problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
3. A right-angled triangle has integer lengths of sides. Its perimeter is the square of a natural number. We also know that one of its legs has a length equal to the square of a prime number. Determine all possible values of this length.
(Patrik Bak) | SOLUTION. We are looking for all prime numbers $p$ for which there exists a described triangle with one leg of length $p^{2}$. Let the length of its other leg be $b$ and the length of the hypotenuse $c$. Then, according to the Pythagorean theorem, the equation $c^{2}=p^{4}+b^{2}$ holds, which we rearrange into a produc... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Natural numbers $a$, $b$, $c$ are placed in a circle as shown in the figure, with each number being a divisor of the sum of the two numbers adjacent to it. How many of the numbers $a$, $b$, $c$ can be different? (Josef Tkadlec)
 Determine the size of angle $B... | SOLUTION. We will describe several solution procedures for both parts a) and b). Without comment, we will use the known equalities $|A S|=|B S|=|C S|=|D S|$. Let us also emphasize that the condition $|A B|>|B C|$ from the problem statement ensures that point $E$ lies on the extension of the diagonal $A C$ beyond point ... | 60 | Geometry | proof | Yes | Yes | olympiads | false |
4. The sequence $\left(a_{n}\right)_{n=0}^{\infty}$ of non-zero integers has the property that for every $n \geqq 0$, $a_{n+1}=a_{n}-b_{n}$, where $b_{n}$ is the number that has the same sign as the number $a_{n}$, but the digits are in reverse order (the representation of $b_{n}$ may start with one or more zeros compa... | SOLUTION. a) To prove that the sequence $\left(a_{n}\right)$ is periodic, it suffices to show that there exist natural numbers $n_{0}$ and $p$ such that $a_{n_{0}+p}=a_{n_{0}}$. Since each subsequent term of the sequence is uniquely determined by the preceding term, it will hold for every $n \geqq n_{0}$ that $a_{n+p}=... | 1012 | Number Theory | proof | Yes | Yes | olympiads | false |
3. How many three-digit numbers have the property that by erasing some digit we get a two-digit number which is a square of some integer? (We do not consider representations like 07 as two-digit numbers.)
(Tomáš Bárta, Tomáš Jurík) | Solution. Let's denote the digit we strike out from the sought three-digit number as $x$ and the resulting two-digit number as $\overline{a b}(a \neq 0)$. Two-digit numbers that are squares of integers are only the numbers in the set
$$
\mathrm{M}=\{16,25,36,49,64,81\}
$$
If we strike out the first digit, i.e., from ... | 156 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. For non-negative real numbers $a, b, c$ it holds that $a+b+c=1$. Find the greatest and the smallest possible value of the expression
$$
(a+b)^{2}+(b+c)^{2}+(c+a)^{2} .
$$ | Solution. Notice that the expression $a+b+c$ contains only the first powers of the variables, and the expression we need to work with further contains their second powers. By expanding and rearranging the expression $V$ from the problem statement, we get
$$
\begin{aligned}
V & =(a+b)^{2}+(b+c)^{2}+(c+a)^{2}=2\left(a^{... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find all natural numbers \( n \) for which the following equality holds:
\[
n + d(n) + d(d(n)) + \cdots = 2021
\]
where \( d(0) = d(1) = 0 \) and for \( k > 1 \), \( d(k) \) is the superdivisor of \( k \) (i.e., its greatest divisor \( d \) with the property \( d < k \)).
(Tomáš Bárta) | SOLUTION. Similarly to the home problem, we use the fact that for the superdivisor $d(m)$ of an integer $m>1$, the formula $d(m)=m / p$ holds, where $p$ is the smallest prime factor of the number $m$.
Let's factorize the sought $n$ into primes: $n=p_{1} p_{2} \ldots p_{k}$, where $p_{1} \geqq p_{2} \geqq \ldots \geqq ... | 1919 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find all eight-digit numbers such that by erasing some quartet of consecutive digits, we obtain a four-digit number that is 2019 times smaller.
\end{abstract}
(Pavel Calábek) | SOLUTION. In the first part of the solution, we will assume that in the sought eight-digit number $N$, we have crossed out the first four digits. These form a four-digit number, which we will denote as $A$. The remaining four digits of the number $N$ form a four-digit number $B$, which is mentioned in the problem, and ... | 10095000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Given a triangle $A B C$ with side $B C$ of length $22 \mathrm{~cm}$ and side $A C$ of length $19 \mathrm{~cm}$, whose medians $t_{a}, t_{b}$ are perpendicular to each other. Calculate the length of side $A B$. | 2. Let $D$ be the midpoint of side $A C$, $E$ the midpoint of side $B C$, and $T$ the centroid of triangle $A B C$ (Fig. 1). If we denote $3 x$ and $3 y$ as the lengths of the medians $t_{a}$ and $t_{b}$, we have $|A T|=2 x,|E T|=x,|B T|=2 y$, $|D T|=y$. From the problem statement, it follows that triangles $A T D, B E... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. A natural number is called wavy if for every three consecutive digits $a, b, c$ in its decimal representation, $(a-b)(b-c)<0$. Prove that it is possible to construct more than 25000 ten-digit wavy numbers using the digits $0,1, \ldots, 9$ (the digit 0 cannot be in the first position) which contain all the digits fro... | 3. Let's call the digits $0,1,2,3$ and 4 small (abbreviated as $m$), and the digits $5,6,7,8$ and 9 large (abbreviated as $v$). Regular alternation of small and large digits always results in a wavy number.
The number of numbers of the form vmvmvmvmvm is (5!)^2, and the number of numbers of the form mvmvmvmvmv is $4 \... | 25920 | Combinatorics | proof | Yes | Yes | olympiads | false |
1. Determine the values that the expression $V=a b+b c+c d+d a$ can take, given that the real numbers $a, b, c, d$ satisfy the following conditions:
$$
\begin{aligned}
& 2 a-5 b+2 c-5 d=4, \\
& 3 a+4 b+3 c+4 d=6 .
\end{aligned}
$$ | 1. For the given expression $V$, we have
$$
V=a(b+d)+c(b+d)=(a+c)(b+d).
$$
Similarly, we can modify both given conditions:
$$
2(a+c)-5(b+d)=4 \quad \text { and } \quad 3(a+c)+4(b+d)=6
$$
If we choose the substitution $m=a+c$ and $n=b+d$, we get the solution of the system (1) as $m=2$ and $n=0$. For the given expres... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On a $7 \times 7$ board, we are playing a game of battleships. There is one ship of size $2 \times 3$ on it. We can ask about any square on the board, and if we hit the ship, the game ends. If not, we ask again. Determine the smallest number of questions we need to ask to be sure to hit the ship. | SOLUTION. According to Fig. 2, we can place 8 disjoint rectangles $2 \times 3$ on the board (the middle cell of the board will remain empty). To definitely hit the ship, we need to ask about at least one cell in each of the eight marked rectangles, so the necessary number of questions is at least 8.
