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0
$$
from which it follows that $a=... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
30. Calculate the value of the expression $\log ^{2} 5+\log 2 \cdot \log 50$.
(8) | First solution. In order, we have:
$$
\begin{aligned}
\log ^{2} 5+\log 2 \cdot \log 50 & =\log ^{2} 5+\log 2 \cdot(\log 25+\log 2) \\
& =\log ^{2} 5+\log 2 \cdot(2 \log 5+\log 2) \\
& =\log ^{2} 5+2 \log 5 \cdot \log 2+\log ^{2} 2 \\
& =(\log 5+\log 2)^{2}=(\log 10)^{2}=1
\end{aligned}
$$
(full solution 8 points) | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Prove that for every natural number $n \geq 2$ the unit digit of the number $2^{2^{n}}$ is 6.
(8) | Solution. We prove the statement by mathematical induction.
Base of induction: For $n=2$ we have
$$
2^{2^{2}}=2^{4}=16
$$
so the statement holds.
Induction hypothesis: Assume that the statement holds for some natural number $n \geq 2$, i.e., the last digit of the number $2^{2^{n}}$ is 6.
Induction step: Prove that... | 6 | Number Theory | proof | Yes | Yes | olympiads | false |
Task B-1.4. (20 points) If $a^{2}+b^{2}=1, c^{2}+d^{2}=1$ and $a c+b d=0$, what is $a b+c d ?$ | First solution. Since $a^{2}+b^{2}=1$, $a$ and $b$ cannot both be 0 at the same time. Let $a \neq 0$.
From the third equality, we have $c=-\frac{b d}{a}$. Then,
$$
1=c^{2}+d^{2}=\frac{b^{2} d^{2}}{a^{2}}+d^{2}=\frac{d^{2}\left(a^{2}+b^{2}\right)}{a^{2}}=\frac{d^{2}}{a^{2}}
$$
from which we have $a^{2}=d^{2}$. Now,
... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.1. (20 points) For complex numbers $z, w$ such that $|z|=|w|=|z-w|$, calculate $\left(\frac{z}{w}\right)^{99}$. | First solution. Let $u=\frac{z}{w}=x+y i$. Then $|u|=1,|u-1|=1$.
Then $x^{2}+y^{2}=1,(x-1)^{2}+y^{2}=1$.
By subtracting these equations, we get $x=\frac{1}{2}, y= \pm \frac{\sqrt{3}}{2}$, so $u=\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$.
Now,
$$
\left(\frac{z}{w}\right)^{99}=u^{99}=\left[\left(\frac{1}{2} \pm \frac{\sqr... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.1. (20 points) In an increasing arithmetic sequence, the product of the second and third term is 3, and the product of the third and fifth term is -3. How many of the first terms of the sequence need to be summed to achieve the minimum sum? What is that sum? | Solution. We will represent the members of the sequence using the first term $a_{1}$ and the difference of the sequence $d$.
According to the condition of the problem, we have these two equations:
\[
\begin{aligned}
\left(a_{1}+d\right)\left(a_{1}+2 d\right) & =3 \\
\left(a_{1}+2 d\right)\left(a_{1}+4 d\right) & =-3
... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.2.
Prove that the value of the expression $\frac{\sqrt[3]{7+5 \sqrt{2}}(\sqrt{2}-1)}{\sqrt{4+2 \sqrt{3}}-\sqrt{3}}$ is a natural number. | ## Solution.
In the numerator of the given fraction, we will place the expression in parentheses under the cube root, and in the denominator, we will recognize the square of a binomial under the first root:
$$
\begin{aligned}
& \frac{\sqrt[3]{7+5 \sqrt{2}}(\sqrt{2}-1)}{\sqrt{4+2 \sqrt{3}}-\sqrt{3}}=\frac{\sqrt[3]{(7+... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
## Task B-2.5.
In a kite where the lengths of the diagonals are $d_{1}=24 \mathrm{~cm}$ and $d_{2}=8 \mathrm{~cm}$, a rectangle is inscribed such that its sides are parallel to the diagonals of the kite. Determine the dimensions of the inscribed rectangle that has the maximum area. | ## Solution.
Let $x$ and $y$ be the sides of the rectangle, and let $y \parallel d_{1}, x \parallel d_{2}$. Notice triangles $B C A$ and $B F E$. According to the K-K theorem,
$$
\triangle... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.1.
In which number system does the equality $\sqrt{2521}-\sqrt{2400}=1$ hold? | ## Solution.
Let the base of the sought number system be denoted by $b$. Clearly, $b>5$.
$$
\begin{aligned}
& \sqrt{2521}-\sqrt{2400}=1 \\
& \Rightarrow \sqrt{2521}=1+\sqrt{2400} /^{2} \\
& \Rightarrow 2521-1-2400=2 \sqrt{2400} \\
& \Rightarrow 120=2 \sqrt{2400}
\end{aligned}
$$
This equality, written in the system ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.1.
The perimeter of a right triangle is 18, and the area is 9. What is the length of the hypotenuse of this triangle? | ## First Solution.
Let $a$ and $b$ be the lengths of the legs, and $c$ be the length of the hypotenuse of the given triangle.
Since $a+b+c=18$, we have
$$
(a+b)^{2}=(18-c)^{2}=18^{2}-36 c+c^{2}
$$
The given triangle is a right triangle, so by the Pythagorean theorem, $a^{2}+b^{2}=c^{2}$. 1 point It follows that
$$... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.2.
Let the complex numbers $a, b$ and $c$ be the solutions of the equation $x^{3}-2 x+2=0$. Determine
$$
\frac{a+1}{a-1}+\frac{b+1}{b-1}+\frac{c+1}{c-1}
$$ | ## Solution.
We use Viète's formulas. Since $a, b$, and $c$ are the solutions of the equation $x^{3}-2 x+2=0$, we have
$$
a+b+c=0, \quad a b+b c+c a=-2, \quad a b c=-2
$$
It remains to express the given expression in terms of $a+b+c, a b+b c+c a$, and $a b c$:
$$
\begin{aligned}
\frac{a+1}{a-1}+\frac{b+1}{b-1}+\fra... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Determine the largest natural number $n$ such that there exists a sequence of $n$ real numbers with the following properties:
(i) the sum of any three consecutive terms of the sequence is positive,
(ii) the sum of any five consecutive terms of the sequence is negative. | Solution.
For $n=6$ there exists a sequence with the desired property, for example,
$$
3,-5,3,3,-5,3
$$
(5 points)
Note. A student who finds only an example for $n=5$, (e.g., $-1-13-1-1$) gets 2 points.
Assume there exists a sequence of length $n \geqslant 7$ with the desired property. Choose any 7 consecutive ter... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.2.
Ivo, Alen, Vanja, Marko, and Saša are chefs in a hotel. Alen and Marko are responsible for preparing breakfast and lunch, Ivo and Vanja work on preparing lunch and dinner, while Saša is available for all three meals. In how many ways can their daily cooking schedule be arranged if each meal is prepared ... | ## Solution.
Lunch can be cooked by all chefs, so we will first choose the person who will cook lunch. There are several cases:
1. If lunch is cooked by Alen or Marko, they must also cook breakfast, so for dinner, we would need to engage two new chefs who would work on only one meal. Therefore, Alen and Marko cannot ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.4.
