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## Task A-1.2.
Let $a$ and $b$ be positive real numbers such that $a^{3}+b^{3}=2 a b(a+b)$. Determine $\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}$. | ## Solution.
By factoring the left side, we get $(a+b)\left(a^{2}-a b+b^{2}\right)=2 a b(a+b)$.
Since the numbers $a$ and $b$ are positive, it follows that $a+b \neq 0$, so we can divide the obtained equation by $a+b$.
Therefore, it follows that $a^{2}-a b+b^{2}=2 a b$, or $a^{2}+b^{2}=3 a b$.
By dividing by $a^{2}... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.7.
In a basketball tournament, each team plays exactly twice against each of the other teams. A win brings 2 points, a loss 0 points, and there are no draws. Determine all natural numbers $n$ for which there exists a basketball tournament with $n$ teams where one team, the tournament winner, has 26 points,... | ## Solution.
Let the total number of teams be denoted by $n$. Let $a$ be the number of teams that scored 24 points, and $b$ be the number of teams that scored 22 points. Then the total number of teams is
$$
n=3+a+b
$$
and the total number of points in all matches is
$$
2 n(n-1)=26+24 a+22 b+40
$$
Substituting $b=n... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.5.
Determine the smallest natural number $n$ such that in every set consisting of $n$ integers, there exist three distinct elements $a, b$, and $c$ such that $a b + b c + c a$ is divisible by 3. | ## Solution.
Consider the maximum number of elements a set can have in which there do not exist three distinct elements $a, b$, and $c$ such that $a b + b c + c a$ is divisible by 3. Let $S$ be such a set with the maximum number of elements.
If the set $S$ contained two numbers that are divisible by 3, let's call the... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.6.
Let $M$ and $N$ be the feet of the altitudes from vertices $A$ and $B$ of an acute-angled triangle $ABC$. Let $Q$ be the midpoint of segment $\overline{M N}$, and $P$ be the midpoint of side $\overline{A B}$. If $|M N|=10$ and $|A B|=26$, determine the length of $|P Q|$. | ## Solution.
Since $\varangle B M A=\varangle B N A=90^{\circ}$, quadrilateral $A B M N$ is cyclic.
The center of the circle circumscribed around quadrilateral $A B C D$ is point $P$.
Segments $\overline{P N}$ and $\overline{P M}$ are radii of this circle, so $|P N|=|P M|=|A P|=13$.
 the period of three digits 347 repeats.
It repeats 668 times, because
$$
2005=668 \cdot 3+1
$$
The number at the 2006th place is the first digit ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task B-1.4. A cattleman has secured food for his 24 cows for 18 weeks. How many cows would he have to sell after 8 weeks so that he would have enough food for another 12 weeks? | ## First solution.
Let $x$ be the number of cows the farmer needs to sell.
The total amount of food is $24 \cdot 18$ weekly required amounts and it does not change, but is only reallocated to the amount needed for 24 cows for 8 weeks and $24-x$ cows for 12 weeks.
Then we have
$$
24 \cdot 18 = 24 \cdot 8 + (24 - x) ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.3. Without solving the quadratic equation $x^{2}+2 x+2=0$, calculate the value of the expression $\frac{x_{1}^{3}+x_{2}^{3}}{\left(x_{1}-x_{2}\right)^{2}}$, where $x_{1}, x_{2}$ are the solutions of the given equation. | ## Solution.
$$
\begin{gathered}
x_{1}+x_{2}=-\frac{b}{a}=-2, \quad x_{1} \cdot x_{2}=\frac{c}{a}=2 \\
\frac{x_{1}^{3}+x_{2}^{3}}{\left(x_{1}-x_{2}\right)^{2}}=\frac{\left(x_{1}+x_{2}\right)^{3}-3 x_{1} x_{2}\left(x_{1}+x_{2}\right)}{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}= \\
=\frac{(-2)^{3}-3 \cdot 2 \cdot(-2)}{... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-3.2. The axial section of a right circular cylinder is a square. Determine the length of the radius of the base so that its surface area and volume have the same numerical value. | ## Solution.
From the condition that the axial section of the cylinder is a square, it follows that $2 r=h$.
Then from $O=V$ we get
$$
\begin{aligned}
2 r^{2} \pi+2 r \pi \cdot h & =r^{2} \pi \cdot h \\
2 r^{2} \pi+2 r \pi \cdot 2 r & =r^{2} \pi \cdot 2 r \\
6 r^{2} \pi & =2 r^{3} \pi
\end{aligned}
$$
From the last... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.1. Determine the natural number $n \geq 2$ such that the equality
$$
\frac{(n-1)^{2} n(n+1)!}{(n+2)!}=\binom{n}{2}
$$
holds. | $$
\begin{aligned}
\frac{(n-1)(n-1) n(n+1)!}{(n+2)!} & =\frac{n!}{2!(n-2)!} \\
\frac{(n-1)(n-1) n(n+1)!}{(n+2)(n+1)!} & =\frac{(n-2)!(n-1) n}{2(n-2)!} \\
\frac{n-1}{n+2} & =\frac{1}{2} \\
2 n-2 & =n+2 \\
n & =4
\end{aligned}
$$ | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.5. In a regular $n$-gon, the radius of the circumscribed circle is $2, S$ is its center, $A, B$ are consecutive vertices of the $n$-gon. Determine $n$ and the interior angle of the regular $n$-gon if the dot product $\overrightarrow{S A} \cdot \overrightarrow{S B}=2 \sqrt{2}$. | Solution. Let $\alpha$ be the central angle of a regular $n$-gon.
From $\overrightarrow{S A} \cdot \overrightarrow{S B}=2 \sqrt{2}$ it follows that
$$
|S A| \cdot|S B| \cos \alpha=2 \sqrt{2}
$$
which means $2 \cdot 2 \cdot \cos \alpha=2 \sqrt{2}$, so
$$
\cos \alpha=\frac{\sqrt{2}}{2}
$$
thus $\alpha=45^{\circ}$.
... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.3.
Each digit of the natural number $n$ (except the first) is strictly greater than the digit immediately to its left. Determine the sum of all digits of the number $9 n$. | ## First solution.
Let $n=\overline{a_{m-1} a_{m-2} \ldots a_{1} a_{0}}$. It holds that $a_{0}>a_{1}>\ldots>a_{m-1}$. Now we have
$$
\begin{aligned}
9 n & =(10-1)\left(10^{m-1} a_{m-1}+10^{m-2} a_{m-2}+\ldots+10 a_{1}+a_{0}\right) \\
& =(10-1) 10^{m-1} a_{m-1}+(10-1) 10^{m-2} a_{m-2}+\ldots+(10-1) 10 a_{1}+(10-1) a_{... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task A-2.3. (10 points)
Determine all natural numbers $n$ for which
$$
5^{n}+2^{n+1} 3^{n}=9^{n}+4^{n}
$$ | ## Solution.
The given equation can be written in the form
$$
\begin{gathered}
5^{n}=3^{2 n}-2 \cdot 3^{n} \cdot 2^{n}+2^{2 n} \\
5^{n}=\left(3^{n}-2^{n}\right)^{2}
\end{gathered}
$$
The right-hand side is a perfect square, so the left-hand side must also be a perfect square.
Therefore, $n$ is an even number, $n=2 ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.3.
Let $a$ and $b$ be positive real numbers such that $a^{2}+b^{2}=8$ and $a^{6}+b^{6}=416$.
Determine $a b$. | ## Solution.
According to the formula for the cube of a binomial, we get:
$$
\begin{aligned}
\left(a^{2}+b^{2}\right)^{3} & =a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6} & & 1 \text { point } \\
& =a^{6}+b^{6}+3 a^{2} b^{2}\left(a^{2}+b^{2}\right) . & & 2 \text { points }
\end{aligned}
$$
By substituting the data from th... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.4.