In Fig. 3, an exam... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Let A, B be sets of positive integers such that the sum of any two different numbers from $A$ belongs to $B$ and the quotient of any two different numbers from $B$ (the larger divided by the smaller) lies in $A$. Determine the maximum possible number of elements in the set $A \cup B$.
| 2. First, we will prove that the set $\mathrm{A}$ can contain at most two numbers. Assume that $\mathrm{A}$ contains three numbers $a < b < c$. Then the set $\mathrm{B}$ contains the numbers $a+b < a+c < b+c$, and thus the set $\mathrm{A}$ must contain the number
$$
\frac{b+c}{a+c}=1+\frac{b-a}{a+c}
$$
which is not a... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Determine the number of all infinite arithmetic sequences of integers that have both numbers 1 and 2005 among their first ten terms. | 1. Let's consider the question for which integer arithmetic sequences $\left(a_{i}\right)_{i=1}^{\infty}$ there exist indices $i, j \in\{1,2, \ldots, 10\}$ such that $a_{i}=1$ and $a_{j}=2005$. Emphasize that such a pair of indices $(i, j)$, if it exists at all, is unique, because in a non-constant arithmetic sequence,... | 68 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Determine all real numbers $s$ for which the equation
$$
4 x^{4}-20 x^{3}+s x^{2}+22 x-2=0
$$
has four distinct real roots, with the product of two of them being equal to -2. | 1. Assume that the number $s$ satisfies the problem statement, and denote the roots $x_{1}, x_{2}, x_{3}, x_{4}$ of the given equation such that
$$
x_{1} x_{2}=-2 \text {. }
$$
From the factorization into root factors
$$
4 x^{4}-20 x^{3}+s x^{2}+22 x-2=4\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\le... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Determine the number of all quadratic polynomials $P(x)$ with integer coefficients such that for every real number $x$ the following holds:
$$
x^{2}+2 x-2023<P(x)<2 x^{2} \text {. }
$$
(Ján Mazák, Michal Rolínek) | SOLUTION. Suitable polynomials $P(x)$ should have the form $P(x)=a x^{2}+b x+c$, where $a$, $b, c$ are integers. If the coefficient of $x^{2}$ were $a>2$. The remaining two cases $a \in\{1,2\}$ will be evaluated individually.
For $a=1$, we rewrite the given inequalities as
$$
(b-2) x+(c+2023)>0 \quad \text { and } \q... | 4042 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. We need to fill a $3 \times 3$ table with nine given numbers such that in each row and column, the largest number is the sum of the other two. Determine whether it is possible to complete this task with the numbers
a) 1, 2, 3, 4, 5, 6, 7, 8, 9;
b) 2, 3, 4, 5, 6, 7, 8, 9, 10.
If yes, find out how many ways the tas... | SOLUTION. The sum of the numbers in each row (and in each column) is even, equaling twice the largest of the three written numbers. Therefore, the sum of all the numbers in the table must also be even. Since in case a) the sum of the numbers is 45, the task cannot be completed.
Since the sum of the numbers in case b) ... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Find the largest natural number $n$ such that the value of the sum
$$
\lfloor\sqrt{1}\rfloor+\lfloor\sqrt{2}\rfloor+\lfloor\sqrt{3}\rfloor+\ldots+\lfloor\sqrt{n}\rfloor
$$
is a prime number. The notation $\lfloor x\rfloor$ denotes the greatest integer not greater than $x$. | SOLUTION. Consider the infinite sequence $\left(a_{n}\right)_{n=1}^{\infty}$ of natural numbers $a_{n}=\lfloor\sqrt{n}\rfloor$. This sequence is clearly non-decreasing, and since
$$
k=\sqrt{k^{2}}6$ the value of the fraction
$$
\frac{(k-1) k(4 k+1)}{6}
$$
remains a natural number even after canceling out six by a na... | 47 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Find the largest possible number of elements in the set $\mathrm{M}$ of integers, which has the following property: for every triple of distinct numbers $z \mathrm{M}$, it is possible to select some two of them whose sum is a power of 2 with an integer exponent. | SOLUTION. Let us note at the outset that the mentioned power as an integer sum must have a non-negative exponent. We will only consider such powers of the number 2 further on.
We will prove that the set M can have at most 6 elements, as does, for example, the suitable (as we will verify immediately) set
$$
\{-1,3,5,-... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Determine the smallest natural number $k$, for which the following holds: If we select any $k$ different numbers from the set $\{1,2,3, \ldots, 2000\}$, then among the selected numbers there exist two whose sum or difference is 667.
School - written part of the first round, category A takes place
## on Tuesday, De... | 3. We will show that the desired $k$ is the number 1001. Let's divide all numbers from the set $\{1,2,3, \ldots, 2000\}$ into 1000 pairs
$$
\begin{gathered}
\{1,666\},\{2,665\},\{3,664\}, \ldots,\{333,334\}, \\
\{667,1334\},\{668,1335\},\{669,1336\}, \ldots,\{1333,2000\} .
\end{gathered}
$$
(The number 667 from the p... | 1001 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. At the party, 10 boys and 10 girls gathered. Each boy likes a different positive number of girls. Each girl likes a different positive number of boys. Determine the largest integer n with the following property: It is always possible to form n disjoint pairs where both individuals like each other.
(Josef Tkadlec) | SOLUTION. We will show that the largest $n$ sought is equal to 1.
In the first part of the solution, we will show that one suitable pair can always be formed. According to the problem statement, the numbers of girls liked by individual boys are all the numbers from 1 to 10 (in some order). The same applies to the numb... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Let $n$ be the sum of all ten-digit numbers that have each of the digits $0,1, \ldots, 9$ in their decimal representation. Determine the remainder when $n$ is divided by seventy-seven. | SOLUTION. First, we will find the value of the number $n$, then it will be easy to determine its remainder when divided by 77. We will not directly add the ten-digit numbers described in the problem. The desired sum can be more easily found by determining how many times each digit appears in all the addends at the unit... | 28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Find the greatest real number $k$ such that the inequality
$$
\frac{2\left(a^{2}+k a b+b^{2}\right)}{(k+2)(a+b)} \geqq \sqrt{a b}
$$
holds for all pairs of positive real numbers $a, b$. | SOLUTION. For $k=2$, the inequality can be simplified to $\frac{1}{2}(a+b) \geqq \sqrt{a b}$, which is the well-known inequality between the arithmetic and geometric means, valid for any positive numbers $a, b$. The largest $k$ we are looking for is therefore at least 2. Let's further examine the given inequality under... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. How many non-empty subsets of the set $\{0,1, \ldots, 9\}$ have the sum of their elements divisible by three?