Fran decided to paint the fence with the help of his friends Tin and Luke. They estimated that it would take Tin 3 hours more than Fran to paint the fence alone, and Luke 2 hours less than Fran to paint the fence alone. Working together, each at their own pace, they would paint the fence in 4 hours. How... | ## First Solution.
Let's denote the number of hours needed for each of the three friends to paint the fence alone.
Fran: $x$ hours, Tin: $x+3$ hours, Luka: $x-2$ hours. Note that $x$ must be greater than 2.
Considering how much each of them paints in one hour, we get the equation
$$
\frac{1}{x}+\frac{1}{x+3}+\frac{... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-1.1. Determine the remainder when the number $\left(7^{2012}\right)^{2014}-\left(3^{12}\right)^{14}$ is divided by 10. | Solution.
$$
7^{1}=7,7^{2}=49,7^{3}=343,7^{4}=2401,7^{5}=16807, \ldots
$$
Therefore, the powers of the number 7 end with the digits $7,9,3,1$.
$$
3^{1}=3,3^{2}=9,3^{3}=27,3^{4}=81,3^{5}=243, \ldots
$$
The powers of the number 3 end with the digits $3,9,7,1$.
The number 2012 is divisible by 4, so the number $7^{201... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.2. Points $H$ and $N$ are the feet of the altitudes from vertex $A$ and vertex $B$ of an acute-angled triangle $ABC$. The length of the altitude from vertex $A$ is $5 \sqrt{3} \mathrm{~cm}$, the length of side $\overline{A B}$ is $14 \mathrm{~cm}$, and the measure of the angle between the altitudes $\overline{... | ## Solution.
Since the given triangle is acute-angled, we conclude that the measure of angle $\angle B O H$ is $60^{\circ}$, so the measure of angle $\angle N B C = 90^{\circ} - 60^{\circ} ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.4. Determine all natural numbers $a$ for which the number $a^{3}+1$ is a power of 3. | ## First solution.
If $a^{3}+1$ is divisible by 3, the number $a$ is of the form $a=3 p-1, p \in \mathbb{N}$. Then,
$$
a^{3}+1=(3 p-1)^{3}+1=27 p^{3}-27 p^{2}+9 p-1+1=9 p\left(3 p^{2}-3 p+1\right)
$$
The expression $3 p^{2}-3 p+1$ is not divisible by 3, and since $a^{3}+1$ is a power of 3, it can only be equal to $3... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.4.
From the starting tram station, three trams depart at 5:00 AM. The first tram takes 1 hour and 30 minutes to return to the starting station, the second tram takes 1 hour, and the third tram takes 40 minutes. At what time will all three trams be at the starting station at the same time again? How many ti... | ## First solution.
The time required for all three trams to return to the starting station at the same time after a joint departure is the least common multiple of the times each tram individually needs to return to the starting station.
Thus, we are looking for the least common multiple of
1 hour and 30 minutes $=9... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.3.
The graphs of two linear functions have slopes of 3 and $\frac{1}{3}$, and intersect at the point $(3,3)$. Determine the area of the triangle bounded by these graphs and the $x$-axis. | ## Solution.
The graphs of the given linear functions are lines with the equation $y=k x+l$.
The slope $k$ is given for both lines, as well as a point that belongs to these lines. Therefore, we can determine the equations of the given lines.
$p_{1} \ldots y=3 x+l$, or $3=9+l$ so $l=-6$, and the equation of the first... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.1.
Solve the equation $\binom{x+1}{x-2}+2\binom{x-1}{3}=7(x-1)$. | ## First Solution.
Due to the definition of binomial coefficients, it must hold that $x-1 \geqslant 3$, i.e., $x \geqslant 4$.
Using the property of binomial coefficients $\binom{n}{k}=\binom{n}{n-k}$, we can rewrite the given equation as:
$\binom{x+1}{x+1-x+2}+2\binom{x-1}{3}=7(x-1)$, or
$\binom{x+1}{3}+2\binom{x-... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.5.
Mate and Roko, while preparing for a competition, were challenging each other with "unsolvable" problems. So Roko asked Mate: Do you know what the sum of the digits of the number $3^{2018}$ is?
To this, Mate replied with a new question:
I don't know, but do you know what the 2018th number is if you co... | ## Solution.
Let's denote the first, second, third, ..., 2018th term of the given sequence of numbers as $a_{1}, a_{2}, a_{3}, \ldots, a_{2018}, \ldots$.
Notice that $a_{1}=3^{2018}=9 \cdot 3^{2016}$.
Since the first term of the sequence is divisible by 9, the sum of its digits is also divisible by 9. Therefore, the... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.1.
In the equation $x^{2}+m-3 x=m x-2$, determine the positive real number $m$ so that the total sum of all solutions of the equation and their squares is 44. | ## Solution.
Given the equation $x^{2}+m-3 x=m x-2$ we write it in the form $x^{2}-(m+3) x+m+2=0. \quad 1$ point The solutions of this quadratic equation must satisfy the condition $x_{1}+x_{2}+x_{1}^{2}+x_{2}^{2}=44$, or $x_{1}+x_{2}+\left(x_{1}+x_{2}\right)^{2}-2 x_{1} \cdot x_{2}=44$.
According to Vieta's formulas... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.6.
Students Marko and Luka worked on a part-time basis, Marko in a tourist office, and Luka in a hotel. Marko earned a total of 200 kn for his work. Luka worked 5 hours less than Marko and earned 125 kn. If Marko had worked as many hours as Luka, and Luka as many hours as Marko, then Luka would have earned... | ## Solution.
$n=$ the number of hours Marko worked, $n>5$.
$x=$ Marko's hourly wage, $y=$ Luka's hourly wage
Then we have the following system of equations
$$
\begin{gathered}
n x=200 \\
(n-5) y=125 \\
n y-(n-5) x=150
\end{gathered}
$$
From the first and second equations, express $x, y$ :
$$
x=\frac{200}{n}, \qua... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.7.
The height $\overline{C D}$ divides the right triangle $A B C$ into two parts in which circles with centers $S_{1}, S_{2}$ are inscribed. Parallels to the height $\overline{C D}$ through the centers $S_{1}, S_{2}$ intersect the legs $\overline{A C}$ and $\overline{B C}$ at points $M$ and $N$, respective... | ## Solution.
$$
\begin{aligned}
& \sin ^{2}\left(x-\frac{2016 \pi}{3}\right)+\sin ^{2}\left(x-\frac{2017 \pi}{3}\right)+\sin ^{2}\left(x-\frac{2018 \pi}{3}\right)= \\
& \sin ^{2}(x-672 \pi)+\sin ^{2}\left(x-\frac{\pi}{3}-672 \pi\right)+\sin ^{2}\left(x-\frac{2 \pi}{3}-672 \pi\right)= \\
& \sin ^{2}(x)+\sin ^{2}\left(x... | 2 | Geometry | proof | Yes | Yes | olympiads | false |
## Task A-1.5. (4 points)
In a bag, there is a sufficiently large number of red, white, and blue balls. Each student randomly takes three balls from the bag. How many students must there be at a minimum to ensure that at least one pair of them has the same combination of balls, i.e., the same number of balls of each c... | ## Solution.
