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7. A survey was conducted involving 420 families. It was found that $95\%$ of the families own a TV, and $80\%$ of the families own a computer. A family is randomly selected. If it is known that each surveyed family has at least one of the mentioned devices, what is the probability that the selected family owns a compu... | First method:
| $\frac{95}{100}$ | 420 = 399, which means 399 families own a TV. | $1 \text{ POINT}$ |
| :---: | :---: | :---: |
| $\frac{80}{100}$ | $420=336$, which means 336 families own a computer. | 1 POINT |
| $399-$ | $-336=735$ | 1 POINT |
| 735 | $420=315$ | 1 POINT |
| $315$ | families own both a TV and a co... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Marko bought $3 \mathrm{~kg}$ of apples, $2 \mathrm{~kg}$ of pears, and $1 \mathrm{~kg}$ of strawberries at the market. He paid 59 kn for all the fruit. Apples are cheaper than pears by $1 \mathrm{kn}$ and 50 lp per kilogram, and they are three times cheaper than strawberries per kilogram. How much does one kilogram... | First method:
Apples are the cheapest, so the following applies:
$8 \square + 3 = 59 \quad 1$ POINT
$8 = 59 - 3 \quad 1$ POINT
$8 \square = 56 \quad 1$ POINT
$\square = 56 : 8 = 7 \quad... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Using the digits 0 and 5, write all ten-digit numbers that are divisible by 9. How many such numbers are there? | 2. The sum of the digits of any ten-digit number composed of the digits 0 and 5 is
a multiple of the number 5.
2 POINTS
Furthermore, since we are looking for numbers that are divisible by 9, the sum of the digits must be divisible by 9.
Since the numbers 5 and 9 are relatively prime, it follows that the desired numbers... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. What is the remainder when the number
$1+1 \cdot 2+1 \cdot 2 \cdot 3+1 \cdot 2 \cdot 3 \cdot 4+1 \cdot 2 \cdot 3 \cdot 4 \cdot 5+\ldots+1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 14+1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 14 \cdot 15$
is divided by 72? | First method:
$1+1 \cdot 2+1 \cdot 2 \cdot 3+1 \cdot 2 \cdot 3 \cdot 4+1 \cdot 2 \cdot 3 \cdot 4 \cdot 5+1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6+\ldots+1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 14+1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 14 \cdot 15=$
$=1+2+6+24+120+1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 ... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Twenty students participating in a math camp have decided to send messages to each other, with each of them sending messages to exactly ten of the remaining students. Determine the minimum possible number of mutual messages, i.e., find an example of a message-sending schedule in which the number of mutual messages i... | 5. A total of $20 \cdot 10=200$ messages were sent, and the number of pairs of students is $\frac{20 \cdot 19}{2}=190$.
Therefore, at least $200-190=10$ messages must be sent between the same pairs of students, so the minimum possible number of mutual messages is at least 10.
Now, let's construct a schedule of sent m... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Throughout the year, Marko took several exams and scored a certain number of points on them. If Marko were to score 89 points on his next exam, his average score on the exams would be 91. However, if he were to score 64 points on the next exam, his average score on the exams would then be 86. How many exams has Mark... | First method:
Let $n$ be the number of exams Marko has written so far, and $x$ the sum of the scores achieved on them.
From the conditions of the problem, it is clear that:
$\frac{x+89}{n+1}=91, \frac{x+64}{n+1}=86$
$1 \text{ POINT}$
From both equations, express $x$:
$x=91 \cdot(n+1)-89 \quad$ and $\quad x=86 \cd... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Determine the remainder when the natural number $n$ is divided by 45, given that the number $(n+2$ 020) gives a remainder of 43 when divided by 45. | 3. The remainder of the division of the number $n$ by 45 is denoted by $\mathrm{s} x$.
The remainder of the division of the number 2020 by 45 is 40.
Since $40<43$, it can be concluded that the remainder of the division of the number ( $n+2020$ ) by 45 is the number $(x+40)$
Now, $x+40=43$, from which it follows that... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. On one of two parallel lines, there are 8 points. How many points are on the other line if all the points together determine 640 triangles?
Each task is scored out of 10 points.
The use of a pocket calculator or any reference materials is not allowed. | first method):
$x^{2}+16 x-10 x-160=0$
$x \cdot(x+16)-10 \cdot(x+16)=0$
$(x+16) \cdot(x-10)=0$
2 POINTS
From $x+16=0$ we get $x_{1}=-16$, and from $x-10=0$ we get $x_{2}=10$.
1 POINT
Since the number of points must be a natural number, there are 10 points on the second line.
1 POINT
TOTAL 10 POINTS
Note 1: Th... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Two towers are of heights $20 \mathrm{~m}$ and $30 \mathrm{~m}$. The top of each tower is connected to the base of the other with a taut wire. At what height above the ground do the two wires cross if the distance between the towers is $40 \mathrm{~m}$? | 8.
According to the K-K theorem on similarity, it follows that $\triangle A C B \sim \triangle F C E$. 2 POINTS
From the similarity, it follows that $20: x = 40 : y$ or $y = 2x$. 2 POINTS
... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. To reduce heating costs in a residential building, the tenants decided to change the facade, and they were granted non-repayable funds from the Environmental Protection and Energy Efficiency Fund, which cover 60% of all total costs. The total cost for the new facade is 1,200,000 kn. The new facade guarantees that th... | 6. The cost for the facade is 1200000 kn. The non-refundable portion of funds is $60\%$, which means that the residents have to cover $40\%$ of the costs, amounting to $480000 \mathrm{kn}$.
1 POINT
The average annual heating cost was 168000 kn, and with a savings of 35%, it will amount to 109200 kn.
1 POINT
Let $n$... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-1.2. Solve the equation in the set of real numbers:
$$
f(x)+f(2-x)=2
$$
where
$$
f(x)= \begin{cases}|x|, & x \leq 1 \\ 2-x, & x>1\end{cases}
$$ | Solution.
$$
f(2-x)= \begin{cases}|2-x|, & \text { for } 2-x \leq 1, \text { i.e. } x \geq 1 \\ x, & \text { for } 2-x>1, \text { i.e. } x<1\end{cases}
$$
1. For $x>1$ the equation becomes
$$
\begin{array}{r}
2-x+|2-x|=2 \\
|2-x|=x
\end{array}
$$
Then we have
$$
2-x=x \quad \text { or } \quad 2-x=-x
$$
The soluti... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Zadatak B-4.1. Ako je
$$
f\left(\log _{3} x\right)=\frac{\log _{3} \frac{9}{x^{4}}}{\log _{0 . \dot{3}} x-\log _{\sqrt{3}} x} \quad \text { i } \quad(f \circ g)(x)=e^{x}
$$
koliko je $g(\ln 2)$ ?
|
Rješenje. Zapišimo izraz
$$
f\left(\log _{3} x\right)=\frac{\log _{3} \frac{9}{x^{4}}}{\log _{0 . \dot{3}} x-\log _{\sqrt{3}} x}
$$
u jednostavnijem obliku.
$$
f\left(\log _{3} x\right)=\frac{\log _{3} 9-\log _{3} x^{4}}{\log _{3^{-1}} x-\log _{3^{\frac{1}{2}}} x}=\frac{2-4 \log _{3} x}{-\log _{3} x-2 \log _{3} x}
... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.4. Determine all values of the real parameter $p$ for which the equation $x^{4}-(3 p+2) x^{2}+p^{2}=0$ has four solutions that form four consecutive terms of an arithmetic sequence. | Solution. Since the equation is biquadratic, its solutions are $x_{1}, -x_{1}, x_{2} \text{ and } -x_{2}$ (symmetric with respect to the origin). Let $x_{1}$ and $x_{2}$ be positive real numbers and let $x_{1} < x_{2}$. Then the order of the terms of the sequence (from the smallest) is $-x_{2}, -x_{1}, x_{1}, x_{2}$.
... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.5. In the race, 100 people participated, and no two people finished the race with the same time. At the end of the race, each participant was asked what place they finished in, and everyone answered with a number between 1 and 100.
The sum of all the answers is 4000. What is the smallest number of incorrect a... | Solution. If all the runners had accurately answered their finishing positions, the sum of their answers would be
$$
1+2+3+\ldots+100=5050
$$
Thus, some people must have exaggerated their results, and the smallest number of incorrect answers is achieved when the runners at the last positions claim they were first.
I... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.5.
Karlo and Lovro are playing the following game. Karlo will cut a paper of size $9 \times 9$ into rectangles of integer dimensions, where at least one dimension is 1. After that, Lovro will choose a natural number $k \in \{1, \ldots, 9\}$, and Karlo will give him as many coins as the total area of all re... | ## Solution.
We claim that Karlo will give Lovro at least 12 coins.
1 point
Assume there is a way for Karlo to cut the paper such that Lovro has to give (strictly) fewer than 12 coins. By cutting the paper in this way, Karlo would create at most 11 rectangles of dimensions $1 \times 1$, five rectangles of dimensions... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.5.
How many integers can a finite set $S$ contain at most such that among any three elements of the set $S$ there are two different numbers whose sum is also in $S$? | ## Solution.
Let $S$ be a finite set of integers such that among any three elements of the set $S$, there exist two distinct numbers whose sum is also in $S$. We will show that $S$ cannot have more than 7 elements.
Assume that $S$ has at least three positive elements. Let $a, b, c$ be the three smallest positive elem... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak B-2.4.
Na nekom natjecanju iz matematike rješava se $n$ zadataka, $10<n<20$. Za svaki točno riješeni zadatak natjecatelj dobiva 5 bodova, a za svaki neriješeni ili netočno riješen zadatak gubi 3 boda. Ako je Sebastijan imao ukupno 0 bodova, koliko je zadataka riješio točno? Koliko je bilo ukupno zadataka?
| ## Drugo rješenje.
Neka je $x$ broj točno riješenih zadataka, a $y$ broj netočno ili neriješenih zadataka.
Vrijedi da je $10<x+y<20, x, y \in \mathbb{N}$, a tada je $x<20$ i $y<20$.
Ukupni broj bodova iznosi $5 x-3 y$.
Budući da Sebastijan ima 0 bodova, vrijedi
$$
5 x-3 y=0, \text { odnosno } x=\frac{3 y}{5}
$$
K... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.1.
Solve the equation $5 \cdot 0.16^{x}+8 \cdot 0.4^{x}=4$. | ## Solution.
Let $t=(0.4)^{x}$. 1 point
After introducing the substitution, the equation takes the form $5 t^{2}+8 t-4=0$.
The solutions of the obtained quadratic equation are:
$t_{1,2}=\frac{-8 \pm \sqrt{64+80}}{10}=\frac{-8 \pm \sqrt{144}}{10}=\frac{-8 \pm 12}{10}$,
which means $t_{1}=\frac{2}{5}$ and $t_{2}=-2$... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.3.
The function is given by $f(x)=A \sin \left(B x+\frac{7 \pi}{10}\right)+D$, where $A, B$, and $D$ are positive real numbers. The point $M\left(\frac{2 \pi}{3},-8\right)$ is the only minimum point, and the point $N\left(\frac{3 \pi}{2}, 12\right)$ is the only maximum point of the function $f$ in the inte... | ## Solution.
Let's plot points $M$ and $N$ in the coordinate system and sketch the graph of the function $f$.
First, let's determine the coefficients $A$ and $D$.
The coefficient $D$ is t... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.5.
A math teacher gave the students ten problems for homework and said that he would select two of them for the next test. How many of the given ten problems does Luka need to solve at minimum to be sure with a probability greater than $\frac{7}{9}$ that at least one of the selected problems will be among ... | ## First Solution.
Let $S$ be a set of ten homework problems. Let $L$ be a subset of $S$, i.e., a set of $x$ problems that Luka has solved. Then $10-x$ is the number of problems that Luka has not solved.
Consider the event $A$ - "the professor has chosen at least one problem from the set $L$."
Then the complementary... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.2.
A quadruple of natural numbers $(a, b, c, d)$ is called green if
$$
b=a^{2}+1, \quad c=b^{2}+1, \quad d=c^{2}+1
$$
and $D(a)+D(b)+D(c)+D(d)$ is odd, where $D(k)$ is the number of positive divisors of the natural number $k$.
How many green quadruples are there whose all members are less than 1000000? | ## Solution.
A natural number has an odd number of divisors if and only if it is the square of some natural number. For a natural number $m$, we have
$$
m^{2}d=c^{2}+1>c^{2}=\left(b^{2}+1\right)^{2}>b^{4}=\left(a^{2}+1\right)^{4}>a^{8}
$$
Therefore, it must be that $a^{8}<10^{6}$. Since $10^{6}=1000^{3}<1024^{3}=2^{... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.5.
On a board, there are 2023 different real numbers. If each number on the board (simultaneously) is replaced by the sum of all the other numbers, the 2023 numbers on the board will be the same as at the beginning.
What values can the product of all the numbers on the board take at some point? | ## First Solution.
Let the numbers initially written on the board be denoted by $x_{1}, x_{2}, \ldots, x_{2023}$, and let $s=x_{1}+x_{2}+\ldots+x_{2023}$.
Since after the first swap the numbers on the board are the same as the initial numbers, it follows that after any other swap, the numbers on the board will also b... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.5.
On the board, there are 2023 different real numbers. If each number on the board (simultaneously) is replaced by the sum of all the other numbers, the 2023 numbers on the board will be the same as at the beginning.
What values can the product of all the numbers on the board take at some point? | ## Second Solution.
As in the previous solution, we conclude that if the numbers on the board after the first move are equal to the opposite values of the numbers written on the board at the beginning, and that the product does not change after each move.
Without loss of generality, let $x_{1}>x_{2}>\ldots>x_{2023}$.... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.3.
Let $p$ and $q$ be prime numbers such that $p+q+4$ and $p q-12$ are also prime numbers. Determine $p+q$. | ## Solution.
Numbers $p$ and $q$ cannot have the same parity because then $p+q+4$ would be an even number strictly greater than 2. Therefore, one of the numbers $p$ or $q$ must be equal to 2.
Without loss of generality, we can assume that $p=2$.
According to the conditions of the problem, $q+6$ and $2q-12$ are prime... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.4.
How many ordered pairs of natural numbers $(a, b)$ satisfy
$$
\log _{2023-2(a+b)} b=\frac{1}{3 \log _{b} a} ?
$$ | ## Solution.
For the equality to be well-defined, it is necessary that
$$
\begin{aligned}
2023-2(a+b) & \in (0,1) \cup (1,+\infty) \\
b & \in (0,1) \cup (1,+\infty) \\
\log _{b} a & \neq 0
\end{aligned}
$$
Since $a$ and $b$ are natural numbers, the first two conditions become $2023-2(a+b) \geqslant 2$ and $b \geqsla... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-1.3. If $a+b=2$ and $a^{2}+b^{2}=6$, what is $a^{-1}+b^{-1}$? | Solution.
$$
a^{-1}+b^{-1}=\frac{1}{a}+\frac{1}{b}=\frac{b+a}{a b}=\frac{2}{a b}
$$
Let's square the expression $a+b=2$. From
$$
a^{2}+2 a b+b^{2}=4
$$
it follows that $2 a b=4-6$, so $a b=-1$.