On a board, the numbers $1,2,3, \ldots, 2021$ are written. Is it possible to erase the numbers one by one until only one number remains, such that after each erasure, the sum of all remaining numbers is a composite number? | ## Solution.
It is possible.
Notice that the sum of the first $n$ natural numbers $1+2+\cdots+n=\frac{n(n+1)}{2}$ is a composite number for all $n \geqslant 3$.
We delete the numbers from the largest to the smallest. First, we will delete the largest number 2021, and then continue deleting all the way down to the nu... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.2.
A 2021-digit number is written. Each two-digit number formed by two consecutive digits of this number (without changing the order) is divisible by 17 or 23. The unit digit of this number is 7. What is its first digit? | ## Solution.
Two-digit multiples of 17 are 17, 34, 51, 68, and 85. Two-digit multiples of 23 are 23, 46, 69, and 92.
If the digit 7 is the last one, the digit before it must be 1. Before the 1 is 5, before the 5 is 8, and before the 8 is necessarily 6. Then, these digits must be preceded by 4, then 3, then 2, then 9,... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.5.
Determine all pairs $\{a, b\}$ of distinct real numbers such that the equations
$$
x^{2} + a x + b = 0 \quad \text{and} \quad x^{2} + b x + a = 0
$$
have at least one common solution in the set of real numbers. | ## First Solution.
Let the common solution be $\mathrm{s} x_{0}$. By equating the equations, we get:
$$
\begin{aligned}
x_{0}^{2}+a x_{0}+b & =x_{0}^{2}+b x_{0}+a \\
x_{0} a-x_{0} b+b-a & =0 \\
x_{0}(a-b)-(a-b) & =0 \\
\left(x_{0}-1\right)(a-b) & =0
\end{aligned}
$$
From the condition $a \neq b$, we get $x_{0}=1$.
... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.2.
For positive real numbers $x, y$ and $z$, if
$$
4^{x+y}=z, \quad z^{1 / x} \cdot z^{1 / y}=1024
$$
determine the value of the expression $\frac{x}{y}+\frac{y}{x}$. | ## Solution.
From the second given equation, we have $z^{\frac{1}{x}+\frac{1}{y}}=1024$.
By raising the first given equation to the power of $\frac{1}{x}+\frac{1}{y}$, we get
$$
\begin{aligned}
4^{(x+y)\left(\frac{1}{x}+\frac{1}{y}\right)} & =z^{\frac{1}{x}+\frac{1}{y}}=1024=4^{5}, & & 2 \text { points } \\
(x+y)\le... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.1.
If $x, y, z$ and $w$ are real numbers such that
$$
x^{2}+y^{2}+z^{2}+w^{2}+x+3 y+5 z+7 w=4
$$
determine the maximum possible value of the expression $x+y+z+w$. | ## Solution.
By completing the square, the given equation can be written in the form
$$
\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{3}{2}\right)^{2}+\left(z+\frac{5}{2}\right)^{2}+\left(w+\frac{7}{2}\right)^{2}=25
$$
By the inequality between the arithmetic and quadratic means, we have
$$
\frac{\left(x+\frac{1}{2}... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.2.
Inside the triangle $A B C$ are points $S$ and $T$. The distances from point $S$ to the lines $A B, B C$, and $C A$ are 10, 7, and 4, respectively. The distances from point $T$ to these lines are 4, 10, and 16, respectively.
Determine the radius of the inscribed circle of triangle $A B C$. | ## First Solution.
Let $a=|B C|, b=|C A|, c=|A B|$. Let $P$ be the area, $s=\frac{1}{2}(a+b+c)$ the semiperimeter, and $r$ the radius of the inscribed circle of triangle $A B C$.
Divide triangle $A B C$ into three smaller triangles by connecting point $S$ to the vertices $A, B$, and $C$.
 of the red ... | ## Solution.
We will prove that the smallest number of intersections Marko can achieve is 2.
Let $A_{1}, A_{2}, \ldots, A_{2022}$ denote the vertices of the given polygon in order.
First, we provide an example of drawing red diagonals such that exactly two intersections are achieved. Draw the diagonals $\overline{A_... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.4.
In the plane of square $A B C D$, but outside it, there is a point $P$. If
$$
|P A|=\sqrt{5}, \quad|P B|=\sqrt{26} \quad \text { and } \quad|P D|=\sqrt{20}
$$
determine the length of the side of the square. | ## Solution.
Place the square in the coordinate system so that its vertices are $A(0,0), B(a, 0), C(a, a)$, and $D(0, a)$, and let $P(-x,-y)$.
The formula for the distance between any two ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.5.
Given is a board of dimensions $2020 \times 2022$. Two fields of this board are said to be adjacent if they share a common side or if they are at the beginning and end of the same row or column. Thus, each field has exactly four adjacent fields.
Viktor, in each step, chooses one field of the board and ... | ## Solution.
We will say that a move has been made on some field if Viktor has chosen that field on the board, and placed a token on that field and the four neighboring fields.
We will prove that the smallest possible $d$ is 5. This number of tokens can be achieved by Viktor making a move on each field exactly once. ... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak A-3.1.
Kolika je najmanja, a kolika najveća vrijednost koju postiže funkcija $f: \mathbb{R} \rightarrow \mathbb{R}$ zadana formulom $f(x)=\cos ^{2} x+\sin x$ ?
| ## Prvo rješenje.
Uvrštavanjem $\cos ^{2} x+\sin ^{2} x=1$ imamo $f(x)=-\sin ^{2} x+\sin x+1$.
Uz supstituciju $t=\sin x$ dobivamo funkciju $g(t)=-t^{2}+t+1$ za $t \in[-1,1]$ koja ima istu najmanju i najveću vrijednost kao i funkcija $f$.
1 bod
Graf ove funkcije je kvadratna parabola s tjemenom u točki $t=\frac{1}{... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.3. In a certain city, there are 10 restaurants and $n$ theaters. A group of tourists spent several days in this city. During their stay, they visited restaurants and theaters. At the end of their stay, they found that each restaurant had been visited by 4 tourists, and each theater had 6 tourists. If each tour... | ## Solution.
Let $t$ be the total number of tourists. The number of visits to restaurants is
$$
t \cdot 5 = 4 \cdot 10
$$
Then $t = 8$.
Similarly, the number of visits to theaters is
$$
t \cdot 3 = 6 \cdot n
$$
It follows that $t = 2n$, or $n = 4$, which means there are exactly 4 theaters in the city. | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.4. In a kite where the lengths of the diagonals are $d_{1}=24 \mathrm{~cm}$ and $d_{2}=8 \mathrm{~cm}$, a rectangle is inscribed such that its sides are parallel to the diagonals of the kite. Determine the dimensions of the inscribed rectangle that has the maximum area. | ## Solution.
Let $x, y$ be the lengths of the sides of the rectangle. Since $y \| d_{1}$ and $x \| d_{2}$, from the similarity of triangles we have
$$
\begin{aligned}
d_{1}: y & =\frac{d_{2... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.2. In one group, every fourth person out of a total of 112 students is a girl. In the second group, there are a total of 53 students, and the ratio of boys to girls is 2:1. In both groups, the number of girls is the same. How is this possible and how many girls are there in each group? | ## Solution.
Determine the base of the system in which the task is possible.
$$
\begin{aligned}
\frac{1}{4} 112_{(b)} & =\frac{1}{3} 53_{(b)} \\
3\left(b^{2}+b+2\right) & =4(5 b+3) \\
3 b^{2}-17 b & -6=0 \\
b & =6
\end{aligned}
$$
The task is possible in a system with base 6.
Let the number of girls be $x$. Then,
... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-1.5. (20 points)
The difference of two odd numbers is divisible by 5. What is the last digit of the difference of the cubes of these numbers? | Solution. Let's denote these numbers as $a=2m-1, b=2n-1$. Their difference is
$$
a-b=2(m-n)
$$
(5 points)
Since the difference is divisible by 5 and by 2, it is divisible by 10, i.e., $a-b=10k$.