(Eliška Macáková) | SOLUTION. We start from the fact that the remainder of the sum of several integers when divided by three is the same as the remainder of the number obtained by summing the remainders of the individual addends.* Therefore, we will divide the 10 given numbers into three sets according to their remainders (when divided by... | 351 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
For real numbers $a, b, c$ it holds that
$$
\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}
$$
Determine all possible values of the expression
$$
\frac{a^{3}+b^{3}+c^{3}}{(b+c)^{3}+(c+a)^{3}+(a+b)^{3}}.
$$
(Michal Rolínek) | SOLUTION. From the problem statement, it follows that the numbers $a, b, c$ satisfy the conditions
$$
b+c \neq 0, \quad c+a \neq 0, \quad a+b \neq 0 .
$$
Under the assumptions $(\mathrm{P})$, we equivalently transform the first of the given equalities:
$$
\begin{gathered}
\frac{a}{b+c}=\frac{b}{c+a}, \\
a(c+a)=b(b+c... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Consider two quadratic equations
$$
x^{2}-a x-b=0, \quad x^{2}-b x-a=0
$$
with real parameters $a, b$. Determine the smallest and the largest values that the sum $a+b$ can take, given that there exists exactly one real number $x$ that satisfies both equations simultaneously. Also, determine all pairs $(a, b)$ of r... | 1. By subtracting the given equations, we obtain the equality $(b-a) x+a-b=0$ or $(b-a)(x-1)=0$, from which it follows that $b=a$ or $x=1$.
If $b=a$, both equations take the form $x^{2}-a x-a=0$. Exactly one solution exists if and only if the discriminant $a^{2}+4 a$ is zero. This holds for $a=0$ and for $a=-4$. Since... | -8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the plane containing the segment $B D$, find the set of all vertices $A$ of convex quadrilaterals $A B C D$ for which the following conditions are simultaneously satisfied:
a) the center $O_{C}$ of the incircle of triangle $B C D$ lies on the circumcircle of triangle $A B D$,
b) the center $O_{A}$ of the incirc... | Solution: Let $\alpha=|\angle B A D|$ and $\gamma=|\angle B C D|$. We have
$$
\left|\angle B O_{C} D\right|=180^{\circ}-\left(\left|\angle O_{C} B D\right|+\left|\angle O_{C} D B\right|\right)=
$$
$$
=180^{\circ}-\frac{1}{2}(|\angle C B D|+|\angle C D B|)=90^{\circ}+\frac{1}{2}
$$
Similarly, $\left|\angle B O_{A} D\... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. When dividing a certain natural number by 19 and 99, the remainders are two prime numbers. The sum of both incomplete quotients equals 1999. Determine the number being divided. | SOLUTION. Both divisions of the sought number $N$ can be expressed by the equations
$$
N=19 a+p \quad \text { and } \quad N=99 b+q
$$
where $a, b$ are the respective incomplete quotients and $p, q$ are the respective remainders. According to the problem, the numbers $p, q$ are prime numbers, and as remainders, they s... | 31880 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The number $a_{n}$ is formed by writing down the first n consecutive natural numbers in sequence, for example, $a_{13}=12345678910111213$. Determine how many numbers divisible by 24 are among the numbers $a_{1}, a_{2}, \ldots, a_{10000}$. | SOLUTION. A natural number is divisible by 24 if and only if it is simultaneously (and relatively prime) divisible by the numbers 3 and 8. For the digit sum of a natural number $k$, let's denote it as $S(k)$. The number $a_{n}$ is divisible by three if and only if its digit sum, i.e., the number $S(1)+S(2)+\ldots+S(n)$... | 834 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Teacher Kadrnozkova was buying tickets at the cash register of the zoo for her students and for herself. The ticket for an adult was more expensive than for a schoolchild, but not more than twice as much. Teacher Kadrnozkova paid a total of 994 CZK. Teacher Hnizdo had three more students with him than his colleague, so... | From the given information, it directly follows that the entrance fee for three students cost $1120 - 994 = 126$ (CZK), so for one student it was $126 : 3 = 42$ (CZK). For 1120 CZK, Teacher Hnízdo could buy tickets for a maximum of 26 students (and 28 CZK would remain), because $1120 : 42 = 26$ (remainder 28). When we ... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
František Nudílek was engaged in writing consecutive natural numbers. He started like this: 1234567891011... After a while, he got bored, finished writing the current number, and critically examined his creation. He noticed that in the sequence of digits he had written, there were five consecutive ones.
1. How many co... | 1. To have five ones in a row, numbers greater than 110 must be written, and the sequence looks like this:
$$
123456789101112 \ldots 110 \underline{111112} \ldots
$$
František wrote at least 112 consecutive natural numbers.
2. For counting the digits, we realize that in the written sequence, there are 9 one-digit nu... | 112 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The highest known volcano on Earth is Mauna Kea in the Hawaiian Islands. Its height from base to summit is even 358 meters greater than the elevation of the world's highest mountain, Mount Everest. However, it does not rise from land but from the bottom of the Pacific Ocean, from a depth of 5000 meters. If the sea leve... | 2. If the sea level dropped by $397 \mathrm{~m}$, the submerged part of the volcano would be $5000-397=4603(\mathrm{~m})$. Therefore, the height of Mauna Kea from its base to the summit would be $4603+4603=$ $=9206(\mathrm{~m})$.
1. It follows that the summit of the volcano is at an elevation of $9206-5000=4206(\mathr... | 4206 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Digital clocks show hours and minutes, such as 14:37.
How many minutes per day does at least one five light up on these clocks? | A 5 cannot light up in the first position. We will first consider the interval of the first 12 hours:
- A 5 lights up in the second position for 60 minutes (from 5:00 to 5:59); for the other positions, we therefore consider only the remaining 11 hours.
- A 5 lights up in the third position every hour for 10 minutes (f... | 450 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Jana had to calculate the product of two six-digit numbers for her homework. When copying from the board, she omitted one digit from one of the numbers, and instead of a six-digit number, she wrote only 85522. When she got home, she realized her mistake. However, she remembered that the number she had copied incorrectl... | An integer is divisible by three if and only if the sum of its digits is divisible by three. The sum of the digits of the given number is
$$
8+5+5+2+2=22
$$
which is not a number divisible by three. Therefore, the missing digit was not 0, and the digit sum of the original number must be greater. The next larger numbe... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the store, they have one type of lollipops and one type of nougats. The price of both lollipops and nougats is given in whole groats.
Barborka bought three lollipops. Eliška bought four lollipops and several nougats - we only know that it was more than one and less than ten nougats. Honzík bought one lollipop and o... | Barborka paid 24 groshen for three lollipops, so one lollipop cost 8 groshen $(24: 3=8)$.
Eliška paid 109 groshen for four lollipops and several nougats. Four lollipops cost 32 groshen $(4 \cdot 8=32)$, so her nougats cost 77 groshen $(109-32=77)$.