Then we have
$$
a^{-1}+b^{-1}=-2
$$ | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-1.7. If in a regular hexagon $A B C D E F$ we draw all six of its shorter diagonals, they determine a hexagon GHIJKL. Prove that the hexagon GHIJKL is also a regular hexagon and calculate how many times smaller its area is compared to the area of the original hexagon. | Solution. Sketch
(2 points)
Let the side length of the hexagon $A B C D E F$ be $a$. Consider, in the regular hexagon $A B C D E F$, the characteristic triangle $S B C$.
Since this is an ... | 3 | Geometry | proof | Yes | Yes | olympiads | false |
Task B-2.5. What is the last digit of the number $2012^{3}+3^{2012}$? | Solution. The number $2012^{3}$ ends with the digit 8.
The powers of the number $3, 3^{1}, 3^{2}, 3^{3}, 3^{4}, 3^{5}, 3^{6}, \ldots$ end with the digits $3,9,7,1,3,9, \ldots$,
that is, every fourth digit repeats.
Therefore, the number $3^{2012}=\left(3^{4}\right)^{503}$ has the same units digit as the number $3^{4}... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task B-3.1. Solve the equation
$$
\log _{5 x-2} 2+2 \cdot \log _{5 x-2} x=\log _{5 x-2}(x+1)
$$ | Solution. Let's determine the conditions for which the given equation makes sense.
$$
\begin{aligned}
& 5 x-2>0, \quad 5 x-2 \neq 1, \quad x>0, \quad x+1>0 . \\
& \log _{5 x-2} 2+2 \cdot \log _{5 x-2} x=\log _{5 x-2}(x+1) \\
& \log _{5 x-2}\left(2 x^{2}\right)=\log _{5 x-2}(x+1) \\
& 2 x^{2}=x+1
\end{aligned}
$$
The ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.1. Calculate the area of a square whose two vertices are at the foci, and the other two at the vertices of the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{b^{2}}=1$. The foci of the ellipse are on the x-axis. | Solution. Sketch
Since it is a square, it must hold that $e=b$,
or $a^{2}=2 b^{2}, b^{2}=4, b=2$.
The area is $P=2 b^{2}=8$ square units. | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.4.
Let $n$ be a natural number. Determine all positive real numbers $x$ for which
$$
\frac{2^{2}}{x+1}+\frac{3^{2}}{x+2}+\cdots+\frac{(n+1)^{2}}{x+n}+n x^{2}=n x+\frac{n(n+3)}{2}
$$ | ## First Solution.
Since $n x=x+x+\cdots+x$, and $\frac{n(n+3)}{2}=(1+2+3+\cdots+n)+n$, the original equality is equivalent to
$$
\sum_{k=1}^{n} \frac{(k+1)^{2}}{x+k}+n x^{2}-\sum_{k=1}^{n} x-\sum_{k=1}^{n} k-n=0
$$
or equivalently,
$$
\sum_{k=1}^{n}\left(\frac{(k+1)^{2}}{x+k}-(x+k)\right)+n x^{2}-n=0
$$
By simpli... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.5.
On a board of dimensions $8 \times 8$, triomino tiles of shape $\exists$ are placed such that each triomino tile covers exactly three squares of the board, and they do not overlap with each other.
How many triomino tiles are needed at a minimum to place on the board so that no additional triomino tile ... | ## Solution.
Divide the board into 16 squares of $2 \times 2$ as shown in the figure.
In each of these squares, at least two fields must be covered, otherwise a tromino-tile can be placed ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.1.
Determine the minimum value of the expression
$$
|x+1|+|x-2|+|x-3|,
$$
where $x$ is a real number. | ## Solution.
Let's calculate the value of the given expression depending on where the number $x$ is located on the real number line.
- For $x \leqslant -1$, the given expression is equal to
$$
-x-1-x+2-x+3=4-3x
$$
We conclude that for real numbers $x \leqslant -1$, the given expression achieves its minimum value wh... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.5.
Ana covered a board of dimensions $2020 \times 2020$ with domino tiles that do not overlap, and each tile covers exactly two squares of the board. Branka wants to color these tiles such that for each tile, among its neighboring tiles, at most two are of the same color. Two tiles are neighbors if they co... | ## Solution.
We will prove that Branka has enough with two colors.
It is obvious that Branka cannot color all the tiles with one color: every tile that is not on the edge has at least four neighboring tiles (which touch it from below, above, left, and right) and which are the same color as the observed one, which is ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.5.
Ana covered a board of dimensions $2020 \times 2020$ with domino tiles that do not overlap, and each tile covers exactly two squares of the board. Branka wants to color these tiles such that for each tile, among its adjacent tiles, at most one is of the same color. Two tiles are adjacent if they cover s... | ## Solution.
We will prove that Branka has enough with three colors.
First, we will prove that Branka does not have enough with two colors (and thus not with fewer). We will find an example of Ana's tiling in which Branka cannot color the tiles with two colors.
Let $a$ be a real number. Determine the sum of all three solutions of the equation
$$
x^{3}-a^{2} x+a x-x+a^{2}-a=0
$$ | ## Solution.
The expression on the left side of the given equation can be factored as follows:
$$
\begin{aligned}
x^{3}-a^{2} x+a x-x+a^{2}-a & =\left(x^{3}-x\right)+a^{2}(-x+1)+a(x-1) & & 1 \text { point } \\
& =(x-1)\left(x^{2}+x-a^{2}+a\right) & & 1 \text { point } \\
& =(x-1)\left(\left(x^{2}-a^{2}\right)+(x+a)\r... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.5. (4 points)
Eleonora has many cubes, all of whose sides are white. First, she separates one cube and puts it in an empty box. Then she takes one cube at a time and paints some of its sides green, but in such a way that this cube differs from all those already in the box, and then she puts this cube in th... | ## Solution.
In the box, there is only one cube without any green sides, and one cube with one green side. Also, there is only one cube with all sides painted and one cube with five painted sides.
Cubes with two green sides are not all the same, as the painted sides can be adjacent or opposite. Therefore, there are t... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task A-2.6. (10 points)
Determine all natural numbers less than 1000 that are equal to the sum of the squares of their digits. | ## Solution.
Let the desired number be $n=\overline{a b c}$. (We will allow leading digits to be 0).
We want $a^{2}+b^{2}+c^{2}=100 a+10 b+c$.
Since $a, b$, and $c$ are digits, $a^{2}+b^{2}+c^{2} \leqslant 3 \cdot 9^{2}=243$, so $a \leqslant 2$.
2 points
Furthermore, we conclude that $a^{2}+b^{2}+c^{2} \leqslant 2... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak A-3.3. (4 boda)
Izračunaj $144^{\log _{5} 1000}: 10^{6 \log _{5} 12}$.
| ## Prvo rješenje.
Vrijedi
$$
\frac{144^{\log _{5} 1000}}{10^{6 \log _{5} 12}}=\frac{\left(12^{2}\right)^{\log _{5} 10^{3}}}{10^{6 \log _{5} 12}}=\frac{12^{6 \log _{5} 10}}{10^{6 \log _{5} 12}}=\left(\frac{12^{\log _{5} 10}}{10^{\log _{5} 12}}\right)^{6}
$$
Općenito je
$$
x^{\log _{a} y}=\left(a^{\log _{a} x}\right)... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.3. (4 points)
Calculate $144^{\log _{5} 1000}: 10^{6 \log _{5} 12}$. | ## Second Solution.
Let's determine the logarithm (base 5) of the desired number:
$$
\begin{aligned}
\log _{5} \frac{144^{\log _{5} 1000}}{10^{6 \log _{5} 12}} & =\log _{5}\left(144^{\log _{5} 1000}\right)-\log _{5}\left(10^{6 \log _{5} 12}\right) \\
& =\log _{5} 1000 \cdot \log _{5} 144-6 \log _{5} 12 \cdot \log _{5... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task A-1.4. (4 points)
Let $a, b$ and $c$ be real numbers such that
$$
a+b+c=3 \quad \text { and } \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0
$$
What is $a^{2}+b^{2}+c^{2}$? | ## Solution.