(5 points)
By factoring the difference of cubes $a^3-b^3=(a-b)(a^2+ab+b^2)$, we get $a^3-b^3=10k(a^2+ab+b^2)$, so $a^3-b... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.1. (20 points) Solve the system of equations (in the set of natural numbers)
$$
\left\{\begin{array}{l}
\binom{x}{y}=\binom{x}{y+2} \\
\binom{x}{2}=153
\end{array}\right.
$$ | Solution. From $\binom{x}{2}=153$ we get $\frac{x(x-1)}{2}=153$ i.e. $x^{2}-x-306=0$. The solutions to this equation are $x_{1}=18, x_{2}=-17$. Since we are only looking for solutions in the set of natural numbers, the only possible solution is $x_{1}=18$.
(10 points)
First method. Now from the first equation we have... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.1.
Determine the non-negative real number $a$ such that the value of the expression
$$
a^{3}-a^{2}-2 \sqrt{a}
$$
is as small as possible. | ## First Solution.
Notice that for $a \geqslant 0$ the given expression is defined.
For $a=1$, the value of the expression is -2. We will prove that this is the minimum possible value.
We have
$$
\begin{aligned}
a^{3}-a^{2}-2 \sqrt{a} & =(a-2 \sqrt{a}+1)+a^{3}-a^{2}-a-1 \\
& =(\sqrt{a}-1)^{2}+a^{2}(a-1)-(a-1)-2 \\
... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.2.
Determine all prime numbers $p$ for which there exist natural numbers $x$ and $y$ such that
$$
\left\{\begin{aligned}
p+1 & =2 x^{2} \\
p^{2}+1 & =2 y^{2}
\end{aligned}\right.
$$ | ## Solution.
We directly verify that $p=2$ is not a solution. It is clear that $y>x$.
By subtracting the given equations, we obtain $p(p-1)=2(y-x)(y+x)$.
From this, we conclude that
$$
p \mid y+x
$$
since otherwise $p$ would be a divisor of the number $y-x$, and $p-1$ would be a multiple of the number $y+x$, which... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.3.
Let $a, b, c$ and $d$ be real numbers different from zero such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=0$ and $a+b+c+d+a b c d=0$. Calculate $\left(\frac{1}{a b}-\frac{1}{c d}\right)\left(\frac{1}{c}+\frac{1}{d}\right)$. | ## First Solution.
The desired expression is equal to
$$
\left(\frac{1}{a b}-\frac{1}{c d}\right)\left(\frac{1}{c}+\frac{1}{d}\right)=\frac{c d-a b}{a b c d} \cdot \frac{d+c}{c d}=\frac{c d^{2}+c^{2} d-a b d-a b c}{a b c^{2} d^{2}}
$$
From $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=0$ we get $b c d+a c d+a b d... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.1.
For complex numbers $z$ and $w$, it is given that $|z+w|=\sqrt{3}$ and $|z|=|w|=1$. Calculate $|z-w|$. | ## First Solution.
Apply the property of the modulus of a complex number $|z|^{2}=z \cdot \bar{z}$ to the number $z+w$:
$$
|z+w|^{2}=(z+w) \cdot \overline{(z+w)}=(z+w)(\bar{z}+\bar{w})=z \cdot \bar{z}+w \cdot \bar{z}+z \cdot \bar{w}+w \cdot \bar{w}. \quad 1 \text { point }
$$
We have $z \cdot \bar{z}=|z|^{2}=1, w \c... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-1.1. How many zeros does the number $N=x^{3}+3 x^{2}-9 x-27$ end with, if $x=999997$? | ## First solution.
We can write the expression in the form
$$
N=x^{3}+3 x^{2}-9 x-27=x^{2}(x+3)-9(x+3)=(x+3)\left(x^{2}-9\right)=(x+3)^{2}(x-3) \text { (3 points) }
$$
Then for $x=999997$ we get
$$
N=(999997+3)^{2}(999997-3)=10^{12} \cdot 999994
$$
which means that the number $\mathrm{N}$ ends with $\mathrm{12}$ z... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.3. Andrea and Mirela are preparing for a math competition. When their teacher asked them how many problems they solved yesterday, they did not answer directly. He only found out that Andrea solved fewer problems than Mirela, and each of them solved at least one problem. They also said that the product of the n... | ## Solution.
Let $x$ be the number of problems solved by Andrea, and $y$ be the number of problems solved by Mirela, where $x<y$. Then the problem reduces to solving the equation
$$
x y + x + y = 59
$$
which can be rewritten as
$$
x y + x + y + 1 = 60
$$
or
$$
(x+1)(y+1) = 60 \text{.}
$$
The prime factorization ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.5. If $x_{1}$ and $x_{2}$ are the solutions of the equation $x^{2}+2013 x+1=0$, and $y_{1}$ and $y_{2}$ are the solutions of the equation $x^{2}+2014 x+1=0$, calculate the value of the expression $\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)\left(x_{1}-y_{2}\right)\left(x_{2}-y_{1}\right)$. | ## Solution.
According to Vieta's formulas, we have
$$
\begin{aligned}
& x_{1}+x_{2}=-2013, x_{1} \cdot x_{2}=1 \\
& y_{1}+y_{2}=-2014, y_{1} \cdot y_{2}=1
\end{aligned}
$$
In the given expression, we will first multiply the first and third, and the second and fourth parentheses, and then apply the previous equaliti... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.7. Determine the value of the parameter $m \in \mathbb{Z}$ for which the quadratic equation
$$
(m-1) x^{2}+(2 m-1) x+3=0
$$
has rational solutions. | ## First Solution.
For the solutions of a quadratic equation to be rational, its discriminant must be the square of some integer. Let this be $k^{2}$.
$$
\begin{aligned}
& D=(2 m-1)^{2}-4(m-1) \cdot 3=4 m^{2}-16 m+13 \\
& 4 m^{2}-16 m+13=k^{2}
\end{aligned}
$$
Let's solve the obtained quadratic equation for $m$.
$$... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.2. Determine the number of sides of a polygon for which the measure of the smallest angle is $132^{\circ}$, and each subsequent angle is $2^{\circ}$ larger. | ## Solution.
The measures of the angles form an arithmetic sequence with $a_{1}=132^{\circ}, d=2$.
The sum of $n$ terms of the arithmetic sequence is $S=\frac{n}{2}\left(2 a_{1}+(n-1) d\right)=n(131+n)$.
The sum of all angles in a polygon is $S=(n-2) \cdot 180$.
Then from $n(131+n)=(n-2) \cdot 180$, it follows that... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak B-1.2.
Izračunajte
$$
\frac{1-\frac{1}{x^{2}}}{1+\frac{6}{x}+\frac{5}{x^{2}}} \cdot \frac{x+5}{x-1}
$$
|
Rješenje.
$$
\begin{aligned}
\frac{1-\frac{1}{x^{2}}}{1+\frac{6}{x}+\frac{5}{x^{2}}} \cdot \frac{x+5}{x-1} & =\frac{\frac{x^{2}-1}{x^{2}}}{\frac{x^{2}+6 x+5}{x^{2}}} \cdot \frac{x+5}{x-1}= \\
& =\frac{(x-1) \cdot(x+1)}{(x+1) \cdot(x+5)} \cdot \frac{x+5}{x-1}=1
\end{aligned}
$$
$x \neq 1,0,-1,-5$
| 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.4.
The orthocenter of an isosceles triangle is located at one of the triangle's vertices. If the lengths of the legs are $3 \sqrt{2} \mathrm{~cm}$, what is the length of the radius of the circumscribed circle around this triangle? | ## Solution.
The orthocenter is the vertex of the triangle if and only if the triangle is a right triangle. We conclude that the given triangle is a right triangle.