Let's denote the red ball with the letter $C$, the white ball with the letter $B$, and the blue ball with the letter $P$. The possible combinations of three balls are:
$C C C, B B B, P P P, C B B, C P P, B C C, B P P, P C C, P B B, C B P$.
There are a total of 10 possible combinations.
(2 points)
To e... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task A-1.6. (10 points)
If $a^{2}+2 b^{2}=3 c^{2}$, show that $\left(\frac{a+b}{b+c}+\frac{b-c}{b-a}\right) \cdot \frac{a+2 b+3 c}{a+c}$ is a natural number. | ## First Solution.
First, we simplify the expression in parentheses:
$$
\begin{aligned}
& \frac{a+b}{b+c}+\frac{b-c}{b-a}=\frac{(a+b)(b-a)+(b-c)(b+c)}{(b+c)(b-a)} \\
& =\frac{b^{2}-a^{2}+b^{2}-c^{2}}{(b+c)(b-a)} \stackrel{(*)}{=} \frac{3 c^{2}-a^{2}-a^{2}-c^{2}}{(b+c)(b-a)} \\
& =\frac{2 c^{2}-2 a^{2}}{(b+c)(b-a)}=\f... | 4 | Algebra | proof | Yes | Yes | olympiads | false |
Task A-2.6. (10 points)
Let $z=\frac{1}{2}+i \frac{\sqrt{3}}{2}$. Calculate
$$
z+z^{2}+z^{3}+z^{4}+\ldots+z^{k}+\ldots+z^{2010}
$$ | ## Solution.
We calculate: $z^{2}=-\frac{1}{2}+i \frac{\sqrt{3}}{2}, z^{3}=-1, z^{4}=-\frac{1}{2}-i \frac{\sqrt{3}}{2}, z^{5}=\frac{1}{2}-i \frac{\sqrt{3}}{2}, z^{6}=1$. (5 points) By adding, we get $z+z^{2}+z^{3}+z^{4}+z^{5}+z^{6}=0$.
Since 2010 is divisible by 6, we have
$$
\begin{aligned}
& z+z^{2}+z^{3}+z^{4}+\l... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.2. (4 points)
Legoplus is a body consisting of seven equal cubes joined in such a way that there is one cube that shares a common face with each of the remaining six cubes.
Each face of the legoplus must be painted with one color. How many colors are minimally needed to do this so that no two adjacent fac... | ## Solution.
Obviously, there are three sides of the Lego brick for which it is true that any two are adjacent. For example, such sides are marked with the letters $A, B$ and $C$ in the image below. From this, we conclude that at least three colors are needed to color the Lego brick in the desired way.
(2 points)
![... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.3. (4 points)
What is the remainder when the number (ABCDE2010) ${ }_{15}$ is divided by 7?
Numbers in base 15 are written using the digits $0,1,2,3,4,5,6,7,8,9$, A, B, C, D, E whose values are respectively $0,1,2,3,4,5,6,7,8,9,10,11,12,13,14$. | ## First Solution.
The number (ABCDE2010) ${ }_{15}$ can be written as
$$
a=10 \cdot 15^{8}+11 \cdot 15^{7}+12 \cdot 15^{6}+13 \cdot 15^{5}+14 \cdot 15^{4}+2 \cdot 15^{3}+1 \cdot 15^{1}
$$
Let $b=10+11+12+13+14+2+0+1+0=63$.
Notice that the number $15^{k}-1$ is divisible by 7 for any natural number $k$ because
$$
1... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.1.
Determine all natural numbers $b$ for which the equality $11 \cdot 22 \cdot 33=13310$ holds in the number system with base $b$. | ## First solution.
Let $b$ be the sought base of the number system.
The equality in the problem is equivalent to the equality
$$
(b+1)(2 b+2)(3 b+3)=b^{4}+3 b^{3}+3 b^{2}+b
$$
We can simplify the expression by factoring out a factor on the left side, and applying the binomial cube formula on the right side:
$$
6(b... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.2.
Let $S=\{0,95\}$. In each step, Lucija expands the set $S$ by choosing some polynomial with coefficients from $S$, different from the zero polynomial, and adding all integer roots of that polynomial to the set $S$. The process continues by selecting another polynomial with coefficients from the thus exp... | ## Solution.
An integer root of a polynomial with integer coefficients must be a divisor of the free term. Since, without loss of generality, we can assume that the free term of the polynomial Lucija chooses is different from zero, we conclude that the set $S$ can only be extended by integer divisors of the number 95.... | 9 | Algebra | proof | Yes | Yes | olympiads | false |
## Task A-4.2.
A Gaussian integer is a complex number whose real and imaginary parts are integers. Determine the largest natural number $n$ for which there exists a set of $n$ Gaussian integers such that the squares of their absolute values are consecutive natural numbers. | ## Solution.
If the complex number $z=x+y i$ is a Gaussian integer, then $|z|^{2}=x^{2}+y^{2}$ is the sum of the squares of two integers.
The square of an even integer is divisible by 4, while the square of an odd integer gives a remainder of 1 when divided by 4. Therefore, the sum of the squares of two integers can ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.4.
For real numbers $a, b$ and $c$, it is given that $a+b+c=0$ and $abc=4$. Determine the value of the expression $a^{3}+b^{3}+c^{3}$. | ## First Solution.
From the first equality, we have $a+b=-c$, or $(a+b)^{3}=-c^{3}$. 1 point
By applying the formula for the sum of cubes, we get
$$
\begin{array}{rlrl}
(a+b)^{3} & =a^{3}+3 a^{2} b+3 a b^{2}+b^{3}=-c^{3} & 1 \text { point } \\
a^{3} & +3 a b(a+b)+b^{3}=-c^{3} & 1 \text { point } \\
a^{3} & +b^{3}+c^... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.2.
Solve the equation $\sqrt{9-5 x}-\sqrt{3-x}=\frac{6}{\sqrt{3-x}}$ in the set of real numbers. | ## Solution.
Let's group the expressions that include $\sqrt{3-x}$ on one side of the equation, then square it:
$$
\begin{array}{rlrl}
\sqrt{9-5 x} & =\sqrt{3-x}+\frac{6}{\sqrt{3-x}}, \quad /^{2} & \\
9-5 x & =3-x+12+\frac{36}{3-x}, & & 1 \text { point } \\
-6-4 x & =\frac{36}{3-x}, \quad / \cdot(3-x) & \\
(-6-4 x)(3... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.1.
For a natural number $n$, let $s(n)$ denote the sum of its positive divisors, and $d(n)$ the number of its positive divisors. Determine all natural numbers $n$ such that
$$
s(n)=n+d(n)+1
$$ | ## Solution.
Notice that $n=1$ is certainly not a solution, so $d(n) \geqslant 2$.
Furthermore, it is not possible for $d(n)=2$ because then the number $n$ would be prime and the given equation would read $1+n=n+2+1$, so $d(n) \geqslant 3$.
Let $1=D_{1}<D_{2}<\cdots<D_{d}=n$ be the divisors of the number $n$. The in... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.2.