Since $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$
from the first given equality, we get
$$
\begin{aligned}
& a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a=9 \\
& a^{2}+b^{2}+c^{2}=9-2(a b+b c+c a)
\end{aligned}
$$
Since from the second given equality it follows that
$$
\frac{b c+c a+a b}{a b c}=0
$$
whic... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task A-1.5. (4 points)
How many elements at least need to be removed from the set $\{2,4,6,8,10,12,14,16\}$ so that the product of the remaining elements is a square of a natural number? | ## Solution.
The product of all elements of the given set is
$$
\begin{aligned}
P & =2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 \\
& =2 \cdot 2^{2} \cdot(2 \cdot 3) \cdot 2^{3} \cdot(2 \cdot 5) \cdot\left(2^{2} \cdot 3\right) \cdot(2 \cdot 7) \cdot 2^{4} \\
& =2^{15} \cdot 3^{2} \cdot 5 \cdot 7
\en... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task A-2.6. (10 points)
Let $a, b$, and $c$ be three distinct real numbers, none of which are equal to zero. We consider the quadratic equations:
$$
a x^{2}+b x+c=0, \quad b x^{2}+c x+a=0, \quad c x^{2}+a x+b=0
$$
If $\frac{c}{a}$ is a solution to the first equation, prove that all three equations have a common solu... | ## First Solution.
Since $\frac{c}{a}$ is a solution to the equation $a x^{2}+b x+c=0$, after substitution we get
$$
a \cdot\left(\frac{c}{a}\right)^{2}+b \cdot \frac{c}{a}+c=0
$$
which simplifies to
$$
c^{2}+b c+a c=0
$$
i.e.,
$$
c(a+b+c)=0
$$
Since $c \neq 0$, it must be that $a+b+c=0$,
so $x_{0}=1$ is a comm... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
## Zadatak A-3.1. (4 boda)
Ako je $\log _{a} x=3$ i $\log _{a b} x=2$, koliko je $\log _{b} x$ ?
| ## Prvo rješenje.
Iz $\log _{a} x=3$ slijedi $x=a^{3}$, a iz $\log _{a b} x=2$ slijedi $x=(a b)^{2}$.
Sada imamo $a^{3}=a^{2} b^{2}$ pa kako $a \neq 0$, vrijedi $a=b^{2}$.
Stoga je $x=\left(b^{2}\right)^{3}=b^{6}$,
pa je $\log _{b} x=\log _{b} b^{6}=6$.
| 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-3.1. (4 points)
If $\log _{a} x=3$ and $\log _{a b} x=2$, what is $\log _{b} x$? | ## Second solution.
Obviously $x \neq 1$, so from $\log _{a} x=3$ it follows that $\log _{x} a=\frac{1}{3}$,
and from $\log _{a b} x=2$ it follows that $\log _{x} a b=\frac{1}{2}$.
Since $\log _{x} a b=\log _{x} a+\log _{x} b$, it follows that $\frac{1}{3}+\log _{x} b=\frac{1}{2}$,
so $\log _{x} b=\frac{1}{6}$ and ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task A-3.3. (4 points)
Point $D$ is the foot of the altitude from vertex $A$, and point $E$ is the foot of the altitude from vertex $B$ of triangle $A B C$. If $|A E|=5,|C E|=3$ and $|C D|=2$, determine the length of $|B D|$. | ## First solution.
Notice that triangles $A C D$ and $B C E$ are similar (two angles).
Therefore, $|A C|:|C D|=|B C|:|C E|$.
Let $|B D|=x$.
Now (*) can be written in the form $\frac{5+3}{2}=\frac{x+2}{3}$,
from which it follows that $x+2=3 \cdot \frac{8}{2}=12$ so the desired length $|B D|=x=10$. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak A-4.2. (4 boda)
Neka je $z$ nultočka polinoma $z^{2}-2 z \cos \frac{\pi}{n}+1$. Odredi sve moguće vrijednosti izraza $z^{n}$.
| ## Rješenje.
Rješenja kvadratne jednadžbe $z^{2}-2 z \cos \frac{\pi}{n}+1=0$ su
$$
z=\frac{2 \cos \frac{\pi}{n} \pm \sqrt{4 \cos ^{2} \frac{\pi}{n}-4}}{2}=\cos \frac{\pi}{n} \pm i \sin \frac{\pi}{n}
$$
pa je
$$
\begin{aligned}
z^{n} & =\left(\cos \frac{\pi}{n} \pm i \sin \frac{\pi}{n}\right)^{n} \\
& =\cos \left(n ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task A-4.6. (10 points)
Let $a$ and $b$ be natural numbers. Which digits can be in the units place of the decimal representation of the number $(a+b)^{5}-\left(a^{5}+b^{5}\right)$? | ## Solution.
The given expression can be expanded as follows:
$$
\begin{aligned}
& (a+b)^{5}-\left(a^{5}+b^{5}\right) \\
& =a^{5}+5 a^{4} b+10 a^{3} b^{2}+10 a^{2} b^{3}+5 a b^{4}+b^{5}-a^{5}-b^{5} \\
& =5\left(a^{4} b+2 a^{3} b^{2}+2 a^{2} b^{3}+a b^{4}\right) \\
& =5 a b\left(a^{3}+2 a^{2} b+2 a b^{2}+b^{3}\right) ... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Using a suitable substitution, determine the number of roots of the equation
$$
8 x\left(1-2 x^{2}\right)\left(8 x^{4}-8 x^{2}+1\right)=1
$$
that lie within the interval $[0,1]$. | 2. We immediately see that $x=0$ and $x=1$ are not solutions, so we introduce the substitution $x=\cos \varphi$, with $\varphi \in\left(0, \frac{\pi}{2}\right)$. The equation becomes
$$
8 \cos \varphi\left(1-2 \cos ^{2} \varphi\right)\left(8 \cos ^{4} \varphi-8 \cos ^{2} \varphi+1\right)=1
$$
5 points
or, because $\... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.1.
If $a=\sqrt{\sqrt{3}-\sqrt{2}}(\sqrt{\sqrt{3}-1}+\sqrt{\sqrt{3}+1}), b=\sqrt{\sqrt{3}+\sqrt{2}}(\sqrt{\sqrt{3}-1}-\sqrt{\sqrt{3}+1})$, what is $a+b$? | ## First Solution.
$$
\begin{aligned}
a^{2} & =(\sqrt{3}-\sqrt{2})(\sqrt{3}-1+2 \sqrt{\sqrt{3}-1} \cdot \sqrt{\sqrt{3}+1}+\sqrt{3}+1) \\
& =2(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=2 \\
b^{2} & =(\sqrt{3}+\sqrt{2})(\sqrt{3}-1-2 \sqrt{\sqrt{3}-1} \cdot \sqrt{\sqrt{3}+1}+\sqrt{3}+1) \\
& =2(\sqrt{3}+\sqrt{2})(\sqrt{3}-\s... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.2.
Determine all even natural numbers $n$ for which
$$
\left|\log _{0.5} 20^{\cos \pi}+\log _{0.5} 30^{\cos 2 \pi}+\log _{0.5} 42^{\cos 3 \pi}+\cdots+\log _{0.5}\left(n^{2}+7 n+12\right)^{\cos n \pi}\right|=1
$$ | ## Solution.