The radius of the circumscribed circle of a right triangle is half the hypotenuse.
$$
r=\frac{c}{2}=\frac{1}{2} \sqrt{2 \cdot(3 \sqrt{2})^{2}}=3
$$ | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.5.
What is the last digit of the sum of all positive divisors of the number $105$? | ## Solution.
$$
105=3 \cdot 5 \cdot 7
$$
The divisors of 105 are 1, 3, 5, 7, 15, 21, 35, 105.
The sum of the divisors is 192. The last digit of the sum is 2. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.4.
If $z$ is a solution to the equation $z^{2}-z+1=0$, what is $z^{2010}-z^{1005}+1$? | ## First solution.
The solution to the given equation is not $z=-1$, so we can multiply that equation by $z+1$. We will get:
$$
\begin{aligned}
(z+1)\left(z^{2}-z+1\right) & =0 \\
z^{3}+1 & =0 \\
z^{3} & =-1
\end{aligned}
$$
Then $z^{1005}=\left(z^{3}\right)^{335}=-1$,
and $z^{2010}=\left(z^{1005}\right)^{2}=1$
so... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak B-4.7.
Riješite jednadžbu
$$
\binom{x}{x-3}+4 \cdot\binom{x+1}{x-2}+\binom{x+2}{3}=125
$$
| ## Rješenje.
Primjenimo li svojstvo simetričnosti binomnih koeficijenata, jednadžbu pišemo u obliku
$$
\begin{gathered}
\binom{x}{3}+4 \cdot\binom{x+1}{3}+\binom{x+2}{3}=125 . \\
\frac{x(x-1)(x-2)}{6}+4 \frac{(x+1) x(x-1)}{6}+\frac{(x+2)(x+1) x}{6}=125
\end{gathered}
$$
Nakon provedenog moženja imamo:
$$
\begin{ali... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.3. (8 points)
Determine the largest integer $n$ for which the inequality $3\left(n-\frac{5}{3}\right)-2(4 n+1)>6 n+5$ holds. | ## Solution.
The inequality $3\left(n-\frac{5}{3}\right)-2(4 n+1)>6 n+5$ is successively equivalent to
$$
\begin{gathered}
3 n-5-8 n-2>6 n+5 \\
-11 n>12 \\
n<-\frac{12}{11}
\end{gathered}
$$
The largest integer that satisfies this inequality is $n=-2$.
For 5 points: at least (*)
For 8 points: correct result and pr... | -2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.8. (20 points)
Determine $a>0$ such that the area of the figure bounded by the graphs of the functions
$$
y=|a x-3|+|a x+3| \quad \text{and} \quad y=10
$$
is equal to 8. | ## Solution.
The function $y=|a x-3|+|a x+3|$ can be written piecewise without absolute values:
$$
y=\left\{\begin{array}{cl}
-2 a x, & x \in\left(-\infty,-\frac{3}{a}\right) \\
6, & x \in\left[-\frac{3}{a}, \frac{3}{a}\right] \\
2 a x, & x \in\left(\frac{3}{a}, \infty\right)
\end{array}\right.
$$
Let's sketch the g... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task A-3.6. (20 points)
Let $\alpha, \beta$ and $\gamma$ be angles such that $\beta=60^{\circ}+\alpha$ and $\gamma=60^{\circ}+\beta$. Prove that the value of the expression
$$
\operatorname{tg} \alpha \operatorname{tg} \beta+\operatorname{tg} \beta \operatorname{tg} \gamma+\operatorname{tg} \gamma \operatorname{tg} \... | ## Solution.
Given $\beta=60^{\circ}+\alpha, \gamma=120^{\circ}+\alpha$, we have
$$
\begin{aligned}
& \tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha = \\
& = \tan \alpha \tan(60^{\circ}+\alpha) + \tan(60^{\circ}+\alpha) \tan(120^{\circ}+\alpha) + \tan(120^{\circ}+\alpha) \tan \alpha = (*)
\... | -3 | Algebra | proof | Yes | Yes | olympiads | false |
Task A-4.2. (8 points)
The third term in the expansion of $\left(2 \cdot \sqrt[n]{2^{-1}}+\frac{4}{\sqrt[4-n]{4}}\right)^{6}$ is 240. Determine $n$. | ## Solution.
The third term in the expansion of the given binomial is $\binom{6}{2} \cdot\left(2^{\frac{n-1}{n}}\right)^{4} \cdot\left(4^{\frac{3-n}{4-n}}\right)^{2}$.
By equating it to 240, we get $15 \cdot 16^{\frac{n-1}{n}} \cdot 16^{\frac{3-n}{4-n}}=240$, which simplifies to $16^{\frac{n-1}{n}+\frac{3-n}{4-n}}=16... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the sides of a square with length 1, isosceles trapezoids are constructed outward such that the vertices of all trapezoids are also vertices of a regular dodecagon. What is the perimeter of this dodecagon?
## City competition:2606, I. Faured,-B wkarijentan-: solutions to problems
Each correctly solved problem i... | Solution.
$$
\frac{469}{1998}=469: 1998=0.2347347 \ldots
$$
Note. The use of a calculator is not allowed!
We observe that (after the digit 2 in the first decimal place) the period of three digits 347 repeats.
(2 points)
It repeats 668 times, because
$$
2005=668 \cdot 3+1
$$
The 2006th place is occupied by the fi... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.3.
Determine all prime numbers $p$ for which $2^{p}+p^{2}$ is also a prime number. | ## Solution.
For $p=2, 2^{p}+p^{2}=8$.
(1 point)
(1 point)
For every odd prime number $p$, the number $2^{p}$ gives a remainder of 2 when divided by 3.
(3 points)
$\left(2^{p} \equiv(3-1)^{p} \equiv(-1)^{p} \equiv-1 \equiv 2(\bmod 3)\right)$
For every prime $p>3$, its square $p^{2}$ gives a remainder of 1 when d... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.5.
If a natural number $n$ when divided by 5 gives a remainder of 2, what remainder does $n^{7}$ give when divided by 5? | ## First solution.
Since $n$ when divided by 5 gives a remainder of 2, it follows that:
$n$ ends in the digit 2 or 7; 1 point
$n^{2}$ ends in the digit 4 or 9; 1 point
$n^{3}$ ends in the digit 8 or 3; 1 point
$n^{4}$ ends in the digit 6 or 1; 1 point
$n^{7}=n^{3} \cdot n^{4}$ ends in the digit 8 or 3.
From the ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## 2nd grade - high school - A category, Primošten, April 7, 2008.
## Solutions
Problem 1. Find all real solutions to the equation
$$
\sqrt{2 x^{2}+3 x+5}+\sqrt{2 x^{2}-3 x+5}=3 x
$$ | Solution. By squaring and rearranging, we get:
$$
\begin{gathered}
2 x^{2}+3 x+5+2 \sqrt{\left(2 x^{2}+5\right)^{2}-(3 x)^{2}}+2 x^{2}-3 x+5=9 x^{2} \\
2 \sqrt{4 x^{4}+11 x^{2}+25}=5 x^{2}-10 \\
16 x^{4}+44 x^{2}+100=25 x^{4}-100 x^{2}+100 \\
144 x^{2}=9 x^{4} \\
9 x^{2}\left(x^{2}-16\right)=0
\end{gathered}
$$
The s... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## 3rd grade - high school - A category, Primošten, April 7, 2008.
## Solutions
Problem 1. The lengths of the sides of a triangle are three consecutive natural numbers, and one of the angles of the triangle is twice as large as one of the remaining two angles. Determine the lengths of the sides of the triangle. | Solution. Let $a=n-1, b=n, c=n+1$. Then $\alpha<\beta<\gamma$. There are three possible cases.