With which digit does the number $2^{2022}+3^{2022}+7^{2022}$ end? | ## Solution.
Powers of 2, starting from $2^{1}$, end with the digits $2,4,8,6$ and then repeat periodically.
Powers of 3, starting from $3^{1}$, end with the digits $3,9,7,1$ and then repeat periodically.
Powers of 7, starting from $7^{1}$, end with the digits $7,9,3,1$ and then repeat periodically.
To determine th... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.6.
The hotel owners at the beginning of the tourist season bought new blankets, towels, and pillowcases for 4000 kn. They paid 120 kn for each blanket, 50 kn for each towel, and 25 kn for each pillowcase. If they bought a total of 100 items, how many blankets, how many towels, and how many pillowcases were... | ## First solution.
Let $x$ be the number of duvets bought, $y$ be the number of blankets bought, and $z$ be the number of pillows bought, $x, y, z \in \mathbb{N}$.
According to the conditions of the problem, we have:
$x+y+z=100 \quad$ and $\quad 120 x+50 y+25 z=4000$.
1 point
Express one unknown from the first equ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.2.
Determine the zeros of the function $f: \mathbf{R} \rightarrow \mathbf{R}, f(x)=\log _{2}\left(18 \cdot 4^{x}-8 \cdot 2^{x}+1\right)-2 x-1$. | ## Solution.
The zeros of the given function are the solutions of the equation $f(x)=0$.
Thus, we need to solve the equation $\log _{2}\left(18 \cdot 4^{x}-8 \cdot 2^{x}+1\right)-2 x-1=0$.
The equation can be written in the following form:
$$
\log _{2}\left(18 \cdot 4^{x}-8 \cdot 2^{x}+1\right)=2 x+1
$$
or
$$
\lo... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak B-4.3.
Ako je $z+z^{-1}=2 \cos 5^{\circ}$, koliko je $\left(z^{2022}-z^{-2022}\right)^{2022}$ ?
| ## Rješenje.
Iz $z+z^{-1}=2 \cos 5^{\circ}$ nakon množenja brojem $z$ dobivamo kvadratnu jednadžbu:
$z^{2}-\left(2 \cos 5^{\circ}\right) z+1=0$.
1 bod
Njezina su rješenja:
$z_{1,2}=\frac{2 \cos 5^{\circ} \pm \sqrt{4 \cos ^{2} 5^{\circ}-4}}{2}=\frac{2 \cos 5^{\circ} \pm 2 \sqrt{\cos ^{2} 5^{\circ}-1}}{2}=\cos 5^{\c... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.5.
Determine all natural numbers $y>1$ that satisfy the equation
$\log _{\sin x} y-3 \log _{y} \sqrt{\sin x}=\frac{1}{2}$, where $x \in\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$. | ## Solution.
Let's write the given equation in the following form:
$\log _{\sin x} y-\frac{3}{2} \cdot \frac{1}{\log _{\sin x} y}=\frac{1}{2}$.
Introduce the substitution $t=\log _{\sin x} y$ so that this equation becomes $t-\frac{3}{2} \cdot \frac{1}{t}=\frac{1}{2}$.
Note that $t=\log _{\sin x} y>0$ due to $y>1$, ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-1.4. Ivo and Mate are clearing the road of snow. First, Ivo cleared $\frac{3}{5}$ of the road, and then Mate cleared the remaining part so that the entire road was cleared in 12 hours. How many hours would it take for them to clear the road together if it is known that it would take Mate 5 hours more to clear th... | ## Solution.
Let $x$ be the number of hours it would take Ivo to clear the road alone.
Then $x+5$ is the number of hours it would take Mate to clear the road alone.
(1 point)
For clearing $\frac{3}{5}$ of the road, Ivo needed $\frac{3 x}{5}$ hours, and for clearing the remaining $\frac{2}{5}$ of the road, Mate need... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-3.2. The lengths of the sides of a triangle are three consecutive odd numbers, and the measure of one of the angles is $\frac{2 \pi}{3}$. Calculate the lengths of the sides of this triangle. | ## Solution.
Let $a=2 n-1, b=2 n+1, c=2 n+3$.
Since $\frac{2 \pi}{3}$ is an obtuse angle, its opposite side is the longest, which is $c=2 n+3$.
From the cosine rule, we get
$$
\begin{aligned}
& \cos \frac{2 \pi}{3}=\frac{(2 n-1)^{2}+(2 n+1)^{2}-(2 n+3)^{2}}{2(2 n-1)(2 n+1)} \\
& \frac{-1}{2}=\frac{4 n^{2}-12 n-7}{2... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-3.4. Helena invited a certain number of children to her birthday party. The age of one child was one eighth of the sum of the ages of the remaining children. This age ratio did not change in later years. How many children were at the party? | ## Solution.
Let $n$ be the number of children at the party, and $x$ be the age of the mentioned child.
Furthermore, let $x_{1}, x_{2}, \ldots, x_{n-1}$ be the ages of the remaining $n-1$ children. Then we have
$$
x=\frac{1}{8}\left(x_{1}+x_{2}+\cdots+x_{n-1}\right)
$$
(2 points)
If we add $t$ years to each child,... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.5. Determine all natural numbers $n$ for which $2^{n}-1$ and $2^{n}+1$ are simultaneously prime numbers. | ## Solution.
For $n=1$ we have $2^{n}-1=1$ and $2^{n}+1=3$, so $n=1$ is not a solution because 1 is not a prime number. (1 point)
For $n=2$ we have $2^{n}-1=3$ and $2^{n}+1=5$.
(1 point)
Let $n>2$. Then the numbers $2^{n}-1, 2^{n}, 2^{n}+1$ are three consecutive natural numbers, one of which is certainly divisible ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.5.
Determine the minimum value of the expression $2 x^{2}+\frac{1}{2} y^{2}$ if $y+2 x=2$. | ## Solution.
From the second equality, express $y=2-2 x$ and substitute into the first. Then we have
$$
\begin{aligned}
2 x^{2}+\frac{1}{2}(2-2 x)^{2} & =2 x^{2}+\frac{1}{2}\left(4-8 x+4 x^{2}\right) \\
& =4 x^{2}-4 x+2 \\
& =4 x^{2}-4 x+1+1 \\
& =(2 x-1)^{2}+1
\end{aligned}
$$
Since $(2 x-1)^{2} \geqslant 0$ for al... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.6.
Solve the equation $i z^{2}+2 \bar{z}=0$ in the set of complex numbers. Calculate the quotient of the sum of the cubes and the product of all solutions that are different from zero. | ## Solution.
Let $z=x+i y, x, y \in \mathbb{R}$. Then we have
$$
\begin{array}{r}
i \cdot(x+i y)^{2}+2(x-i y)=0 \\
i\left(x^{2}+2 x y i-y^{2}\right)+2 x-2 y i=0 \\
i x^{2}-2 x y-i y^{2}+2 x-2 y i=0
\end{array}
$$
By equating the real and imaginary parts, we get the system of equations
$$
\begin{array}{r}
-2 x y+2 x... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.5.