Simplify the given expression.
$$
\begin{array}{cc}
\left|\log _{0.5}\left(20^{\cos \pi} \cdot 30^{\cos 2 \pi} \cdot 42^{\cos 3 \pi} \cdots\left(n^{2}+7 n+12\right)^{\cos n \pi}\right)\right|=1 & \\
\left|\log _{0.5}\left(20^{-1} \cdot 30^{1} \cdot 42^{-1} \cdots\left(n^{2}+7 n+12\right)^{1}\right)\right... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak B-2.5.
Kružnice $k_{1}$ i $k_{2}$ sijeku se u točkama $A$ i $B$ kao što je prikazano na slici. Točka $C$ nalazi se na kružnici $k_{1}$, a točka $D$ na kružnici $k_{2}$ tako da vrijedi $\varangle A C B=60^{\circ} \mathrm{i} \varangle B D A=30^{\circ}$. Ako su središta kružnica $k_{1}$ i $k_{2}$ udaljena $4 \... | ## Prvo rješenje.
Označimo s $r_{1} \mathrm{i} r_{2}$ polumjere kružnica $k_{1} \mathrm{i} k_{2}$. Neka je $E$ polovište tetive $\overline{A B}$ i $x=|B E|$.
Kut $\varangle A C B$ je obodni nad lukom $\widehat{A B}$ pa je njegov središnji kut $\varangle A S_{1} B=120^{\circ}$, odnosno $\varangle E S_{1} B=60^{\circ}$... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.5.
Circles $k_{1}$ and $k_{2}$ intersect at points $A$ and $B$ as shown in the figure. Point $C$ is on circle $k_{1}$, and point $D$ is on circle $k_{2}$ such that $\varangle A C B=60^{\circ} \text{ and } \varangle B D A=30^{\circ}$. If the centers of circles $k_{1}$ and $k_{2}$ are $4 \sqrt{3} \text{ cm}$... | ## Second Solution.
We use the same sketch and labels as in the first solution.
Angle $\varangle A C B$ is an inscribed angle over the arc $\overparen{A B}$, so its central angle is $\varangle A S_{1} B=120^{\circ}$, or $\varangle E S_{1} B=60^{\circ}$.
Angle $\varangle B D A$ is an inscribed angle over the arc $\wi... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.3.
The sum of all 1002-digit numbers that in their representation have a thousand zeros and two ones is $S$. Determine the remainder obtained when the number $S$ is divided by 3. | ## First solution.
The number 1 must be in the first position, from which it follows that there are 1001 numbers with the described property.
We can add these numbers by writing one below the other:
$$
\begin{array}{r}
1100000 \ldots 00 \\
1010000 \ldots 00 \\
1001000 \ldots 00 \\
1000100 \ldots 00 \\
\ldots \ldots ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.2.
Let $n$ be a natural number and let
$$
A=0.1^{n} \cdot 0.01^{n} \cdot 0.001^{n} \cdot 0.0001^{n} \cdot \ldots \cdot 0 . \underbrace{00 \ldots 0}_{199 \text { zeros }} 1^{n} .
$$
The number $A^{-1}$ has 140701 digits. Determine the number $n$. | ## Solution.
$A=10^{-n} \cdot 10^{-2 n} \cdot 10^{-3 n} \cdot 10^{-4 n} \cdot \ldots \cdot 10^{-200 n}=10^{-n(1+2+3+\ldots+200)} \quad 1 \text{ point}$
Then $A^{-1}=10^{n \cdot \frac{200 \cdot 201}{2}}=10^{20100 n}$. 2 points
This number has a leading digit of one and 20100n zeros in its decimal representation.
It ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.5.
Determine the last two digits of the number $7^{2016^{2017}}$. | ## Solution.
The number $7^{2016^{2017}}$ can be written as follows:
$$
7^{2016^{2017}}=7^{2016 \cdot 2016 \cdot 2016 \cdots \cdots \cdot 2016}
$$
The exponent of the given number is divisible by 8 because 2016 is divisible by 8, so we have:
$$
7^{2016^{2017}}=7^{8 x}=\left(7^{4}\right)^{2 x}
$$
where $x$ is some ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.1.
How many integers $a$ are there for which both solutions of the equation $(x-20)(x+17)=\frac{1}{4} a$ are positive real numbers? | ## First solution.
$(x-20)(x+17)=\frac{1}{4} a \Leftrightarrow x^{2}-3 x-340-\frac{1}{4} a=0$.
Both solutions of the equation will be positive real numbers if the following conditions are met:
1. $D \geqslant 0 \Leftrightarrow 9+4\left(340+\frac{1}{4} a\right) \geqslant 0 \Rightarrow a \geqslant-1369$.
2 points
2. ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.7.
Let $A B C D$ be a parallelogram with side lengths $|A B|=a \mathrm{~cm}$ and $|B C|=b \mathrm{~cm} (a>b)$ and an acute angle $\alpha$. The area of the quadrilateral formed by the intersection of the angle bisectors of the internal angles of the parallelogram is $48 \mathrm{~cm}^{2}$, and $\sin \frac{\a... | ## Solution.
In triangle $A V D$, $\angle V A D = \frac{\alpha}{2}$ and $\angle A D V = \frac{180^\circ - \alpha}{2} = 90^\circ - \frac{\alpha}{2}$. Therefore, triangle $A V D$ is a right t... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.1. Draw in the complex plane the set of points $(x, y)$, associated with complex numbers $z=x+yi$, for which
$$
\operatorname{Re} z \cdot \operatorname{Im} z<0 \quad \mathrm{and} \quad|z|^{2} \leq \operatorname{Im}\left(z^{2}\right)+1
$$
Calculate the area of this set of points. | ## Solution.
From the first inequality, it follows that $x \cdot y < 0$, which means that the desired points lie in the second or fourth quadrant.
From the second inequality, it follows that
$$
x^{2} + y^{2} \leq 2 x y + 1
$$
or $(x-y)^{2} \leq 1$ or $|x-y| \leq 1$.
From this, we have $-1 \leq x-y \leq 1$, or $y \... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.2. Rowing evenly, a fisherman traveled upstream as many kilometers as the solution to the equation $\sqrt{5 x}-\frac{4}{\sqrt{3 x+1}}=\sqrt{3 x+1}$, and returned to the harbor without stopping in 1 hour and 20 minutes. At what speed did he row if the river's speed is $2 \mathrm{~km} / \mathrm{h}$? | ## Solution.
First, let's solve the given equation
$$
\sqrt{5 x}-\frac{4}{\sqrt{3 x+1}}=\sqrt{3 x+1}
$$
Multiply both sides by the denominator.
$$
\begin{aligned}
\sqrt{5 x(3 x+1)}-4 & =3 x+1 \\
\sqrt{5 x(3 x+1)} & =3 x+5
\end{aligned}
$$
If we square both sides, we get
$$
\begin{aligned}
15 x^{2}+5 x & =9 x^{2}+... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.1.
If $x y z=a b c$, what is $\frac{b x}{x y+a b+b x}+\frac{c y}{y z+b c+c y}+\frac{a z}{z x+c a+a z}$? | ## Solution.
First, extend the first fraction by $z$, and the second by $a$.
$$
\begin{aligned}
& \frac{b x}{x y+a b+b x} \cdot \frac{z}{z}+\frac{c y}{y z+b c+c y} \cdot \frac{a}{a}+\frac{a z}{z x+c a+a z}= \\
& \frac{b x z}{x y z+a b z+b x z}+\frac{c a y}{a y z+a b c+a c y}+\frac{a z}{z x+c a+a z}
\end{aligned}
$$
... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.2.
Natural numbers are sequentially written to form a spiral of numbers as shown in the figure:
| 21 | 22 | | 23 | | 24 | | 25 | 26 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | | | | | | | | $\vdots$ |
| 20 | 7 | $\rightarrow$ | 8 | $\rightarrow$ | 9 | $\right... | ## Solution.