$1^{\circ} \beta=2 \alpha$
Using the sine and cosine rules, we get:
$$
\begin{gathered}
\cos \alpha=\frac{\sin 2 \alpha}{2 \sin \alpha}=\frac{\sin \beta}{2 \sin \alpha}=\frac{b}{2 a}=\frac{n}{2(n-1)} \\
\cos \alpha=\frac{b... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-3.4. Determine all values of the positive real number $k$ for which the equation
$$
\frac{\log (k x)}{\log (x+1)}=2
$$
has exactly one solution. | ## Solution.
First, let's write down the conditions for which the given equation makes sense.
1. $x+1>0, x>-1$;
2. $k x>0$, and since $k>0$ it follows that $x>0$;
3. $\log (x+1) \neq 0, x \neq 0$.
From conditions (1), (2), and (3), we get $x>0$.
(1 point)
The given equation can be written in the form
$$
\begin{al... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.2.
Solve the equation: $\frac{1}{\sqrt[3]{x+1}+\sqrt[3]{2 x-1}}-\frac{1}{\sqrt[3]{12 x}}=0$. | ## Solution.
The conditions for the equation to have a solution are as follows: $\sqrt[3]{x+1}+\sqrt[3]{2 x-1} \neq 0$ and $12 x \neq 0$, i.e., $x \neq 0$. From the first condition, $\sqrt[3]{x+1} \neq -\sqrt[3]{2 x-1}$, and by cubing, we get $x+1 \neq -2 x+1$, hence $x \neq 0$. If we multiply the given equation by th... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.1.
Calculate
$$
\frac{20182019^{2}-20182018^{2}}{20182018 \cdot 20182020-20182017 \cdot 20182019}
$$ | ## Solution.
Let's introduce the substitution $x=20182018$.
Then we have
$$
\begin{array}{cc}
\frac{(x+1)^{2}-x^{2}}{x \cdot(x+2)-(x-1)(x+1)} & 2 \text { points } \\
=\frac{2 x+1}{x^{2}+2 x-x^{2}+1} & 2 \text { points } \\
=\frac{2 x+1}{2 x+1}=1 . & 1 \text { point }
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.3.
Solve the equation
$$
\frac{2 x}{x-1}-\frac{x}{x+2}-\frac{5 x}{(x-1)(x+2)}=\frac{4}{x^{2}+x-2}-\frac{2}{x-1}
$$ | ## Solution.
Let's write the trinomial $x^{2}+x-2$ in the form of a product:
$$
x^{2}+x-2=(x+2)(x-1)
$$
Then the given equation becomes
$$
\frac{2 x(x+2)-x(x-1)-5 x}{(x-1)(x+2)}=\frac{4-2(x+2)}{(x-1)(x+2)}, \quad x \neq 1, x \neq-2
$$
Then we have
$$
\begin{array}{cc}
2 x^{2}+4 x-x^{2}+x-5 x=4-2 x-4 & 1 \text { p... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.7.
From point $D$ on side $\overline{A B}$, where $|A D|>|B D|$, of an equilateral triangle $A B C$, a perpendicular is drawn to side $\overline{B C}$ with its foot at $E$. Then, from $E$, a perpendicular is drawn to $\overline{C A}$ with its foot at $F$, and from $F$, a perpendicular is drawn to $\overlin... | ## Solution.
Each of the triangles $D B E, E C F$, and $F A G$ is half of an equilateral triangle.
Therefore, we have
$$
\begin{gathered}
|B E|=\frac{1}{2}|B D|=\frac{x}{2} \quad \text { ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.3.
One year, January 1 and April 1 were both on a Thursday. How many months in that year have five Fridays? Justify your answer. | ## First solution.
From January 1 to April 1, there are 90 or 91 days. Since both dates fell on the same day, the number of days between them must be divisible by 7.
Therefore, there are 91 days between them, indicating a leap year.
A leap year has 52 weeks and 2 days. January 1 was a Thursday, which means January 2... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak B-2.4.
Neka su $a$ i $b$ rješenja jednadžbe $z^{2}+z+1=0$. Izračunajte $a^{2018}+b^{2018}$.
| ## Prvo rješenje.
Ako su $a$ i $b$ rješenja jednadžbe $z^{2}+z+1=0$, prema Vieteovim formulama je
$$
a+b=-1, \quad a b=1
$$
Kako su $a$ i $b$ rješenja dane jednadžbe (uočimo da su različita od 1 ), vrijedi
$$
\begin{aligned}
z^{2}+z+1 & =0 / \cdot(z-1) \\
z^{3}-1 & =0 \quad \Longrightarrow \quad a^{3}=1, b^{3}=1
\e... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.6.
At a height of $x$ meters above the stage, a spotlight is directed vertically down onto the stage, illuminating a circle of area $P \mathrm{~m}^{2}$. If we raise the spotlight by 1 meter, the illuminated area would increase by $2.25 \mathrm{~m}^{2}$. If we lower the spotlight by 1 meter, the illuminated... | ## First Solution.
A beam of light that emerges from a spotlight and illuminates a circle on the stage defines a cone.
The bases of these cones are concentric circles, and the vertices (de... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.2.
Determine the maximum value of the function
$$
\begin{aligned}
f(x)= & \left(\sin \frac{\pi}{3}+\sin \frac{2 \pi}{3}+\cdots+\sin \frac{k \pi}{3}+\cdots+\sin \frac{2018 \pi}{3}\right)^{2} \cdot \cos x \\
& +\left(\cos \frac{\pi}{3}+\cos \frac{2 \pi}{3}+\cdots+\cos \frac{k \pi}{3}+\cdots+\cos \frac{2018 ... | ## Solution.
The values that the sines in the first parenthesis take are $\sin \frac{k \pi}{3} \in\left\{\frac{\sqrt{3}}{2}, 0,-\frac{\sqrt{3}}{2}\right\}, k \in \mathbb{Z}$.
The sum of the first six addends is
$\sin \frac{\pi}{3}+\sin \frac{2 \pi}{3}+\sin \pi+\sin \frac{4 \pi}{3}+\sin \frac{5 \pi}{3}+\sin 2 \pi=\fr... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.5.
Christmas ornaments are packed in two types of boxes, red and green. In red boxes, the ornaments are packed in five rows with four ornaments each, and in green boxes, in three rows with six ornaments each. In how many different ways can we choose the number of red and green boxes to pack 2018 Christmas ... | ## First solution.
Red boxes are packed with 20 balls each, and green boxes with 18 balls each. Let $x$ be the number of red boxes, and $y$ be the number of green boxes. We need to determine $x, y \in \mathbb{N}_{0}$ such that $20 x + 18 y = 2018$, or solve the equation $10 x + 9 y = 1009, x, y \in \mathbb{N}_{0}$.
O... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.7.
A cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with edge $a$ is intersected by a plane passing through points $E \in \overline{A B}$, $F \in \overline{B C}$, and $G \in \overline{B_{1} C_{1}}$ such that $|A E|=\frac{1}{4}|A B|$, $|B F|=\frac{2}{3}|B C|$, and $\left|B_{1} G\right|=\frac{1}{3}\left|B_{1} C_{1}\... | ## First Solution.
Notice that the resulting smaller geometric body is a truncated pyramid with height $a$.
Its lower base is a right triangle with legs $|E B|=\frac{3}{4} a$ and $|B F|=\f... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.1.
Solve the equation
$$
\binom{x+1}{x-2}+2\binom{x-1}{3}=7(x-1)
$$ | ## Solution.
If we apply the rule $\binom{n}{k}=\binom{n}{n-k}$ to the expression $\binom{x+1}{x-2}=\binom{x+1}{x+1-x+2}=\binom{x+1}{3}$, the given equation transforms into the form $\binom{x+1}{3}+2\binom{x-1}{3}=7(x-1)$.
The condition for the problem to be well-defined is $x+1 \geqslant 3 \Longrightarrow x \geqslan... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.3.