The lengths of the sides of triangle $A B C$ are $|B C|=4 \mathrm{~cm}$ and $|A C|=5 \mathrm{~cm}$, and the length of the part of the angle bisector of $\varangle A C B$ that lies within the triangle is $s=\frac{10}{3} \mathrm{~cm}$. Calculate the length of side $\overline{A B}$. | ## Solution.
The bisector of an angle in a triangle intersects the opposite side in the ratio of the remaining sides (The Angle Bisector Theorem of a Triangle), i.e., $\frac{x}{y}=\frac{5}{4}$, where $x=|A D|, y=|D B|$, and point $D$ is the intersection of the given bisector and side $\overline{A B}$. Then, $x=5 k, y=... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak B-4.2.
Dane su beskonačne sume
$$
\begin{aligned}
& a=2020^{x+3}+2020^{x+2}+2020^{x+1}+2020^{x}+\ldots \\
& b=2019^{x+2}+2019^{x+1}+2019^{x}+2019^{x-1}+\ldots
\end{aligned}
$$
Odredite $x \in \mathbb{R}$ tako da vrijedi $\frac{a}{b}=2018$.
| ## Rješenje.
Uočimo da su beskonačne sume $a$ i $b$ geometrijski redovi. Njihovi kvocijenti su redom $\frac{1}{2020} \mathrm{i} \frac{1}{2019}$.
Kako su kvocijenti manji od 1 redovi su konvergentni, pa vrijedi
$$
\begin{aligned}
& a=2020^{x+3}+2020^{x+2}+2020^{x+1}+2020^{x}+\cdots=\frac{2020^{x+3}}{1-\frac{1}{2020}}... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.5.
Spouses Ana and Tomislav came to a party where four other couples were also present. Upon arrival, a certain number of handshakes occurred. No one shook hands with their own spouse or with themselves. Later, when Tomislav asked everyone how many people they had shaken hands with, he received nine differ... | ## Solution.
No one dealt with more than eight people, so the answers Tomislav received were: "0", "1", "2", "3", "4", "5", "6", "7", "8".
Ana definitely did not deal with eight people. If Ana had dealt with eight people, everyone else (except Tomislav) would have dealt with her, so no one would have answered "0".
L... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.3.
Determine all values of the parameter $a$ for which the system
$$
\begin{aligned}
2^{|x|}+|x| & =x^{2}+y+a \\
x^{2}+y^{2} & =1
\end{aligned}
$$
has exactly one solution $(x, y) \in \mathbb{R}^{2}$. | ## Solution.
If the pair $(x, y)$ is a solution to the given system, then the pair $(-x, y)$ is also a solution.
We conclude that the unique solution of this system must be of the form $(0, y)$.
Substituting $x=0$ into the given system, we get
$$
\begin{aligned}
1 & =y+a \\
y^{2} & =1
\end{aligned}
$$
so $y=1$ or ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak B-1.3.
Ako je $\frac{a}{b}=\frac{c}{d}, b \neq 0, d \neq 0$ koliko je
$$
\frac{(a+c)(b+d)}{a+b+c+d}-\frac{a b}{a+b}-\frac{c d}{c+d} ?
$$
| ## Prvo rješenje.
Ako je $\frac{a}{b}=\frac{c}{d}$, postoji neki $k \in \mathbb{R}, k \neq 0$, tako da je $a=k c$ i $b=k d$.
Tada je
$$
\begin{aligned}
\frac{(a+c)(b+d)}{a+b+c+d}-\frac{a b}{a+b}-\frac{c d}{c+d} & =\frac{c(1+k) d(1+k)}{c(1+k)+d(1+k)}-\frac{k c k d}{k(c+d)}-\frac{c d}{c+d} & & 1 \text { bod } \\
& =\f... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.4.
How many right triangles are there with the lengths of the legs being integers $a, b$, and the length of the hypotenuse $b+1$, where $b<100$? | ## Solution.
For a right-angled triangle, we have $(b+1)^{2}=a^{2}+b^{2}$, which means $b^{2}+2 b+1=a^{2}+b^{2}$, from which it follows that $2 b=a^{2}-1, \text{i.e., } b=\frac{a^{2}-1}{2}$.
Since $b<100$, we have $\frac{a^{2}-1}{2}<100, a^{2}<201$.
The squares of integers less than 201 are 4, 9, 16, 25, 36, 49, 64,... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.6.
Luka and Pero decided to visit a friend and set off from place $A$ to place $B$, which is $170 \mathrm{~km}$ away. Luka started walking, while Pero was driven by a motorcyclist. After some time, the motorcyclist returned to pick up Luka, and Pero continued on foot, so that both arrived at place $B$ at t... | ## Solution.
Since both arrived at the same time traveling at the same speed, it means they walked the same part of the path on foot and the same part on the motor.
Let's introduce some nota... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.2.
In a rectangular hall, there is a carpet in the shape of a rhombus. The area of the carpet is $54 \mathrm{~m}^{2}$, and the perimeter is $30 \mathrm{~m}$. The vertices of the rhombus are exactly at the midpoints of the sides of the rectangle. Determine the dimensions of the hall. | ## Solution.
Notice that the dimensions of the hall (the sides of the rectangle $a$ and $b$) are also the diagonals of the rhombus.
If $P=\frac{a \cdot b}{2}=54$, and $o=4 d=30$, then the ... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.4.
Determine the last digit of the product of the first hundred natural numbers that give a remainder of 3 when divided by 5. | ## Solution.
All natural numbers that give a remainder of 3 when divided by 5 can be written in the form $5k + 3, k \in \mathbb{N}_{0}$.
Let's write the product of the first hundred such numbers:
$$
\begin{aligned}
& 3 \cdot 8 \cdot 13 \cdot 18 \cdot 23 \cdot 28 \cdot 33 \cdots (5k + 3) \cdots 498 \\
& = (3 \cdot 8)... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.3.
Each of the 28 students visited at least one of the three countries. The number of students who visited only Italy and Spain is equal to the number of students who visited only Italy. No student visited only Spain or only Greece. Six students visited Greece and Italy, but not Spain. The number of studen... | ## Solution.
We solve the problem using Venn diagrams. Let $x$ be the number of students who visited only Italy, and $y$ be the number of students who visited all three countries.
In set $I... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.5.
Ivo and Ana both drank lemonade at the cinema and watched a movie. Ivo took a medium size, and Ana a large one, which is $50\%$ larger than the medium. After both had drunk $\frac{3}{4}$ of their lemonade, Ana gave Ivo one third of what was left to her and an additional 0.5 dl. After the movie ended and... | ## Solution.
Let $x$ be the average amount of lemonade Ivo took at the beginning. Ana took $1.5x$.
Ivo received from Ana $\frac{1}{3} \cdot \frac{1}{4} \cdot 1.5 x + 0.5 = \frac{1}{8} x + 0.5$,
which means he drank a total of $x + \frac{1}{8} x + 0.5 = \frac{9}{8} x + 0.5 \text{ dl}$ of lemonade.
Ana drank a total ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.7.
Determine the value of the real parameter $m$ so that the solutions of the equation
$$
(m x-1) \cdot x=m x-2
$$
represent the lengths of the legs of a right triangle with a hypotenuse of length $\frac{5}{6}$. | ## Solution.