This price is the product of the number of nougats and the price of o... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A tourist group organized a three-day cycling trip. On the first day, they wanted to cover $\frac{1}{3}$ of the entire route, but unfortunately, they fell short by $4 \mathrm{~km}$. On the second day, they aimed to cover half of the remaining distance, but it ended up being $2 \mathrm{~km}$ less. On the third day, howe... | We proceed by reasoning "backwards".
$\frac{1}{11}$ of the remaining distance after the second day is equal to $4 \text{ km}$, so the entire remainder was $11 \cdot 4 = 44 (\text{ km})$. If on the second day they had traveled (as planned) $2 \text{ km}$ more, the remainder after the second day would have been $2 \text... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The organizer of the exhibition "I build, you build, we build" divided the exhibition into two parts. Since he was interested in the visitors' reactions to the exhibition, every visitor filled out a simple questionnaire upon leaving. The following interesting facts emerged from it:
- $96 \%$ of visitors who liked the ... | Let $n$ be the number of all people who visited the exhibition, $p$ the number of visitors who liked the first part of the exhibition, and $d$ the number of visitors who liked the second part of the exhibition. We are looking for a relationship between $n$ and $p$, from which we can easily derive the answer to the ques... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Along the beach, parking spaces of the same length are marked. Vehicles park in a row, and all drivers observe the marked spaces. The parking lot is reserved only for buses, and each bus occupies exactly three parking spaces.
At a certain moment, only two buses were standing in the parking lot - one occupied the fifth... | We need to determine the maximum number of spaces on the parking lot, which means we need to find out how inefficiently the buses can be parked. We consider such arrangements where the gaps between buses are as large as possible, and at the same time, no additional bus can fit into these gaps. From the problem statemen... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A circle with center $S$ and radius $39 \, \text{mm}$ is given. We need to inscribe a triangle $ABC$ in the circle such that the length of side $AC$ is $72 \, \text{mm}$ and point $B$ lies in the half-plane determined by the line $AC$ and point $S$.
From the given data, calculate the height of triangle $ABC$ from vert... | The length of side $AC$ is indeed smaller than the diameter of the circle, so this side does not contain the center $S$. The height of triangle $ABC$ from vertex $B$ can be arbitrarily small, but it cannot be arbitrarily large: this height is the largest when it contains the center $S$. In such a case, the problem has ... | 54 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
If we walk along the fence from north to south, the distances between its posts are initially the same. From a certain post, the distance decreases to 2.9 meters and remains so until the southern end of the fence. Between the 1st and 16th posts (counted from the north), the distances do not change, and the distance bet... | In the section between the 1st and 16th columns, there are 15 spaces, and according to the problem, their width does not change. Each of them therefore measures $48: 15=3.2$ meters, and is thus 0.3 meters wider than the space at the southern end of the fence.
If all the spaces between the 16th and 28th columns measure... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Mr. Kutil wanted to paint star ornaments on 20 tiles in the bathroom. On the can of paint, it was written that the paint would cover $750 \mathrm{~cm}^{2}$. How many cans of paint did Mr. Kutil have to buy at least, if one square of the grid has an area of $1 \mathrm{~cm}^{2}$? The ornament on one tile is shown in the ... | The tile is a square with a side of $12 \mathrm{~cm}$, its area is $(12 \mathrm{~cm})^{2}=144 \mathrm{~cm}^{2}$.
Let's calculate the area of the parts that will not be colored:
4 pentagons in the corners have a total area of $4 \cdot 15 \mathrm{~cm}^{2}=60 \mathrm{~cm}^{2}$,
4 isosceles triangles in the middle of th... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Anička has 50 Kč, Anežka has 46 Kč, and they want to buy snacks for a family celebration with all their money. They are deciding between cupcakes and windmills: a windmill is 4 Kč more expensive than a cupcake, and with all their money, they could buy a third more cupcakes than windmills.
How much does each snack cost... | Anička and Anežka together have 96 Kč. This amount can only be spent in a few ways, which are derived from expressing the number 96 as a product of two natural numbers:
$$
96=1 \cdot 96=2 \cdot 48=3 \cdot 32=4 \cdot 24=6 \cdot 16=8 \cdot 12 \text {. }
$$
The price of a kite and the price of a cake should correspond t... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Determine for how many natural numbers greater than 900 and less than 1001 the digital sum of the digital sum of their digital sum is equal to 1.
(E. Semerádová)
Hint. What is the largest digital sum of numbers from 900 to 1001? | Among the numbers 900 and 1001, the number with the greatest digit sum is 999, which is 27; we don't need to consider larger sums.
Among the numbers 1 and 27, the number with the greatest digit sum is 19, which is 10; we don't need to consider larger sums.
Among the numbers 1 and 10, the digit sum of 1 is only found ... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
King Pan was distributing ducats to his sons. He gave a certain number of ducats to his eldest son, one ducat less to the next younger son, and continued this pattern until he reached the youngest. Then he returned to the eldest son, giving him one ducat less than he had just given to the youngest, and continued distri... | For a specific number of sons, one can vividly test the king's method of distributing ducats. It's enough to proceed from the back: the youngest received one ducat in the second round, the second youngest received two ducats, etc. For example, for two, three, and four sons, the number of ducats in individual rounds wou... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Vojta started writing the number of the current school year in his notebook: 2019202020192020... and continued doing so indefinitely. When he wrote 2020 digits, he got bored and stopped.
How many twos did he write?
(L. Růžičková)
Hint. How many twos would Vojta write if he only wrote 20 digits? | The school year number 20192020 is eight digits long and contains three twos. Since $2020=8 \cdot 252+4$, Vojta wrote the entire number 20192020, 252 times, and on the remaining four places, the number 2019.
In total, Vojta wrote $252 \cdot 3+1=757$ twos. | 757 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On each wall of a regular octahedron, one of the numbers $1,2,3,4,5,6,7$ and 8 is written, with different numbers on different walls. For each wall, Jarda determined the sum of the number written on it and the numbers of the three adjacent walls. Thus, he obtained eight sums, which he also added together.
What values ... | The number on each face is counted in a total of four partial sums (each face is counted once as a middle and three times as an adjacent). Therefore, each of the numbers is also counted four times in the final sum. The resulting sum thus has the value
$$
4 \cdot(1+2+3+4+5+6+7+8)=4 \cdot 36=144
$$
and this is independ... | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Mom brought 10 snacks of three types: there were fewer coconut snacks than lollipops, and the most were caramel chews. Josef chose two different types of snacks, Jakub did the same, and for Jan, only snacks of the same type were left.
How many coconut snacks, lollipops, and caramel chews did Mom bring?
Hint. Which ty... | When Jan received $\mathrm{k}$ snacks, there were 6 of the same kind, and they were caramel chews - if they were coconuts or laskonky, there would have to be more than 6 chews and more than 10 snacks in total. Therefore, there were at least 6 caramel chews originally, and mom brought
- either 1 coconut, 3 laskonky, an... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Each brick in the following pyramid contains one number. Whenever possible, the number in each brick is the least common multiple of the numbers in the two bricks directly below it.