If $z_{1}, z_{2} \in \mathbb{C}$ are solutions to the equation $z+\frac{1}{z}=2 \sin \frac{\pi}{7}$, what is $z_{1}^{770}+z_{2}^{770}$? | ## Solution.
If we multiply $z+\frac{1}{z}=2 \sin \frac{\pi}{7}$ by $z$, we get the equation $z^{2}-2 \sin \frac{\pi}{7} \cdot z+1=0$.
The solutions to the equation are:
$$
\begin{aligned}
z_{1,2} & =\frac{2 \sin \frac{\pi}{7} \pm \sqrt{4 \sin ^{2} \frac{\pi}{7}-4}}{2}=\frac{2 \sin \frac{\pi}{7} \pm 2 \sqrt{-\left(1... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.1.
If $60^{a}=3, 60^{b}=5$ and $x=\frac{1-a-b}{2(1-b)}$, prove that $12^{x}$ is a natural number. | ## Solution.
Given $a=\log _{60} 3$ and $b=\log _{60} 5$. Therefore,
$$
\begin{array}{rlr}
x & =\frac{1-a-b}{2(1-b)}=\frac{1-\log _{60} 3-\log _{60} 5}{2\left(1-\log _{60} 5\right)} & 1 \text { point } \\
& =\frac{\log _{60} \frac{60}{3 \cdot 5}}{2 \log _{60} \frac{60}{5}} & 1 \text { point } \\
& =\frac{\log _{60} 4... | 2 | Algebra | proof | Yes | Yes | olympiads | false |
## Task A-3.6.
Calculate the product
$$
\left(1-\frac{\cos 61^{\circ}}{\cos 1^{\circ}}\right)\left(1-\frac{\cos 62^{\circ}}{\cos 2^{\circ}}\right) \ldots\left(1-\frac{\cos 119^{\circ}}{\cos 59^{\circ}}\right)
$$ | ## Solution.
We observe that all factors are of the same form, so we first calculate
$$
\begin{array}{rlrl}
1-\frac{\cos \left(60^{\circ}+k^{\circ}\right)}{\cos k^{\circ}} & =\frac{\cos k^{\circ}-\cos \left(60^{\circ}+k^{\circ}\right)}{\cos k^{\circ}} & & 1 \text{ point} \\
& =\frac{2 \sin 30^{\circ} \sin \left(30^{\... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.6.
All four intersections of the parabola $y=x^{2}+p x+q$ and the lines $y=x$ and $y=2 x$ are located in the first quadrant. Consider the two segments of the parabola that lie between these lines. Prove that the difference in the lengths of their orthogonal projections onto the $x$-axis is 1. | ## First Solution.
The abscissas ($x$-coordinates) of the intersection points of the line $y=x$ and the given parabola are the solutions of the equation
$$
x^{2}+(p-1) x+q=0
$$
The absci... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
## Task A-4.7.
Determine all natural numbers $n$ such that $\log _{2}\left(3^{n}+7\right)$ is also a natural number. | ## Solution.
Let the given expression be equal to $m$. Instead of $\log _{2}\left(3^{n}+7\right)=m$, we can write
$$
3^{n}+7=2^{m}
$$
(*) 1 point
The expression on the left side of this equation gives a remainder of 1 when divided by 3.
Powers of the number 2 give remainders of 2 and 1 alternately when divided by ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.2.
Determine the minimum and maximum value of the expression
$$
\frac{1}{\sin ^{4} x+\cos ^{2} x}+\frac{1}{\sin ^{2} x+\cos ^{4} x}
$$
Determine all real numbers $x$ for which these values are achieved. | ## First Solution.
Notice that
$$
\begin{aligned}
\sin ^{4} x+\cos ^{2} x & =\sin ^{2} x \cdot\left(1-\cos ^{2} x\right)+\cos ^{2} x \\
& =\sin ^{2} x+\cos ^{2} x \cdot\left(1-\sin ^{2} x\right) \\
& =\sin ^{2} x+\cos ^{4} x
\end{aligned}
$$
Therefore, we need to determine the minimum and maximum values of the expre... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.5.
The base of the pyramid is a regular $n$-sided polygon. Each side of the base is painted black, while each diagonal of the base and each lateral edge of the pyramid is painted either red or blue. Determine the smallest natural number $n \geqslant 4$ for which there necessarily exists a triangle whose ve... | ## Solution.
The answer is $n=9$.
Let the apex of the pyramid be $V$. If $n=9$, then by the Pigeonhole Principle, we conclude that at least five lateral edges are painted the same color. Without loss of generality, let this color be red and let these edges be $\overline{V A}, \overline{V B}, \overline{V C}, \overline... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task 4.
Determine all natural numbers $n$ for which there exist natural numbers $a$ and $b$ such that
$$
\left(n^{2}+2\right)^{a}=(2 n-1)^{b}
$$ | ## Solution.
Notice that $n^{2}+2 \geqslant 3, n^{2}+2>2 n-1$ and that $n^{2}+2$ and $2 n-1$ have the same prime factors. Let $p$ be a prime factor of $n^{2}+2$ and $2 n-1$. Then
$$
p \mid n^{2}+2 \quad \text { and } \quad p|2 n-1 \Longrightarrow p|\left(n^{2}+2\right)+(2 n-1)=(n+1)^{2} \Longrightarrow p \mid n+1
$$
... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task 4.
Determine all natural numbers $n \geqslant 3$ for which the product of the first $n$ natural numbers divides the product of all sums of distinct pairs of prime numbers not greater than $n$, i.e., for which
$$
n! \mid \prod_{\substack{p<q \leqslant n \\ p, q \text{ prime }}}(p+q)
$$ | ## Solution.
Notice that 3, 4, 5, and 6 are not solutions to the problem because in these cases, the left side is divisible by 3, while the right side is not. Furthermore, for $n=7$, the left side is equal to $2^{4} \cdot 3^{2} \cdot 5 \cdot 7$, and the right side is $5 \cdot 7 \cdot 9 \cdot 8 \cdot 10 \cdot 12 = 7 \c... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task 2.
Ante wrote down a sequence $a_{1}, a_{2}, \ldots, a_{2020}$ in which each of the numbers $1,2, \ldots, 2020$ appears exactly once. Barbara wants to find out this sequence, and Ante gives her information through a series of exchanges.
In one exchange, Barbara writes down the numbers from 1 to 2020 on a piec... | ## Solution.
After one exchange, Barbara cannot determine the Antin sequence with certainty. Indeed, if the numbers $a_{i}$ and $a_{j}$ are connected on Antin's paper, Barbara cannot know which number is $a_{i}$ and which is $a_{j}$.
We will show that Barbara can always determine the Antin sequence with certainty aft... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task 2.
An equilateral triangle is given, with 9 points marked on each side, dividing each side into 10 congruent segments. These points are connected by a total of 27 lines parallel to the sides of the triangle. In this way, the triangle is divided into 100 small equilateral triangles. The area between two adjacen... | ## Solution.
Let $A B C$ be the initial triangle and let $D, E$ and $F$ be the midpoints of sides $\overline{B C}, \overline{A C}$ and $\overline{A B}$, respectively. Then the triangles $A E F, D B F, D E C$ and $D E F$ are congruent. Let $a, b, c, d$ be the number of small triangles chosen in each of them, respective... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task 4.
Let $a_{1}, a_{2}, a_{3}, \ldots$ be an infinite sequence of numbers from the set $\{1,2,3,4,5,6,7,8\}$ such that for every pair of natural numbers $(m, n)$, the conditions $a_{n} \mid n$ and $a_{m} \mid m$ are satisfied if and only if $a_{m+n}=a_{m}+a_{n}-1$.
Determine all possible values that $a_{5555}$ ... | ## Solution.
Consider a sequence $a_{1}, a_{2}, \ldots$ that satisfies the conditions of the problem. For a natural number $j$, we say that $j$ is beautiful if $a_{j}=1$.