Notice that the numbers on the diagonal to the right and up are squares of odd natural numbers $1, 9, 25, 49, \ldots$
The number $9=(2 \cdot \mathbf{1}+1)^{2}$ and its position... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.3.
Each member of Marića's family drank 4 deciliters of a coffee and milk mixture. The amount of coffee and milk is different in each cup, but it is never zero. Marića drank one quarter of the total amount of milk and one sixth of the total amount of coffee. How many members are there in Marića's family? | ## Solution.
Let $k$ be the amount of coffee, and $m$ be the amount of milk that Marica drank (in dl). Let the family have $n$ members. Since each member of the family drank $4 \mathrm{dl}$ of the coffee and milk mixture, a total of $4 n$ of the mixture was consumed, so the following equations hold:
$$
\begin{aligned... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.1.
Determine the real parameter $k$ so that the system of equations
$$
\left[\operatorname{Im}\left(z+\frac{1}{2} i\right)\right]^{2}=|z+2|^{2}+\frac{5}{4} \quad \text { and } \quad \operatorname{Im}(z)-2 \operatorname{Re}(z)=k, z \in \mathbb{C}
$$
has only one solution? | ## Solution.
Let $z=x+y i ; x, y \in \mathbb{R}$.
Then the first of the given equations becomes
$$
\left[\operatorname{Im}\left(x+y i+\frac{1}{2} i\right)\right]^{2}=|x+y i+2|^{2}+\frac{5}{4}
$$
or
$$
\left(y+\frac{1}{2}\right)^{2}=(x+2)^{2}+y^{2}+\frac{5}{4}
$$
By simplifying, we get the set of points $(x, y)$ t... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.1.
Ordered pairs $(a, b)$ and $(c, d)$ are solutions to the system of equations
$$
\begin{aligned}
\log _{225} x+\log _{64} y & =4 \\
\log _{x} 225-\log _{y} 64 & =1
\end{aligned}
$$
Calculate $\log _{30}(a b c d)$. | ## Solution.
The solutions $x$ and $y$ of the problem must be positive real numbers different from 1. Notice that
$$
\log _{x} 225=\frac{1}{\log _{225} x} \quad \text{and} \quad \log _{y} 64=\frac{1}{\log _{64} y}
$$
If we introduce the substitutions $\log _{225} x=m$ and $\log _{64} y=n$, the system transforms into... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak B-2.1.
Odredite vrijednost brojevnog izraza $\sqrt{2}+\left(\frac{3}{\sqrt[3]{4}-\sqrt[3]{2}+1}-1\right)^{3}-\frac{\sqrt{2}}{2+\sqrt{2}}$.
| ## Rješenje.
Racionalizirajmo najprije nazivnike zadanih razlomaka.
$\frac{3}{\sqrt[3]{4}-\sqrt[3]{2}+1}=\frac{3}{\sqrt[3]{4}-\sqrt[3]{2}+1} \cdot \frac{\sqrt[3]{2}+1}{\sqrt[3]{2}+1}=\frac{3 \cdot(\sqrt[3]{2}+1)}{2+1}=\sqrt[3]{2}+1$
$\frac{\sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}}{2+\sqrt{2}} \cdot \frac{2-\sqrt{2}}{2-\... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.1.
If $\operatorname{ctg} x=\frac{1}{2}$, what is $\frac{(\cos x-1+\sin x)(\cos x+1-\sin x)}{\cos x-\cos x \sin x}$ ? | ## First Solution.
If in the numerator of the given expression we notice a difference of squares or multiply the parentheses, and in the denominator we factor out $\cos x$, we get successively:
$$
\begin{aligned}
& \frac{(\cos x-1+\sin x)(\cos x+1-\sin x)}{\cos x-\cos x \sin x}=\frac{[\cos x-(1-\sin x)][\cos x+(1-\si... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.7.
It has been established that the number of student absences in the third grade during the school year can be described by the function $f(t)=2^{t+a}+b$, where $t$ is the number of months that have passed since the start of the school year on September 1st, $f(t)$ is the total number of absences in the f... | ## Solution.
a) First, let's determine the constants $a$ and $b$ in the function $f(t)=2^{t+a}+b$.
Given that students were absent for 128 hours in December, we write:
$$
f(3)-f(2)=128
$$
According to the given rule, we have:
$$
\begin{gathered}
2^{3+a}+b-\left(2^{2+a}+b\right)=128 \\
8 \cdot 2^{a}-4 \cdot 2^{a}=1... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.5.
If $z=-\sin \frac{2022 \pi}{5}-i \cos \frac{2023 \pi}{5}$, what is $z^{10}$? | ## Solution.
By applying the periodicity properties of sine and cosine functions and reduction formulas, we have the following sequence of equalities:
$$
\begin{aligned}
& z=-\sin \frac{2022 \pi}{5}-i \cos \frac{2023 \pi}{5} \\
& z=-\sin \left(404 \pi+\frac{2 \pi}{5}\right)-i \cos \left(404 \pi+\frac{3 \pi}{5}\right)... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.3.
In how many ways can the number 455 be written as the sum of an increasing sequence of two or more consecutive natural numbers? | ## Solution.
Let $n$ be the first number in a sequence of $k+1$ consecutive numbers, $k \geqslant 1$. Then,
$$
\begin{aligned}
455=n+(n+1)+(n+2)+\cdots+(n+k) & =(k+1) n+(1+2+3+\cdots+k) \\
& =(k+1) n+\frac{k(k+1)}{2}=\frac{(k+1)(2 n+k)}{2}
\end{aligned}
$$
which implies $910=(2 n+k)(k+1)$.
The prime factorization o... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Zadatak B-2.1.
Koliko je $1+z^{2}+z^{4}+\cdots+z^{2 \cdot 2019}$, ako je $z=\frac{1+\sqrt{3} i}{2}$ ?
| ## Prvo rješenje.
Izračunajmo redom
$$
\begin{aligned}
& z^{2}=\left(\frac{1+\sqrt{3} i}{2}\right)^{2}=\frac{2 \sqrt{3} i-2}{4}=\frac{\sqrt{3} i-1}{2} \\
& z^{3}=\frac{1+\sqrt{3} i}{2} \cdot \frac{\sqrt{3} i-1}{2}=\frac{-3-1}{4}=-1 \\
& z^{4}=-z \\
& z^{6}=z^{4} \cdot z^{2}=-z^{3}=1 \\
& z^{8}=z^{6} \cdot z^{2}=z^{2}... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.2.
All natural numbers from 1 to some number were written on a board. After one of the numbers was erased, the arithmetic mean of the remaining numbers on the board is $\frac{673}{18}$. Which number was erased? | ## First Solution.
Let the largest written natural number be $n$, and the erased number be $k$. The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.
The condition of the problem can be written as
$$
\frac{\frac{n(n+1)}{2}-k}{n-1}=\frac{673}{18}
$$
If the erased number $k$ were $n$, we would get the small... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.3.
A billiard table has the shape of a rectangle $A B C D$ with dimensions $|A B|=2 \mathrm{~m}$ and $|B C|=1 \mathrm{~m}$. A billiard ball moves straight on the table until it reaches the edge of the rectangle, at which point it bounces off such that the path of the ball before and after the bounce forms ... | ## First Solution.
Let $E, F,$ and $G$ be the points where the billiard ball bounces off the sides $\overline{C D}, \overline{B C},$ and $\overline{A B}$, respectively. The condition of the... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.1.
Determine all real solutions of the equation
$$
\sqrt{x^{2}+x+3}-\sqrt{x^{2}-2 x+4}=1
$$ | ## Solution.