The given equation can be written in the form $m x^{2}-(m+1) x+2=0$.
The solutions must be real, so the discriminant of the equation must be greater than or equal to zero,
$$
\begin{aligned}
& (m+1)^{2}-8 m \geqslant 0 \\
& m^{2}-6 m+1 \geqslant 0 \\
& m \in(-\infty, 3-2 \sqrt{2}] \cup[3+2 \sqrt{2}, \in... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.1.
Determine the value of the expression $A=1+\operatorname{tg} 35^{\circ}+\operatorname{tg} 10^{\circ}+\operatorname{tg} 35^{\circ} \cdot \operatorname{tg} 10^{\circ}$. | ## Solution.
From $35^{\circ}+10^{\circ}=45^{\circ}$ it follows:
$$
\begin{aligned}
& 1=\tan 45^{\circ}=\tan\left(35^{\circ}+10^{\circ}\right)=\frac{\tan 35^{\circ}+\tan 10^{\circ}}{1-\tan 35^{\circ} \cdot \tan 10^{\circ}} \\
& \Rightarrow \tan 35^{\circ}+\tan 10^{\circ}=1-\tan 35^{\circ} \cdot \tan 10^{\circ} \\
& \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.1.
Solve the equation
$$
\binom{n}{n-2}+2\binom{n-1}{n-3}=\binom{n+1}{n-1}+\binom{n-2}{n-3}
$$ | ## Solution.
From the symmetry of binomial coefficients, it follows that
$$
\binom{n}{2}+2\binom{n-1}{2}=\binom{n+1}{2}+\binom{n-2}{1}
$$
The solution can be a natural number \( n \geqslant 3 \).
By expanding the binomial coefficients, we get:
$$
\begin{aligned}
& \frac{n(n-1)}{2}+2 \cdot \frac{(n-1)(n-2)}{2}=\fra... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.2.
Determine the natural number $n$, so that the ratio of the seventh term counting from the beginning to the seventh term counting from the end, in the expansion of the binomial $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n}$, is equal to $\frac{1}{6}$. | ## Solution.
In the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n}=\left(2^{\frac{1}{3}}+3^{-\frac{1}{3}}\right)^{n}$, the seventh term from the beginning is
$$
\binom{n}{6}\left(2^{\frac{1}{3}}\right)^{n-6}\left(3^{-\frac{1}{3}}\right)^{6}
$$
and the seventh term from the end is
$$
\binom{n}{6}\l... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.3.
The lengths of the sides of a triangle are three consecutive natural numbers, not less than 3. Calculate the difference in the lengths of the segments that the height to the middle side divides on that side. | ## Solution.
Let the lengths of the sides of the triangle be $|A C|=n-1,|A B|=n$ and $|B C|=n+1$, as shown in the figure.
$\overline{C N}$ is the altitude to the side $\overline{A B}$, the... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.6.
Determine the value of the real parameter $a$ for which each of the equations
$$
2 a-1=\frac{3-3 a}{x-1} \quad \text { and } \quad a^{2}(2 x-4)-1=a(4-5 x)-2 x
$$
has a unique solution and their solutions are equal. | ## Solution.
First, let's solve the equation $2 a-1=\frac{3-3 a}{x-1}$. Note that $x \neq 1$ must hold. After multiplying the equation by $x-1$ and simplifying the resulting algebraic expressions, we get:
$$
\begin{gathered}
2 a x-2 a-x+1=3-3 a \\
x(2 a-1)=2-a \\
x=\frac{2-a}{2 a-1}, a \neq \frac{1}{2}
\end{gathered}... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.2.
Let $A=1202^{2}+2^{2021}$. Determine the units digit of the number $A^{2021}$. | ## Solution.
The unit digit of the square of the number 1202 is 4.
1 point
Powers with a base of 2 have unit digits in the sequence $2,4,8$, 6, depending on the remainder when the exponent is divided by 4.
Therefore, the unit digit of the number $2^{2021}$ is 2.
Then the unit digit of the number $A$ is $4+2=6$.
T... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.1.
Calculate $\left(\frac{6+4 \sqrt{2}}{\sqrt{2}+\sqrt{6+4 \sqrt{2}}}+\frac{6-4 \sqrt{2}}{\sqrt{2}-\sqrt{6-4 \sqrt{2}}}\right)^{2}$. | ## Solution.
Notice that the following holds:
$$
\begin{aligned}
& \sqrt{6+4 \sqrt{2}}=\sqrt{4+4 \sqrt{2}+2}=\sqrt{(2+\sqrt{2})^{2}}=|2+\sqrt{2}|=2+\sqrt{2} \\
& \sqrt{6-4 \sqrt{2}}=\sqrt{4-4 \sqrt{2}+2}=\sqrt{(2-\sqrt{2})^{2}}=|2-\sqrt{2}|=2-\sqrt{2}
\end{aligned}
$$
By substituting the obtained equalities into the... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.6.
Solve the equation
$$
\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+4 x-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)=112-4 \sqrt{x}
$$ | ## First Solution.
First, note that $x>0, x \neq 1$ must hold.
1 point
By simplifying the expression in parentheses, we get
$$
\begin{array}{cc}
\sqrt{x}-\frac{1}{\sqrt{x}}=\frac{x-1}{\sqrt{x}}, & 1 \text { point } \\
\frac{\sqrt{x}+1}{\sqrt{x}-1}+4 x-\frac{\sqrt{x}-1}{\sqrt{x}+1}=4 x+\frac{(\sqrt{x}+1)^{2}-(\sqrt{... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.1.
Let $a>1$. The points $(x, y)$ for which the coordinates satisfy $|x|+y=a, x^{2}+y=a|x|$ determine a figure in the coordinate plane with an area of 120 square units. Calculate the real number $a$. | ## Solution.
From the first equation of the system, we have $y=-|x|+a$, which, when substituted into the second equation, gives
$$
x^{2}-|x|+a=a|x| \text {. }
$$
This leads to a series of equivalent equations
$$
\begin{aligned}
& x^{2}-|x|+a-a|x|=0 \\
\Longleftrightarrow & |x|^{2}-|x|+a-a|x|=0 \\
\Longleftrightarro... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.4.
For real numbers $x, y$ if the equality $|x+y|+|x-y|=2$ holds, determine the maximum value of the expression $x^{2}-6 x+y^{2}$. | ## Solution.
We have the following cases:
1) $x+y \geqslant 0, x-y \geqslant 0$, that is, $y \geqslant-x, y \leqslant x$. Then we have
$$
x+y+x-y=2 \quad \Rightarrow \quad x=1, y \geqslant-1, y \leqslant 1 \quad \Rightarrow \quad x=1,|y| \leqslant 1
$$
In this case, the maximum value of the expression $x^{2}-6 x+y^... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.3.
If $f(x)=4 \sin ^{2} \frac{3 x}{2}-4 \cos ^{2} \frac{3 x}{2}$, determine $f\left(\frac{2020 \pi}{9}+2021 k \pi\right)$ depending on the integer $k$. | ## Solution.