What number can be in the bottom brick? Determine all possibilities.
(A. Bohiniková)
The number $C$ is the least common multiple of 13 and $A$, the number $E$ is the least common multiple of $... | 2730 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\mathrm{Na}$ our planned cottage, we brought the cat Vilda. On Monday, she caught $\frac{1}{2}$ of all the mice, on Tuesday $\frac{1}{3}$ of the remaining, on Wednesday $\frac{1}{4}$ of those left after Tuesday's hunt, and on Thursday only $\frac{1}{5}$ of the remainder. On Friday, the remaining mice preferred to move... | On Monday, the cat caught $\frac{1}{2}$ of all the mice.
On Tuesday, it caught $\frac{1}{3}$ of the remaining $\frac{1}{2}$, i.e., $\frac{1}{6}$ of all the mice; $\frac{1}{2}-\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$ of all the mice remained.
On Wednesday, it caught $\frac{1}{4}$ of $\frac{1}{3}$, i.e., $\frac{1}{12}$ of ... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Adam and Eve were playing chess.
Adam won and comforted Eve: "You know, I've been playing chess for a long time, twice as long as you!" Eve got upset: "But last time you said you've been playing three times longer!"
Adam was surprised: "Did I say that? And when was that?"
"Two years ago!"
"Well, then I was telling ... | Let's assume that Eva has been playing chess for $x$ years. Then the time data appearing in the problem can be briefly expressed in the following table:
| | last year | today |
| :---: | :---: | :---: |
| Eva | $x-2$ | $x$ |
| Adam | $2 x-2$ | $2 x$ |
Last year, Adam played chess three times longer than Eva, which w... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The average age of all the people who gathered for the family celebration was equal to the number of those present. Aunt Beta, who was 29 years old, soon excused herself and left. Even after Aunt Beta left, the average age of all the people present was equal to their number.
How many people were originally at the cele... | The average age of all people who gathered for the celebration is equal to the ratio of the sum of the ages of all present (denoted as $s$) and their number (denoted as $n$). According to the problem, we have
$$
\frac{s}{n}=n \quad \text { or } \quad s=n^{2}
$$
After Aunt Betty left, the number of people present decr... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Robots Robert and Hubert assemble and disassemble coffee grinders. Each of them assembles a coffee grinder four times faster than they can disassemble it. When they came to the workshop in the morning, several coffee grinders were already assembled there.
At 7:00, Hubert started assembling and Robert started disassemb... | In the five-hour morning shift, 70 coffee grinders were added, which corresponds to $70: 5=14$ coffee grinders per hour. In the nine-hour afternoon shift, 36 coffee grinders were added, which corresponds to $36: 9=4$ coffee grinders per hour. If the robots worked one hour in the morning mode and one hour in the afterno... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the picture, you can see a pool with a long step on one of its walls. We started filling the empty pool with a constant flow and observed the water level. After
8 min, the water level had ... | Rise rate during the time $0-8 \mathrm{~min}: 20: 8=2.5(\mathrm{~cm} / \mathrm{min})$.
Rise rate during the time $23-35.5 \mathrm{~min}:(80-55):(35.5-23)=25: 12.5=2(\mathrm{~cm} / \mathrm{min})$.
The time it takes for the water level to rise to height $h$ is denoted as $t$ and is calculated using the equation:
$$
t ... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Grandma had a square garden. She bought a few neighboring plots, thus obtaining a square plot whose side was three meters longer than the side of the original garden. The area of the new plot was nine square meters larger than twice the original area.
How long was the side of the original garden?
(K. Buzáková) | Let $a$ be the length of the side of the original square garden. After the purchase, a new square plot was formed, with the side length of $a+3$. According to the areas of the plots given in the problem, we have
$$
(a+3)^{2}=2 a^{2}+9 \text {. }
$$
By equivalent transformations, we get
$$
\begin{aligned}
a^{2}+6 a+9... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Karel, Mirek, and Luděk were comparing their stamp collections. When they checked the numbers, they found that Karel and Mirek had 101 stamps together, Karel and Luděk 115 stamps, and Mirek and Luděk 110. When they verified what they could exchange, they found that none of them had the same stamp, but Karel and Mirek h... | Let's denote the number of stamps owned by Karel, Mirek, and Luděk as $K$, $L$, and $M$ respectively. According to the problem, we have:
$$
K+M=101, \quad K+L=115, \quad M+L=110
$$
By summing these three equations and performing further manipulations, we get:
$$
\begin{aligned}
2 K+2 M+2 L & =326, \\
K+M+L & =163 \\... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the picture, there is a structure composed of twelve identical cubes. To how many different places can we move the dark cube so that the surface area of the assembled body does not change?
... | Originally, on the surface of the assembled body, there are three dark cube walls, and the other three walls touch light cubes. After removing the dark cube, the total surface area of the body does not change (three dark walls are replaced by three light ones).
In order for the surface area of the body to remain uncha... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The waiter at the restaurant U Šejdíře always adds the date to the bill: he increases the total amount spent by the paying customer by as many crowns as the day of the month it is.
In September, a trio of friends met at the restaurant twice. The first time, each of them paid separately, so the waiter added the date to... | If on the day when friends uncovered the waiter's fraud, each of them paid separately, the waiter would have said to each of them $168+4=172$ (CZK). In total, they would have paid $3 \cdot 172=516$ (CZK). By paying for everyone at once, the price was reduced by an amount corresponding to two dates. According to the pro... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Three musicians, Janek, Mikeš, and Vávra, usually divide their shared honorarium in the ratio $4: 5: 6$, with Janek receiving the least and Vávra the most. This time, Vávra did not play well and thus waived his share. Janek suggested that they split Vávra's share equally between himself and Mikeš. However, Mikeš insist... | Given the sequential ratio $4: 5: 6$, we will expand it so that we can appropriately divide its third term according to both Janek's proposal (division in the ratio $1: 1$) and Mikš's proposal (division in the ratio $4: 5$). Therefore, we will expand the ratio so that its third term is divisible by both two and nine, e... | 1800 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Majka, Vašek, and Zuzka solved problems over the weekend. Majka and Vašek solved a total of 25 problems. Zuzka and Vašek solved a total of 32 problems. Moreover, Zuzka solved twice as many problems as Majka.
How many problems did Vašek solve?
(M. Dillingerová) | Zuzka calculated twice as many problems as Majka. Thus, the total number of problems calculated by Zuzka and Vašek is the same as the total number of problems calculated by Majka and Vašek, increased by the number of problems calculated by Majka.