By moving one expression from the left side of the equation to the right and squaring, we get
$$
\begin{array}{cc}
-\sqrt{x^{2}-2 x+4}=1-\sqrt{x^{2}+x+3} & \\
x^{2}-2 x+4=1-2 \sqrt{x^{2}+x+3}+x^{2}+x+3 & 1 \text { point } \\
-3 x=-2 \sqrt{x^{2}+x+3} & 3 \text { points } \\
3 x=2 \sqrt{x^{2}+x+3} &
\end{a... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.4.
A semicircle with diameter $\overline{P Q}$ is inscribed in rectangle $A B C D$ and touches its sides $\overline{A B}$ and $\overline{A D}$. Point $P$ is located on side $\overline{B C}$, and point $Q$ is on side $\overline{C D}$. If $|B P|=2$ and $|D Q|=1$, determine $|P Q|$. | ## First Solution.
Let $O$ be the center and $r$ the radius of the given semicircle. Let points $K, L, M$, and $N$ be the orthogonal projections of point $O$ onto sides $\overline{C D}, \ov... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.2.
Let $x_{1}, x_{2}$, and $x_{3}$ be distinct roots of the polynomial $P(x)=x^{3}-3 x+1$. Determine
$$
\frac{1}{x_{1}^{2}-3}+\frac{1}{x_{2}^{2}-3}+\frac{1}{x_{3}^{2}-3}
$$ | ## First Solution.
For the root $x_{1}$ of the polynomial $P$, we have
$$
\begin{aligned}
x_{1}^{3}-3 x_{1}+1 & =0 \\
x_{1}^{3}-3 x_{1} & =-1 \\
x_{1}^{2}-3 & =-\frac{1}{x_{1}} \\
\frac{1}{x_{1}^{2}-3} & =-x_{1} .
\end{aligned}
$$
Similarly, this holds for $x_{2}$ and $x_{3}$, so for the desired expression we have
... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.4.
Determine all prime numbers $p$ for which there exist exactly five natural numbers $n$ such that $n+p \mid n^{3}+p^{2}$. | ## First Solution.
The sum of cubes $n^{3}+p^{3}=(n+p)\left(n^{2}-n p+p^{2}\right)$ is divisible by $n+p$. Since, according to the problem's condition, $n^{3}+p^{2}$ is divisible by $n+p$, it is equivalent to say that
$$
\left(n^{3}+p^{3}\right)-\left(n^{3}+p^{2}\right)=p^{3}-p^{2}=p^{2}(p-1)
$$
is divisible by $n+p... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.5.
For real numbers $a, b, c$, which are not equal to zero, it holds that $a+b+c=0$. What is
$$
\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b} ?
$$ | ## Solution.
If $a+b+c=0$, then $c=-(a+b)$.
The given expression is reduced to a common denominator and we substitute $c=-(a+b)$. We get
$$
\begin{array}{rlrl}
\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b} & =\frac{a^{3}+b^{3}-(a+b)^{3}}{-a b(a+b)} & & 1 \text { point } \\
& =\frac{(a+b)\left(a^{2}-a b+b^{2}... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.6.
A student has passed 31 exams over the course of a five-year study. Each year, he took more exams than the previous year, and in the fifth year, he took three times as many exams as in the first year. How many exams did the student take in the fourth year? | ## Solution.
Let's denote the number of exams the student passed in the $i$-th year by $x_{i}, i \in\{1,2,3,4,5\}$.
Then $x_{5}=3 x_{1}, x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$ and $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=31$.
1 point
If $x_{1}=1$, then $x_{5}=3$. This is impossible because there are no 3 different natural numbers be... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.6.
If
$$
f(n)=\left(\frac{1+i}{\sqrt{2}}\right)^{n}+\left(\frac{1-i}{\sqrt{2}}\right)^{n}, \quad \text { where } n \in \mathbb{N}, i^{2}=-1
$$
what is $f(n+2016)-f(n-2016)$? | ## Solution.
We calculate
$$
\begin{aligned}
f(n+2016) & =\left(\frac{1+i}{\sqrt{2}}\right)^{n+2016}+\left(\frac{1-i}{\sqrt{2}}\right)^{n+2016} \\
& =\left(\frac{1+i}{\sqrt{2}}\right)^{2016}\left(\frac{1+i}{\sqrt{2}}\right)^{n}+\left(\frac{1-i}{\sqrt{2}}\right)^{2016}\left(\frac{1-i}{\sqrt{2}}\right)^{n}
\end{aligned... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-4.5.
For real numbers $x_{1}, x_{2}, \ldots, x_{2016}$, the following equalities hold:
$$
\frac{x_{1}}{x_{1}+1}=\frac{x_{2}}{x_{2}+2}=\cdots=\frac{x_{2016}}{x_{2016}+2016}, \quad x_{1}+x_{2}+x_{3}+\cdots+x_{2016}=2017
$$
Calculate $x_{1008}$. | ## Solution.
Notice
$$
\frac{x_{1}}{x_{1}+1}=\frac{x_{i}}{x_{i}+i}, \text { for all } i \in\{1,2, \ldots, 2016\}
$$
Then $x_{1}(x_{i}+i)=x_{i}(x_{1}+1)$, so $x_{i}=i \cdot x_{1}$ for all $i \in\{1,2, \ldots, 2016\}$. It follows that
$$
\begin{array}{cc}
x_{1}+x_{2}+\cdots+x_{2016}=2017 & \\
x_{1} \cdot(1+2+\cdots+2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-1.8. The perimeter of a right triangle is $14 \mathrm{~cm}$. A square is constructed outward on each side. The sum of the areas of all the squares is $72 \mathrm{~cm}^{2}$. What is the area of the given right triangle? | ## Solution.
Let $a$ and $b$ be the lengths of the legs, and $c$ be the length of the hypotenuse of a right triangle. It is given that
$$
\begin{aligned}
a+b+c & =14 \\
a^{2}+b^{2}+c^{2} & =72
\end{aligned}
$$
Since, by the Pythagorean theorem, $a^{2}+b^{2}=c^{2}$, the last equation becomes $2 c^{2}=72$ or $c=6 \tex... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-2.2. If the graph of the quadratic function $f(x)=x^{2}-(m-n) x+n, m, n \in \mathbb{R}$, has its vertex at point $T(2,3)$, determine the value of $f(m-n)$. | ## Solution.
The $x$-coordinate of the vertex of the quadratic function $f$ is $\frac{m-n}{2}$. Then $\frac{m-n}{2}=2$, so
$$
m-n=4
$$
(1 point)
Furthermore, the $y$-coordinate of the vertex is $\frac{4 n-(m-n)^{2}}{4}=3$. From this and (1), we get
$$
4 n-(m-n)^{2}=12
$$
or $4 n-16=12$, i.e., $n=7$.
Then $f(m-n)... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.1. Solve the equation:
$$
2^{\binom{n+1}{2}}-4 \cdot 2^{\binom{n-1}{2}}=7 \cdot 2^{\binom{n}{2}}
$$ | Solution. We solve the equation under the condition that $n \geq 3$.
$$
2^{\frac{(n+1) n}{2}}-4 \cdot 2^{\frac{(n-1)(n-2)}{2}}=7 \cdot 2^{\frac{n(n-1)}{2}}
$$
After dividing the equation by $2^{\frac{n(n-1)}{2}}$ we get
$$
2^{\frac{(n+1) n}{2}-\frac{n(n-1)}{2}}-4 \cdot 2^{\frac{(n-1)(n-2)}{2}-\frac{n(n-1)}{2}}=7
$$
... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.5. If we draw a tangent $t_{1}$ to the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$ at an angle of $45^{\circ}$ to the positive direction of the $x$-axis, its intercept on the $y$-axis is 4. If we draw a tangent $t_{2}$ at an angle of $60^{\circ}$, the intercept on the $y$-axis will increase by 2. Determine t... | Solution. Let's write the equations of the given tangents:
$$
\begin{gathered}
t_{1} \ldots k_{1}=\operatorname{tg} 45^{\circ}=1, l_{2}=4, \quad t_{2} \ldots k_{2}=\sqrt{3}, l_{2}=6 \\
t_{1} \ldots y=x+4 \\
t_{2} \ldots y=\sqrt{3} x+6
\end{gathered}
$$
From the condition of tangency $a^{2} k^{2}+b^{2}=l^{2}$, it foll... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.7. The base of a triangular pyramid is a triangle with sides $a, b$, and $c$. The length of side $c$ is $7 \, \text{cm}$, $a - b = 5 \, \text{cm}$, and the angle opposite side $c, \gamma = 60^\circ$. The lateral face containing the longest base edge is perpendicular to the base plane and congruent to the base.... | ## Solution.