The function $f$ can be written in the following form:
$$
f(x)=4 \sin ^{2} \frac{3 x}{2}-4 \cos ^{2} \frac{3 x}{2}=-4\left(\cos ^{2} \frac{3 x}{2}-\sin ^{2} \frac{3 x}{2}\right)=-4 \cos 3 x
$$
Therefore,
$$
f\left(\frac{2020 \pi}{9}+2021 k \pi\right)=-4 \cos \left(3\left(\frac{2020 \pi}{9}+2021 k \pi\r... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.2.
Students wrote posts for Instagram in pairs to invite people to the "Math Evening." The pair that collects the most likes expects a reward. Maja collected 3731, Matko 2754, but they claim that together they have 6705, which makes other pairs suspicious of cheating.
If we allow the possibility that the ... | ## Solution.
Let $b$ be the base of the number system used by Maja and Marko.
$$
\begin{array}{cc}
3731_{(b)}+2754_{(b)}=6705_{(b)} & 1 \text { point } \\
\left(3 b^{3}+7 b^{2}+3 b+1\right)+\left(2 b^{3}+7 b^{2}+5 b+4\right)=6 b^{3}+7 b^{2}+0 \cdot b+5 & 1 \text { point } \\
b^{3}-7 b^{2}-8 b & \\
b(b-8)(b+1)=0 & 2 \... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.7.
Leda and Una are playing with clay in the shape of a cylinder whose height is 6 times greater than the diameter of its base. Leda took a part of that clay and made a larger ball, while Una made a smaller ball from the remainder. How many times greater is the volume of Leda's ball compared to the volume ... | ## Solution.
Let $r$ be the radius of the base of the cylinder. The height of the cylinder is 6 diameters, or 12 radii of the cylinder, so the volume of the cylinder is $V_{V}=r^{2} \pi v=r^{2} \pi 12 r=12 r^{3} \pi$.
Let $R_{L}$ be the radius of the sphere made by Leda, and $R_{U}$ be the radius of the sphere made b... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.2.
Let $a$ and $b$ be positive real numbers such that
$$
\frac{a}{b}+\frac{b}{a}=3 \quad \text { and } \quad \frac{a^{2}}{b}+\frac{b^{2}}{a}=10
$$
Determine $\frac{1}{a}+\frac{1}{b}$. | ## Solution.
By multiplying each of the equalities by $a b$ we have
$$
a^{2}+b^{2}=3 a b \quad \text { and } \quad a^{3}+b^{3}=10 a b
$$
Notice, $a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$.
Substituting the first equality into the second, we get $10 a b=(a+b)(3 a b-a b)=2 a b(a+b)$, from which dividing by $2 a ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.6.
Borna wants to color each of the numbers $2,3, \ldots, 32$ with one of $k$ colors $(k \in \mathbb{N})$ such that no number is a multiple of another number of the same color. Determine the smallest natural number $k$ for which Borna can achieve this. | ## Solution.
Notice that the numbers $2, 4, 8, 16,$ and 32$ must be colored with different colors according to the condition of the problem.
Therefore, at least 5 colors are needed.
It remains to prove that $k=5$ colors are sufficient. One possible coloring is:
- Color 1: numbers 2, 3
- Color 2: numbers $4, 5, 6, 7... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.5.
How many different bracelets can be made consisting of four black and four white beads arranged in a circle? Two bracelets are considered different if they cannot be rotated to have the same sequence of beads. | ## Solution.
Let's consider how long the longest string consisting only of black balls is.
If it is 4 in length, then the remaining four balls are obviously white, and this is the only possibility in this case.
^{2018}
$$
and $2 \cdot 9^{2017}$ is not divisible by 10, we see that the decimal representation of the given number will end with 2018 zeros.
The last digit different from zero is equal to the last digit of the number 2 . $9^{20... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.7.
Marko places coins on some fields of a $3 \times 3$ board, and then records how many coins are in each individual row and column. How many coins at least must Marko place on the board if he wants these six numbers to be mutually distinct? | ## Solution.
Let $Z$ be the sum of all six numbers (coins in rows and columns). 1 point
If all six numbers are different, the sum of these numbers $Z$ is at least $0+1+2+3+4+5=15. \quad 3$ points
Notice that $Z$ is an even number because it is equal to twice the number of coins on the board.
Due to its evenness, $Z... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.1.
The product of the second and fourth terms of an arithmetic sequence with difference $d$ is $-d^{2}$. Determine the product of the third and fifth terms of this sequence. | ## Solution.
Let's denote the first five terms of the given arithmetic sequence as $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$.
Since the common difference is $d$, we have $a_{2}=a_{1}+d, a_{3}=a_{1}+2 d, a_{4}=a_{1}+3 d, a_{5}=a_{1}+4 d$.
According to the condition of the problem, the product of the second and fourth terms... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.3.
For real numbers $x_{1}, x_{2}, \ldots, x_{30}$, the following holds:
$$
\begin{aligned}
& 20^{3} x_{1}+21^{3} x_{2}+\cdots+49^{3} x_{30}=13 \\
& 21^{3} x_{1}+22^{3} x_{2}+\cdots+50^{3} x_{30}=1 \\
& 22^{3} x_{1}+23^{3} x_{2}+\cdots+51^{3} x_{30}=19
\end{aligned}
$$
What is the value of $21 x_{1}+22 x... | ## First solution.
It is known that for real numbers $a$ and $b$, $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$.
If $a-b=1$, then $a^{3}-b^{3}=a^{2}+a b+b^{2}$. Therefore, by subtracting the first equation from the second, we get:
$\left(21^{2}+21 \cdot 20+20^{2}\right) x_{1}+\left(22^{2}+22 \cdot 21+21^{2}\right)... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.5.
Martin assigns each of the 12 edges of a cube a number, either 1 or -1. Then he assigns to each of the six faces of the cube the product of the 4 numbers on the edges of that face. Finally, Martin sums all 18 numbers assigned to the edges and faces of the cube.
What is the smallest sum Martin can achie... | ## First Solution.
Consider the five numbers associated with one face and the four edges that bound it. Note that their sum can be at least -3.
Indeed, if all the numbers associated with the edges are -1, then their product is 1, so the sum is -3. If at least one of the numbers on the edges is not -1, then the sum is... | -12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.4.
Determine all natural numbers $m$ for which $2^{m}-63$ is the cube of some integer. | ## Solution.
We seek all $m \in \mathbb{N}$ for which $2^{m} - 63 = n^{3}$ for some $n \in \mathbb{Z}$.
Consider the remainders that both sides of this equation give when divided by 7. 1 point
By listing all possible remainders for $n$ when divided by 7, we see that the cube of a natural number $n^{3}$ gives remaind... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.2.
Determine the smallest value that the expression
$$
a^{2}+b^{2}+|a b-1|+|a b-4|
$$
can take for some real numbers $a$ and $b$. | ## Solution.
Let $S:=a^{2}+b^{2}+|a b-1|+|a b-4|$. By considering the sign of the expressions $a b-1$ and $a b-4$, we distinguish three cases.
In the first case, let $a b \geqslant 4$. Then the expressions in both absolute value terms of the initial expression are non-negative, so the expression becomes
$$
S=a^{2}+b... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.4.