Zuzka and Vašek calculated a total of 32 problems, Majka and Vašek calcu... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Insert two digits into the number 2019 so that the resulting six-digit number
- starts with 2 and ends with 9,
- is composed of six different digits,
- is divisible by three,
- its first three-digit number is divisible by three,
- its first four-digit number is divisible by four,
- the sum of the inserted digits is od... | To form a new number consisting of six different digits, we can insert two different digits from the digits
$$
3,4,5,6,7,8
$$
The sum of the digits of the number 2019 is 12, which means the number is divisible by three. To ensure the newly formed number is also divisible by three, we can only insert digits whose sum ... | 69180 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Martin decided to spend all his savings on sweets. He found out that he could buy three lollipops and $3 \mathrm{dl}$ of cola or $18 \mathrm{dkg}$ of yogurt-covered raisins or $12 \mathrm{dkg}$ of yogurt-covered raisins and half a liter of cola. In the end, he bought one lollipop and 6 dl of cola. How many grams of yog... | Let's denote:
| 1 windmill | $\ldots v$ |
| :--- | :--- |
| 1 deciliter of cola | $\ldots k$ |
| 1 dekagram of raisins | $\ldots r$ |
| the unknown amount of raisins (in dekagrams) | $\ldots x$ |
Since Martin wanted to spend all his savings each time, we can symbolically represent the individual relationships using s... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A box of chocolates in the shape of a cuboid was full of chocolates arranged in rows and columns. Misha ate some of them and rearranged the remaining ones so that they filled (without gaps) up to one place three whole rows. The remaining chocolates from the next incomplete row were eaten. Then the remaining chocolates ... | Given that in the end only four columns were filled, which was only a third of the original number of candies, the box must contain 12 columns. (1 point)
Now we know that each row contains 12 candies. It is therefore clear that before the second rearrangement, there were 24 candies in the box. (1 point)
Since one was... | 25 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
U ohně seděli náčelníci tří indiánských kmenů se třemi stejnými dýmkami. Měli válečnou poradu a kouřili. První z nich vykouří celou dýmku za deset minut, druhý za půl hodiny a třetí za hodinu. Jak si mají náčelníci mezi sebou měnit dýmky, aby se mohli radit co nejdéle.
(Bednářová)
#
By the fire sat the chiefs of thr... | 1. way
| 1. chief | $\ldots 10$ minutes |
| :--- | :--- |
| 2. chief | $\ldots 30$ minutes |
| 3. chief | ... 60 minutes |
They will pass the pipe every $x$ minutes, so the total duration of the council is $3 x$ minutes. Then we have:
$$
\begin{gathered}
\frac{x}{10}+\frac{x}{30}+\frac{x}{60}=1, \\
x \cdot\left(\fra... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The digits $3, 4, 5, 7$ and 9 were used to form all possible three-digit numbers such that each digit appears in each number at most once.
Determine the number of such numbers and their total sum.
(M. Volfová) | When creating three-digit numbers with the given properties, the following procedure can be followed:
The first place can be occupied by any of the five given digits, i.e., 5 possibilities. For each of these placements, the second place can be occupied by any of the four remaining digits, i.e., a total of $5 \cdot 4=$... | 37296 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
When admitting students to the university, each applicant is assigned a cover code consisting of five digits. The thorough, yet superstitious associate professor decided to eliminate from all possible codes (i.e., 00000 to 99999) those that contained the number 13, meaning the digit 3 immediately following the digit 1.... | First, we will determine how many codes contain the number 13 once:
Codes of type 13***: On the third, fourth, and fifth positions, any of the ten digits can appear, so there are $10 \cdot 10 \cdot 10=1000$ codes.
Codes of type *13**: Similarly, we find that there are 1000 codes.
Codes of type **13*: 1000 codes.
Co... | 3970 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\mathrm{Na}$ in the zoo, they opened a maze for the children with six stations where candies were being distributed. At one station, 5 candies were given out with each entry, at two stations, 3 candies were given out, and at three stations, 1 candy was given out. Jirka first entered the station marked with an arrow an... | The most candies Jirka can get is when he walks each path exactly once, and the stations he visits most frequently distribute the most candies. It is possible to walk the maze in the specified way, for example, the path
$$
\mathrm{A}-\mathrm{B}-\mathrm{C}-\mathrm{D}-\mathrm{E}-\mathrm{C}-\mathrm{A}-\mathrm{E}-\mathrm{... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
How many small cubes, each having a surface area of $54 \mathrm{~cm}^{2}$, are needed to build a large cube with a surface area of $864 \mathrm{~cm}^{2}$?
(M. Krejčová) | A small cube has an edge length of $3 \mathrm{~cm}$. A large cube has an edge length of $12 \mathrm{~cm}$. To build the large cube, we need 64 small cubes.
# | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
We have four containers. The first three contain water, the fourth is empty. In the second one, there is twice as much water as in the first, and in the third one, there is twice as much water as in the second. We pour half of the water from the first container, a third of the water from the second container, and a qua... | From the text, it follows that in the first container there is one part of water, in the second container there are two parts of water, and in the third container there are four parts of water. To make it easier to remove water from each container, we will divide each part into six sixths. So, in the first container th... | 84 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Eva wrote down consecutive natural numbers: 1234567891011. . What digit did she write on the 2009th place?
(M. Volfová) | There are 9 single-digit numbers (1 to 9) and it takes 9 digits to write them. There are 90 two-digit numbers (10 to 99) and it takes 180 digits to write them. There are 900 three-digit numbers (100 to 999) and it takes a total of 2700 digits to write them. To write all single-digit and two-digit numbers, 189 digits ar... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
When Bořek was painting the garage door, he accidentally painted over the scale of the outdoor wall thermometer. However, the mercury tube remained undamaged, so Bořek covered the original scale with a homemade strip. On it, he carefully marked divisions, all of the same size and labeled with numbers. His division, how... | At $11^{\circ} \mathrm{C}$, the thermometer shows 2 boroks. When the temperature drops to $-4{ }^{\circ} \mathrm{C}$, which is a decrease of $15^{\circ} \mathrm{C}$, the thermometer shows -8 boroks, which is 10 boroks less than in the first case. A change of 10 boroks corresponds to a change of $15^{\circ} \mathrm{C}$,... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A beginning songwriter always sold CDs with his music after performances. On Thursday, he sold eight identical CDs. The next day, he also offered his new CD, so people could buy the same as on Thursday or the new one. On Saturday, all the listeners wanted the new CD, and the songwriter sold six of them that day. He ear... | First, we will try to assign the individual amounts to the days. The Thursday revenue must be a multiple of eight, and the Saturday revenue must be a multiple of six. The numbers 720 and 840 are both multiples of six and eight. The number 590 is not a multiple of six or eight. Therefore, the singer must have earned a) ... | 90 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Vojta wrote the number 2010 a hundred times in a row without spaces. How many four-digit and how many five-digit palindromic numbers are hidden in this sequence? (A palindromic number is a number that reads the same backward as forward, e.g., 39193.)