Assume that $m$ and $n$ are beautiful. Then $a_{m} \mid m$ and $a_{n} \mid n$, so from the condition of the problem it follows that
$$
a_{m+n}=a_{... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task 4.
Determine all natural numbers $n$ such that
$$
\frac{n^{3 n-2}-3 n+1}{3 n-2}
$$
is an integer. | ## First Solution.
Assume that $\frac{n^{3 n-2}-3 n+1}{3 n-2}$ is an integer.
Then $3 n-2$ divides $n^{3 n-2}-3 n+1$, and thus divides $n^{3 n-2}-1$. Therefore, we have $n^{3 n-2} \equiv 1(\bmod 3 n-2)$.
Notice that $n$ is odd (otherwise, an even number would divide an odd number, which is impossible).
Let $p>2$ be... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task 3.
On the side $\overline{A B}$ of the cyclic quadrilateral $A B C D$, there is a point $X$ with the property that the diagonal $\overline{B D}$ bisects the length of $\overline{C X}$, and the diagonal $\overline{A C}$ bisects the length of $\overline{D X}$.
What is the smallest possible ratio $|A B|:|C D|$ i... | ## First Solution.
Let $M$ and $N$ be the midpoints of segments $\overline{C X}$ and $\overline{D X}$, and let $\alpha=\varangle B A C=\varangle B D C$ and $\beta=\varangle C A D=$ $\varangle C B D$.
=(k, 1)$, we get
$$
\frac{m+n}{m^{2}-k m n+n^{2}}=\frac{k+1}{k^{2}-k^{2}+1}=k+1
$$
By substituting $(m, n)=\left(k^{2}+k-1, k+1\right)$, we get
$$
\frac{m+n}{m^{2}-k m n+n^{2}}=\frac{k^{2}+2 k}{\left(k^{2}+k-1\right... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task 4.
Determine all natural numbers $n$ for which there exist distinct divisors $a$ and $b$ of $n$ such that there are no other divisors of $n$ between them and that
$$
n=a^{2}-b
$$ | ## First solution.
Let $n, a, b$ be numbers that satisfy the conditions of the problem.
Since $a \mid n$ and $a \mid a^{2}$, it must be true that $a \mid b$, i.e., $b=m a$ for some natural number $m>1$. Since $b=m a \mid n$, we can write $n=k m a$ for some natural number $k$. From the conditions, we know that there i... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak 4.
Odredi sve prirodne brojeve $n$ za koje postoje različiti djelitelji $a$ i $b$ od $n$ takvi da između njih nema drugih djelitelja od $n$ i da vrijedi
$$
n=a^{2}-b
$$
| ## Drugo rješenje.
Kao i u prvom rješenju dobijemo da je $b=m a$ za neki prirodan broj $m>1$. Jednadžba postaje
$$
n=a(a-m) \text {. }
$$
Pretpostavimo da je $n$ neparan. Tada su $a$ i $b$ također neparni kao njegovi djelitelji, pa je $a^{2}-b$ paran, kontradikcija. Dakle, $n$ je paran.
Ako je $a$ neparan, onda je ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task 2.
On each field of an $n \times n$ board, a dragon is sleeping. Dragons are neighbors if they are on fields that share a side. Every hour, Mini wakes up one dragon that has at least one living neighbor, and Maks directs that dragon towards one of its living neighbors. The awakened dragon breathes fire and des... | ## Solution.
(a) Mini cannot choose which neighbor the awakened dragon will destroy, but if she selects the same dragon enough times, she can achieve that all its neighbors are destroyed. We claim that on a $4 \times 4$ board, four dragons will remain.
In the image, the positions of four dragons that are not adjacent... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In triangle $A B C$, we have
$$
|\angle B A C|=40^{\circ},|\angle C B A|=20^{\circ} \text { and }|A B|-|B C|=10
$$
The angle bisector of $\angle A C B$ intersects the segment $\overline{A B}$ at point $M$. Determine the length of the segment $\overline{C M}$. | ## First solution.
Let $E$ be a point on the segment $\overline{A B}$ such that $|B E|=|B C|$.
We have $|A B|-|B C|=|A B|-|B E|=|A E|$, so $|A E|=10 \mathrm{~cm}$.
Two angles in triangle $A B C$ are known, so the third angle is $|\angle A C B|=120^{\circ}$.
Since $C M$ is the angle bisector of that angle, we have $... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In a hat, there are 11 slips of paper with numbers $0,1,2,3,4,5,6,7,8,9,10$ (one with each number). Each of the eleven students takes one slip. Then they announce in alphabetical order the sentence "so far, at least $k$ students have told a lie," where $k$ is the number on the drawn slip. How many of these 11 statem... | ## Solution.
If the statement of a student who drew $k \geq 6$ is true, then at least 6 students before that student gave false statements, so the maximum number of students who gave true statements is $11-6=5$.
If all statements of students who drew $k \geq 6$ are false, the maximum number of true statements is 6.
... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## Task 2.
A natural number $n$ is good if we can color each side and diagonal of a regular $n$-gon in some color so that for any two vertices $A$ and $B$, there is exactly one vertex $C$, different from $A$ and $B$, such that the segments $\overline{A B}, \overline{B C}$, and $\overline{C A}$ are colored the same col... | ## Solution.
The numbers 8, 10, 11, and 12 are not good, while 7 and 9 are.
First, we show that even numbers are not good. Assume that $n$ is a good number and consider a fixed vertex $A$ of a regular $n$-gon colored in the prescribed manner. For each other vertex $B$, there is a unique vertex $C$ such that the sides... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task 3.
Let $A B C$ be a right triangle with a right angle at vertex $C$. Circle $k$ with diameter $\overline{A C}$ intersects side $\overline{A B}$ at point $D$, and the tangent to circle $k$ at point $D$ intersects side $\overline{B C}$ at point $E$. The circumcircle of triangle $C D E$ intersects side $\overline... | ## Solution.
According to Thales' theorem, $\varangle CDA = 90^{\circ}$, i.e., $\overline{CD}$ is the altitude to the hypotenuse $\overline{AB}$.
Let $P$ be the midpoint of side $\overline{AC}$. Then $\overline{DP}$ is the radius of circle $k$, so it is perpendicular to the tangent $DE$. We have $\varangle PAE + \var... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task 1.
Let $a, b, c,$ and $d$ be real numbers such that
$$
a^{2}+b^{2}+c^{2}+d^{2}=a b+b c+c d+d a+8
$$
Determine the smallest possible value of the largest number in the set $\{|a-b|,|b-c|,|c-d|,|d-a|\}$. | ## Solution.
If we multiply the original equation by 2, we get
$$
2 a^{2}+2 b^{2}+2 c^{2}+2 d^{2}=2 a b+2 b c+2 c d+2 d a+16
$$
which is successively equivalent to
$$
a^{2}-2 a b+b^{2}+b^{2}-2 b c+c^{2}+c^{2}-2 c d+d^{2}+d^{2}-2 d a+d^{2}=16
$$
or
$$
(a-b)^{2}+(b-c)^{2}+(c-d)^{2}+(d-a)^{2}=16
$$
If we assume tha... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Suppose you have 9 evenly spaced dots in a circle on a piece of paper. You want to draw a 9-pointed star by connecting dots around the circle without lifting your pencil, skipping the same number of dots each time.
Determine the number of different stars that can be drawn, if the regular nonagon does not count as a st... | 
There are two different stars that can be drawn, by skipping 1 dot each time (as in the first star shown above) or skipping 3 dots each time (as in the third star shown above). Skipping 2 dots... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Calculate, with proof, the last digit of
$$
3^{3^{3^{3^{3}}}}
$$
Remark. Note that this means $3^{\wedge}\left(3^{\wedge}\left(3^{\wedge}\left(3^{\wedge} 3\right)\right)\right)$, not $\left(\left(\left(3^{\wedge} 3\right)^{\wedge} 3\right)^{\wedge} 3\right)^{\wedge} 3$. | When 3 is raised to the successive powers $1,2,3,4, \ldots$, the units digits are $3,9,7,1, \ldots$ From then on, since the digit 1 has been reached, the units digits will repeat in this cycle of four elements. So it is necessary to find the remainder when
$$
n_{4}=3^{3^{3^{3^{3}}}}
$$
is divided by 4.