Using the Law of Cosines, we will calculate the lengths of sides $a$ and $b$.
$$
\begin{aligned}
& c^{2}=a^{2}+b^{2}-2 a b \cdot \cos \gamma \\
& 49=(b+5)^{2}+b^{2}-2(b+5) b \c... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.8. Determine the sum of the reciprocals of all positive divisors of the number $n$, including 1 and $n$, if $n=2^{p-1}\left(2^{p}-1\right)$, and $2^{p}-1$ is a prime number. | Solution. The divisors of the number $n$ are the numbers
$$
1,2,2^{2}, 2^{3}, \ldots, 2^{p-1}, 2^{p}-1,2 \cdot\left(2^{p}-1\right), 2^{2} \cdot\left(2^{p}-1\right), \ldots, 2^{p-1} \cdot\left(2^{p}-1\right)
$$
The sum of their reciprocals is
$$
S=1+\frac{1}{2}+\frac{1}{2^{2}}+\ldots+\frac{1}{2^{p-1}}+\frac{1}{2^{p}-... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Riješite jednadžbu
$$
\log _{2}^{2}(x+y)+\log _{2}^{2}(x y)+1=2 \log _{2}(x+y)
$$ | 1. Zapišimo jednadžbu u obliku
$$
\log _{2}^{2}(x y)+\left(\log _{2}(x+y)-1\right)^{2}=0
$$
Time dobivamo sustav jednadžbi
$$
\log _{2}(x y)=0, \quad \log _{2}(x+y)=1
$$
Odavde dobivamo jednadžbe
$$
x y=1, \quad x+y=2
$$
čije rješenje je $x=y=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task B-1.5.
Two friends, Esma and Ljerka, came up with a game: each letter of their names is replaced by a digit from 1 to 5 according to a given table. Thus, the name Esma has the code 4431, and Ljerka has 24351.
The digits of the obtained code $a_{1} a_{2} \ldots a_{k}$ are used for moving in steps of the same l... | ## Solution.
Without loss of generality, assume that Esma and Ljerka started from the point $(0,0)$.
Esma's movement looks like this:
First loop: $(0,0),(0,4),(-4,4),(-4,1),(-3,1)$
Secon... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## Task B-2.5.
The math test consisted of three problems. Professor Max informed the students of their test results in a somewhat vague manner, setting a new problem:
The first and second problems were correctly solved by 7 students, the second and third problems by 9 students, and the first and third problems by as ... | ## Solution.
Let the number of students who solved only the first and second problems be denoted by $a$, only the first and third by $b$, only the second and third by $c$. Let $d$ be the number of students who correctly solved all three problems, $e$ the number who correctly solved only the first problem, $f$ only the... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task B-3.4.
In a square $A B C D$, a circle $k$ is inscribed. From a point $T$ which lies on the circle $k$, the diagonal $\overline{A C}$ is seen at an angle $\alpha$, and the diagonal $\overline{B D}$ at an angle $\beta$.
Prove that $\operatorname{tg}^{2} \alpha+\operatorname{tg}^{2} \beta=8$. | ## First Solution.
Let the radius of the circle $k$ be $r$, and its center be the origin of the coordinate system $O$. The coordinates of the vertices of the square are $A(-r,-r), B(r,-r), C(r, r), D(-r, r)$, and the coordinates of point $T$ are $(x, y)$.
Point $T$ is at a distance $r$ from the center, so $\sqrt{x^{2... | 8 | Geometry | proof | Yes | Yes | olympiads | false |
## Task B-1.4.
A lively hedgehog is hopping through a meadow, first moving 4 meters to the east, then 2 meters southwest, and finally 4 meters to the south. How far is the hedgehog from its starting position? | ## Solution.
For the sketch.
The angle $\measuredangle E B C=45^{\circ}$, so $|E B|=|E C|=\sqrt{2}$.
$$
\begin{aligned}
|A E| & =4-\sqrt{2},|E D|=4+\sqrt{2} \\
|A D| & =\sqrt{|A E|^{2}+|E ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-3.6. A given isosceles triangle has a base length of $6 \mathrm{~cm}$ and legs of length 5 cm. A circle whose diameter is one of the legs of the triangle intersects the other two sides of the triangle at points $E$ and $F$. Determine the length of the segment $|E F|$. | ## Solution.
Sketch.
$|A C|=|B C|=5 \text{ cm}, |A B|=6 \text{ cm}$
$\measuredangle C E B=90^{\circ}$ (angle subtended by the diameter of the circle)
(1 point)
We conclude that point E ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task B-4.3. Find all solutions to the equation
$$
\frac{1}{\binom{4}{x}}-\frac{1}{\binom{5}{x}}=\frac{1}{\binom{6}{x}}
$$ | Solution. From the definition of the binomial coefficient, we have
$$
\binom{4}{x}=\frac{4!}{x!(4-x)!},\binom{5}{x}=\frac{5!}{x!(5-x)!},\binom{6}{x}=\frac{6!}{x!(6-x)!}
$$
so the given equation becomes
$$
\frac{x!(4-x)!}{4!}-\frac{x!(5-x)!}{5!}=\frac{x!(6-x)!}{6!}, \text { with the condition } x \leq 4
$$
The equat... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.2.
Let $x$ and $y$ be real numbers such that $x+y=1$ and $x^{3}+y^{3}=13$. What is $x^{2}+y^{2}$? | ## First Solution.
From the formula for the cube of a sum
$$
(x+y)^{3}=x^{3}+3 x^{2} y+3 x y^{2}+y^{3}=x^{3}+y^{3}+3 x y(x+y)
$$
it follows that
$$
3 x y(x+y)=(x+y)^{3}-\left(x^{3}+y^{3}\right)
$$
By substituting $x+y=1$ and $x^{3}+y^{3}=13$, we have
$$
3 x y=1^{3}-13=-12
$$
which means $x y=-4$.
Now, from the ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-1.3.
Dino, Pino, and Tino go to the same kindergarten. For the game, each boy needs two dice of the same color, but it is not necessary that the dice of different boys be of different colors. The caregiver has red, blue, and green dice in one box. If she draws without looking, how many dice does she need to ... | ## First Solution.
Two dice of the same color form a pair. Therefore, for each color, the caregiver draws some number of pairs and at most one additional die. If the caregiver were to draw dice among which there are no three pairs, the number of pairs could be at most two, and with those two pairs, there could be at m... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## Task A-2.7.
Two teams are playing handball. Neither team has scored 30 or more goals. The recorder at the beginning of the game and after each goal scored records the score and calculates the sum of all digits in the score. For example, with a score of 15 : 6, the sum of the digits is 12. How many times during the ... | ## Solution.
At each change of the result, exactly one of the numbers $\overline{a b}$ and $\overline{c d}$ increases by one, while the other remains unchanged.
Consider some (fixed) $a$ and $c$. Let $b$ and $d$ be such that $a+b+c+d=10$. With each subsequent change of the result, one of them will increase, and the o... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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