(L. Hozová) | Vojta's notation starts like this: 2010201020102010 ... If Vojta wrote 2010 twice in a row, there would be one five-digit symmetrical number 20102 and one five-digit symmetrical number 10201 in the notation. If he wrote 2010 three times in a row, each of the above symmetrical numbers would appear twice. If we continue ... | 198 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the quadrilateral $A B C D$ (with internal angles $\alpha, \beta, \gamma, \delta$ it holds: $\alpha=\gamma=90^{\circ},|A D|=|C D|$ and $\beta=\delta+100^{\circ}$. Let $M$ be the intersection of the angle bisectors of $\angle D A C$ and $\angle A C D$. What is the measure of the internal angle at vertex $M$ in triang... | First, we will calculate the sizes of the remaining angles. The sum of the interior angles in any quadrilateral is equal to $360^{\circ}$, so $\alpha+\beta+\gamma+\delta=360$. Substituting the given values, we get $90+(\delta+100)+90+\delta=360$, from which $\delta=40^{\circ}$ and $\beta=140^{\circ}$.
Now we will focu... | 125 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Dominik observed a chairlift. First, he found that one chair passed the lower station every 8 seconds. Then he picked one chair, pressed his stopwatch, and wanted to measure how long it would take for the chair to return to the lower station. After 3 minutes and 28 seconds, the cable car was sped up, so chairs passed t... | In 3 minutes and 28 seconds (which is 208 seconds), 208 / 8 = 26 seats passed by Dominik. From the moment of acceleration until Dominik's seat passed by, 11 minutes and 13 seconds - 3 minutes and 28 seconds, which is 7 minutes and 45 seconds (465 seconds) elapsed. During the time the cable car was moving faster, 465 / ... | 119 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
We divided a rectangle into four triangles as shown in the picture. We measured all the internal angles in these triangles and obtained the following values: $15^{\circ}, 20^{\circ}, 20^{\circ}, 50^{\circ}, 55^{\circ}, 70^{\circ}, 75^{\circ}, 75^{\circ}, 90^{\circ}, 90^{\circ}, 130^{\circ}$, and one value that we forgo... | First, we will fill in the angles that are "obvious."
The sum of the remaining angles in triangle $E C F$ is $50^{\circ}$. From the angles available, 50 cannot be obtained, so the missing an... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Jirka drew a square grid with 25 squares, see the image. Then he wanted to color each square so that squares of the same color do not share any vertex.
How many colors did Jirka need at least?
(M. Dillingerová)
 | We will gradually examine the possibilities based on the number of children:
- If she had 1 child, then the dough should yield $1 \cdot 3 + 2 = 5$ and simultaneously $1 \cdot 4 - 1 = 3$ rolls. These values are different, so she did not have 1 child.
- If she had 2 children, then the dough should yield $2 \cdot 3 + 2 =... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Majka was exploring multi-digit numbers in which odd and even digits alternate regularly. She called those starting with an odd digit "funny" and those starting with an even digit "cheerful." (For example, the number 32387 is funny, and the number 4529 is cheerful.)
Majka created one three-digit funny number and one t... | The product of two three-digit numbers is at least a five-digit number and at most a six-digit number. For further consideration, let's denote Majčina's numbers and indicate the product calculation:
=6 n+3
$$
for some natural number $n$. The side lengths (vm) of Jakub's and David's triangles correspond to triples of natural numbers that sum to $k$ and satisfy the triangle inequali... | 2019 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Tourists planned a long three-day tour, with the intention of walking a third of the entire route each day. They only kept this plan on the first day. On the second day, they only walked a third of the remaining distance, and on the third day, exhausted, they only managed a quarter of the remaining distance. The last $... | On the first day, the tourists walked $\frac{1}{3}$ of the journey, leaving $\frac{2}{3}$ of the total distance to the destination.
On the second day, they walked $\frac{1}{3}$ of the remaining distance from the previous day, which is $\frac{1}{3} \cdot \frac{2}{3}=\frac{2}{9}$ of the entire journey; the remaining dis... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Zuzka wrote a five-digit number. When she appended a one to the end of this number, she got a number that is three times larger than the number she would get if she wrote a one before the original number.
Which five-digit number did Zuzka write? | A five-digit number written as $\overline{a b c d e}$ represents the value $10000 a + 1000 b + 100 c + 10 d + e$, which we denote as $x$. When we write a one in front of this five-digit number, i.e., $\overline{1 a b c d e}$, we get a number that is 100000 greater, i.e., the number $x + 100000$. When we write a one aft... | 42857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Inside the rectangle $A B G H$ are two identical squares $C D E F$ and $I J K L$. The side $C F$ of the first square lies on the side $B G$ of the rectangle, and the side $I L$ of the second square lies on the side $H A$ of the rectangle. The perimeter of the octagon $A B C D E F G H$ is $48 \, \text{cm}$, and the peri... | Significant parts of the perimeters of the octagon $A B C D E F G H$ and the dodecagon $A B C D E F G H I J K L$ are common to both figures. On the other hand, the segments $I J, J K$, and $K L$ are part of only the perimeter of the dodecagon, and the segment $I L$ is part of only the perimeter of the octagon. These fo... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On the segment $P Q$, a square $M N O P$ is placed with one side, see the figure. The line $P Q$ is successively flipped over the sides of the square $M N O P$, with point $Q$ leaving a trail on the paper. After the first flip, this trail is $5 \mathrm{~cm}$ long, and after five flips, point $Q$ coincides with one of t... | At each flip, point $Q$ describes a quarter-circle with the center at one of the vertices of the square and with a radius that decreases by the side length of the square at each step. For point $Q$ to coincide with one of the vertices of the square after five flips, the segment $M Q$ must be five times the side length ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A cat broke into a wine store at night. It jumped onto a shelf where bottles of wine were lined up in a long row - the first third of the bottles at the edge cost 160 Kč each, the next third cost 130 Kč each, and the last third cost 100 Kč each. First, the cat knocked over a 160 Kč bottle at the very beginning of the r... | In the problem statement, it is not specified in which third of the row the cat stopped knocking over bottles. We will consider each third as the one where the cat ended, and in each case, we will conclude whether it could have ended there or not.
If the cat stopped in the first third of the row, the damage from knock... | 36 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the quadrilateral $K L M N$, we know the marked angles and that $|K N|=|L M|$. What is the measure of angle $K N M$?
(L. Hozová) | Since the sum of the interior angles in any triangle is $180^{\circ}$, the measure of angle $L K M$ is $180^{\circ}-75^{\circ}-30^{\circ}=75^{\circ}$. Therefore, triangle $K L M$ is isosceles, i.e., $|L M|=|K M|$. According to the problem, $|L M|=|K N|$, thus $|K M|=|K N|$ and triangle $K M N$ is also isosceles. The me... | 65 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Father played chess with uncle. For a won game, the winner received 8 crowns from the opponent, and for a draw, nobody got anything. Uncle won four times, there were five draws, and in the end, father earned 24 crowns.