When powers o... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $a \leq b \leq c \leq d$ be real numbers such that
$$
a+b+c+d=0 \quad \text { and } \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=0
$$
Prove that $a+d=0$. | . Let us rearrange the first equation to read
$$
a+b=-(c+d)
$$
and the second to give
$$
\begin{aligned}
\frac{1}{a}+\frac{1}{b} & =-\frac{1}{c}-\frac{1}{d} \\
\frac{a+b}{a b} & =-\frac{c+d}{c d}
\end{aligned}
$$
If $a+b \neq 0$, then we can divide (1) by (2) to yield $a b=c d$. So we conclude that
$$
a+b=0 \quad ... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
Given an $n \times n$ matrix whose entries $a_{i j}$ satisfy $a_{i j}=\frac{1}{i+j-1}, n$ numbers are chosen from the matrix no two of which are from the same row or the same column. Prove that the sum of these $n$ numbers is at least 1 . | Suppose that $a_{i j}$ and $a_{k l}$ are among the chosen numbers and suppose that $i<k$ and $j<l$. It is straightforward to show that $a_{i j}+a_{k l} \geqslant a_{i l}+a_{k j}$. Hence, whenever $a_{i j}$ and $a_{k l}$ with $i<k$ and $j<l$ are among chosen numbers, we can lower the sum by replacing these two numbers w... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
Define the digitlength of a positive integer to be the total number of letters used in spelling its digits. For example, since "two zero one one" has a total of 13 letters, the digitlength of 2011 is 13 . We begin at any positive integer and repeatedly take the digitlength. Show that after some number of steps, we must... | We first claim that if a number has at least two digits, then the digitlength is less than the number itself. To see this, note that any digit has at most five letters; thus a 2 -digit number, which is at least 10 , has digitlength at most $5 \cdot 2=10$; a 3 -digit number, which is at least 100 , has digitlength at mo... | 4 | Number Theory | proof | Yes | Yes | olympiads | false |
Let $A$ and $B$ be two points on the plane with $A B=7$. What is the set points $P$ such that $P A^{2}=P B^{2}-7$ ? | If we let $K$ be the point on $A B$ with $A K=4, B K=3$, then the answer is the line through $K$ perpendicular to $A B$. To see this, set $A=(0,0)$ and $B=(7,0)$. Then the points $P=(x, y)$ are exactly those satisfying
$$
(x-0)^{2}+(y-0)^{2}=(x-7)^{2}+(y-0)^{2}-7
$$
which rearranges to $9=-14 x+42$, id est $x=3$. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Find the minimum of the function
$$
f(x, y)=\sqrt{(x+1)^{2}+(2 y+1)^{2}}+\sqrt{(2 x+1)^{2}+(3 y+1)^{2}}+\sqrt{(3 x-4)^{2}+(5 y-6)^{2}} \text {, }
$$
defined for all real $x, y>0$. | Note that $\sqrt{(x+1)^{2}+(2 y+1)^{2}}$ is the distance in the coordinate plane from $(0,0)$ to $(x+1,2 y+1) ; \sqrt{(2 x+1)^{2}+(3 y+1)^{2}}$ is the distance from $(x+1,2 y+1)$ to $(3 x+2,5 y+2)$; $\sqrt{(3 x-4)^{2}+(5 y-6)^{2}}$ is the distance from $(3 x+2,5 y+2)$ to $(6,8)$. Thus, $f(x, y)$ is the distance along t... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $A$ and $B$ be diagonally opposite vertices of a cube. An ant is crawling on a cube starting from $A$, and each second it moves at random to one of the three vertices adjacent to its current one. Find the expected number of steps for the ant to get to vertex $B$. | Let $C_{1}, C_{2}, C_{3}$ be the vertices of the cube adjacent to $A$, and $D_{1}, D_{2}, D_{3}$ the vertices adjacent to $B$. Let $x$ be expected time to get to $B$ starting from $A$. From $A$, you have to go to $C_{1}, C_{2}$, or $C_{3}$, so the expected time to get to $B$ from one of these vertices is $x-1$.
Let $y... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Determine, with proof, the value of $\log _{2} 3 \log _{3} 4 \log _{4} 5 \ldots \log _{255} 256$. | We use the fact that $\log _{b} a=\frac{\log a}{\log b}$. Thus, the product equals
$$
\frac{\log 3}{\log 2} \frac{\log 4}{\log 3} \ldots \frac{\log 256}{\log 255}=\frac{\log 256}{\log 2}=\log _{2}(256)=8
$$ | 8 | Algebra | proof | Yes | Yes | olympiads | false |
If $a<b<c<d$ are distinct positive integers such that $a+b+c+d$ is a square, what is the minimum value of $c+d$ ? | The answer is 11 , which comes from $2+3+5+6$ or $1+4+5+6$.
The minimum possible value of $a+b+c+d$ for $a<b<c<d$ is $1+2+3+4=10$. The smallest square at least 10 is $4^{2}$.
Now assume for sake of contradiction that $c+d \leq 10$. Then, $c<\frac{c+d}{2} \leq 5$, so $a<b<$ $c \leq 4$ and $a+b \leq 2+3=5$. Therefore, ... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all prime numbers $p$ such that $p^{2}+2007 p-1$ is prime as well.
Hint. Except for 3, prime numbers are not divisible by 3 . Hence if $p$ is not equal to 3 then either $p=3 k+1$ or $p=3 k-1$ for some integer $k$. If you wish you may use lists of prime numbers from the internet (e.g. www.imomath.com/primes | If $p=3$, then $p^{2}+2007 p-1=6029$ which is a prime. For $p \neq 3$, we know that $p=3 k \pm 1$ hence $p^{2}+2007 p-1=$ $9 k^{2} \pm 6 k+1+2007 p-1=9 k^{2} \pm 6 k+2007 p$ which is divisible by 3 and can't be prime. Thus the only such prime number is $p=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
If $x$ is a positive real number, find the smallest possible value of $2 x+\frac{18}{x}$. | The answer is 12 . This is achieved when $x=3$; to see it is optimal, note that $2 x+\frac{18}{x} \geq 6 \Longleftrightarrow x^{2}-6 x+9 \geq 0$, which is obviously true since the left-hand side is $(x-3)^{2} \geq 0$.
Alternatively, those who know the so-called AM-GM inequality may apply it directly. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
When the number
$$
N=1^{1} \times 2^{2} \times 3^{3} \times \cdots \times 9^{9}
$$
is written as a decimal number, how many zeros does it end in? | The number $N$ ends with five zeros. Indeed, $10^{5}$ divides $N$ since $4^{4} \times 5^{5}=$ 800,000 divides $N$. But $10^{6}$ does not divide $N$, since the only terms in the product which are divisible by 5 are $5^{5}$, and hence $5^{6}$ does not divide $N$. So the answer is 5 . | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the diagram at right, $A B C D$ is a parallelogram. Given that the distances from $A, B$, and $C$ to line $\ell$ are, respectively, 3,1 , and 5 , find the distance from $D$ to $\ell$ and prove that your answer is correct.
Remark. In the contest, the labels on the points $A$ and $C$ were switched. This makes the fig... | As shown, label the feet of the perpendiculars from $A, B, C, D$ to $\ell$ by $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$. Also, drop perpendiculars $B X$ and $C Y$ to $A A^{\prime}$ and $D D^{\prime}$ respectively. We find that triangles $A B X$ and $D C Y$ are similar since their corresponding sides are parallel... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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