Determine the smallest natural number $n$ for which there exist real numbers $x_{1}, \ldots, x_{n} \in [1,4]$ that satisfy the inequalities:
$$
\begin{aligned}
x_{1}+x_{2}+\ldots+x_{n} & \geqslant \frac{7}{3} n \\
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}} & \geqslant \frac{2}{3} n
\end{ali... | ## First Solution.
Assume that for some natural number $n$ there exist real numbers $x_{1}, x_{2}, \ldots, x_{n}$ that satisfy the given conditions. Since $1 \leqslant x_{i} \leqslant 4$, the inequality holds:
$$
\left(4-x_{i}\right)\left(1-\frac{1}{x_{i}}\right) \geqslant 0
$$
where equality holds if and only if $x... | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.2.
Let $S$ be the set of all natural numbers less than 1000 whose all digits in decimal notation are even. Let $\omega$ be a complex number such that $\omega^{2}+\omega+1=0$.
Calculate the sum $\sum_{k \in S} \omega^{k}$, i.e., the sum of the values $\omega^{k}$ for all $k$ in the set $S$. | ## First Solution.
Let $Z$ be the desired sum. Multiplying the equation $\omega^{2}+\omega+1=0$ by $\omega-1$ we get $\omega^{3}=1$. In particular, the number $\omega^{n}$ is equal to the number $1, \omega$ or $\omega^{2}$ respectively if $n$ gives a remainder of 0, 1 or 2 when divided by 3.
Consider any three three-... | -2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task B-1.1. Calculate $\frac{1234321234321 \cdot 2468642468641-1234321234320}{1234321234320 \cdot 2468642468641+1234321234321}$. | Solution. Let $n=1234321234321$.
Then the given expression is equivalent to the expression
$$
\begin{aligned}
& \frac{n \cdot(2 n-1)-(n-1)}{(n-1) \cdot(2 n-1)+n}= \\
& \frac{2 n^{2}-n-n+1}{2 n^{2}-2 n-n+1+n}= \\
& \frac{2 n^{2}-2 n+1}{2 n^{2}-2 n+1}=1
\end{aligned}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Zadatak B-2.1. Izračunajte:
$$
\left[\frac{(1+\sqrt{3})+(1-\sqrt{3}) i}{\sqrt{6}+\sqrt{2} i}\right]^{2012}
$$
| ## Rješenje.
$$
\begin{aligned}
{\left[\frac{(1+\sqrt{3})+(1-\sqrt{3}) i}{\sqrt{6}+\sqrt{2} i}\right]^{2012} } & =\left[\frac{(1+\sqrt{3})+(1-\sqrt{3}) i}{\sqrt{6}+\sqrt{2} i} \cdot \frac{\sqrt{6}-\sqrt{2} i}{\sqrt{6}-\sqrt{2} i}\right]^{2012} \\
& =\left[\frac{4 \sqrt{2}-4 \sqrt{2} i}{6+2}\right]^{2012} \\
& =\left[\... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.3. Calculate the area of the figure determined in the complex plane by the set of all complex numbers $z$ for which $|\operatorname{Re} z| \leq 1,|\operatorname{Im} z| \leq 1$.
翻译完成，但根据您的要求，最后一句是额外的，因此在翻译结果中未包含。如果需要任何进一步的修改或有其他要求，请告知。 | Solution. Let $z=x+y i$. Then, according to the conditions of the problem, $|x| \leq 1$ and $|y| \leq 1$.
Sketch
(2 points)
The side of the square is 2, so its area is 4 square units.
(1... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.4. Let $f(x)=x^{2}+b x+c, b, c \in \mathbb{R}$. If $f(0)+f(1)=\frac{1}{2}$, calculate $f\left(\frac{1}{2}\right)$. | ## Solution.
$$
f(0)=c, \quad f(1)=1+b+c
$$
From $f(0)+f(1)=c+1+b+c=2 c+b+1=\frac{1}{2}$
it follows that $2 c+b=-\frac{1}{2}$,
or $c+\frac{1}{2} b=-\frac{1}{4}$. Then,
$$
f\left(\frac{1}{2}\right)=\frac{1}{4}+\frac{1}{2} b+c=\frac{1}{4}-\frac{1}{4}=0
$$ | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.7. Today is Valentine's Day, and Valentino wants to go out with his girlfriend Lorna. With his 120 kuna, he can buy several roses and two lemonades. The price of one rose is equal to the price of one lemonade. Since they have a math test tomorrow, they will postpone the outing until Saturday. On Saturday, the ... | Solution. Let $x$ be the number of roses, and $y$ be the price of one rose on Valentine's Day. From the conditions of the problem, we have
$$
\begin{aligned}
x y+2 y & =120 \\
x(y-7)+78 & =126
\end{aligned}
$$
We solve the system
$$
\left\{\begin{array}{l}
x y+2 y=120 \\
x y-7 x=48
\end{array}\right.
$$
If we expre... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-3.4. How many isosceles triangles have side lengths that are integers, and a perimeter of $30 \mathrm{~cm}$? | Solution. Let $a$ be the length of the base, and $b$ the length of the legs of the isosceles triangle. From $a+2 b=30$ it follows that $a$ must be an even number, and from the triangle inequality $a<b+b=2 b$.
(2 points)
Therefore, $a \leq 14$ and is an even number, so there are only these 7 cases:
$$
(a, b) \in\{(2,... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-3.6. The area of a rectangle is $12 \mathrm{~cm}^{2}$, and if the acute angle between its diagonals is halved, the area is $7.5 \mathrm{~cm}^{2}$. What is the length of the diagonal of the rectangle? | ## Solution.
If the angle between the diagonals is $\alpha$, the area is
$$
\begin{array}{r}
P=4 \cdot \frac{\frac{d}{2} \frac{d}{2} \sin \alpha}{2}=\frac{1}{2} d^{2} \sin \alpha=12 \\
d^{... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.1. Solve the equation
$$
\frac{1}{(2 n-4)!}=\frac{1}{(2 n-3)!}+\frac{8}{(2 n-2)!}
$$ | Solution. The number $n$ must satisfy the condition $n \geq 2$.
If we multiply the equation
$$
\frac{1}{(2 n-4)!}=\frac{1}{(2 n-3)!}+\frac{8}{(2 n-2)!}
$$
by $(2 n-2)!$,
we get
$$
(2 n-2)(2 n-3)=2 n-2+8
$$
i.e., $n^{2}-3 n=0$.
The solutions to the last equation are $n=0$ and $n=3$.
However, only the solution $n... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.5. Let $z=-\cos \frac{9 \pi}{8}+i \sin \frac{7 \pi}{8}$ be a complex number. Determine the smallest natural number $n$ such that the real part of the number $z^{n}$ is 0. | Solution. Let's write the number $z=-\cos \frac{9 \pi}{8}+i \sin \frac{7 \pi}{8}$ in trigonometric form:
$$
z=\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}
$$
Then $z^{n}=\cos \frac{n \pi}{8}+i \sin \frac{n \pi}{8}$.
Equate the real part to zero: $\cos \frac{n \pi}{8}=0$.
We get $\frac{n \pi}{8}=\frac{\pi}{2}+k \pi, k \i... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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