How many games did father play with uncle?
(M. Volfová) | Father lost four times, so he had to pay his uncle $4 \cdot 8=32$ crowns.
However, Father won so many times that even after paying these 32 crowns, he still gained 24 crowns. His total winnings were $32+24=56$ crowns, so he won $56: 8=7$ games.
Father won seven times, lost four times, and drew five times, so he playe... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
I want to construct a triangle $ABC$ with sides $|AB|=3 \text{~cm}$ and $|BC|=4 \text{~cm}$. Furthermore, I want to construct all circles, each of which will have its center at one of the vertices of the triangle and will pass through another vertex.
How long must the side $AC$ be so that there are exactly five such c... | If sides $AB$ and $AC$ were of the same length, then the circle centered at point $A$ passing through point $B$ would also pass through point $C$. In such a case, Ema would construct only one circle centered at point $A$.
If sides $AB$ and $AC$ were of different lengths, then the circle centered at point $A$ passing t... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Pat and Mat dug a well. On the first day, Pat dug a hole $40 \mathrm{~cm}$ deep. The second day, Mat continued and dug to three times the depth. On the third day, Pat dug as much as Mat did the previous day and hit water. At that moment, the ground was $50 \mathrm{~cm}$ above the top of his head.
Determine how tall Pa... | On the first day, Pat dug $40 \mathrm{~cm}$.
On the second day, Mat dug to $3 \cdot 40=120(\mathrm{~cm})$, so he dug $120-40=80(\mathrm{~cm})$.
On the third day, Pat also dug $80 \mathrm{~cm}$, reaching a depth of $120+80=200(\mathrm{~cm})$. At that moment, the hole was $50 \mathrm{~cm}$ deeper than Pat, so Pat measu... | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Three gardeners invited a ploughman to plough their plots. The ploughman charged the same price per square meter for all of them. In total, the gardeners paid 570 Kč.
Two of the plots were square, one was rectangular. The area of the rectangular plot was 6 ares and one of its sides measured $20 \mathrm{~m}$. The side ... | The area of the rectangular plot was 6 ares, i.e., $600 \mathrm{~m}^{2}$. One of its sides measured $20 \mathrm{~m}$, so the other side measured $600: 20=30(\mathrm{~m})$. The side of one square plot was $20 \mathrm{~m}$ long, its area was $20 \cdot 20=400\left(\mathrm{~m}^{2}\right)$, i.e., 4 ares. The side of the sec... | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Mrs. Teacher needed to come up with examples of equations for a written test. Therefore, she wrote down all equations of the form
$$
a \cdot x + b = 13
$$
where $a$ and $b$ are one-digit natural numbers. From all of them, she selected the equations whose root $x$ was 3. She put one equation into each group. What is t... | We know that $x=3$ is a solution to the given equation, so the equality
$$
a \cdot 3 + b = 13 \text{.}
$$
holds. For $a$ and $b$ to be natural numbers, $a$ must be either $1, 2, 3$, or $4$ (for $a=5$, we get $5 \cdot 3 = 15 > 13$ and $b$ would have to be negative, which is not possible). Now we substitute the individ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Students at our school travel in various ways. Those living nearby walk. The ratio of local students to commuting students is $3: 1$. Among the commuting students, the ratio of those who use public transportation and those who ride a bike or are driven by their parents is $3: 2$. Among those using public transportation... | Those who use public transportation are 24 and form 3 parts of the number of commuters. The remaining 2 parts, which belong to non-public transportation, thus correspond to 16 students $\left(\frac{2}{3}\right.$ of 24 is 16). The total number of commuters is $24+16=40$. Commuters form 1 part of all the students in the ... | 160 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
We received a cube, the length of whose edge was expressed in centimeters as an integer greater than 2. We painted all its faces yellow and then cut it into smaller cubes, each with an edge length of $1 \mathrm{~cm}$. We sorted these small cubes into four piles. The first pile contained cubes with one yellow face, the ... | Let the length of the edge of the original cube in centimeters be denoted by $a+2$, where $a$ is a natural number. Each face of the original cube corresponds to $a^{2}$ small cubes with exactly one colored face, so there are a total of $6 a^{2}$ such small cubes. On each edge of the original cube, we have $a$ small cub... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the garden, four times as many cabbages grew as broccoli and three times as many radishes as cabbages. The total weight of the broccoli was $5 \mathrm{~kg}$. How many pieces of vegetables grew in the garden if each broccoli weighed 250 g? (No other vegetables grew there.) (L. Černíček) | All broccoli weigh $5 \mathrm{~kg}$, one broccoli weighs $250 \mathrm{~g}$, so we first find the number of broccolis: $5 \mathrm{~kg}=5000 \mathrm{~g}, 5000: 250=20$. If there are 20 broccolis, there are four times as many radishes, i.e., $4 \cdot 20=80$ pieces. There are three times as many carrots as radishes, i.e., ... | 340 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Daddy goes running with his daughter Jana. While Jana ran three times around the small circuit around the school, Daddy ran four times around the larger circuit around the adjacent park. This way, Daddy ran twice the distance Jana did. The small circuit measures 400 meters.
How many meters longer is the large circuit ... | Jana ran the small circuit three times, which means she ran a total of $3 \cdot 400=1200$ meters.
Daddy ran twice the distance Jana did, which is a total of $2 \cdot 1200=2400$ meters. This distance corresponds to four large circuits; the large circuit measures $2400: 4=600$ meters.
The difference in length between t... | 200 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Petr brought a bag of candies to ten friends and distributed them so that each got the same amount. Then he realized that the bag had the smallest possible number of candies that could also be distributed in such a way that each of the friends would get a different (but non-zero) number.
Determine how many candies wer... | According to how Petr distributed the candies, we know that their total number was divisible by ten. To be able to distribute the candies in the second way, he would need at least
$$
1+2+3+4+5+6+7+8+9+10=55 .
$$
The smallest number satisfying both conditions is 60. This number can be distributed in the second way, fo... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Two pairs of parallel lines $A B \| C D$ and $A C \| B D$ are given. Point $E$ lies on line $B D$, point $F$ is the midpoint of segment $B D$, point $G$ is the midpoint of segment $C D$, and the area of triangle $A C E$ is $20 \mathrm{~cm}^{2}$.
Determine the area of triangle $D F G$.
(V. Semeráková)
Hint. Compare t... | The area of a triangle depends on the length of its side and the height to this side. Since lines $A C$ and $B D$ are parallel and point $E$ lies on line $B D$, the area of triangle $A C E$ is always the same for any chosen point $E$. In particular, the area of triangle $A C E$ is the same as the area of triangle $A C ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The zoo offered discounted admission for school groups: every fifth student gets a free ticket. Mr. Teacher of 6.A calculated that if he bought tickets for the children in his class, he would save on four tickets and pay 1995 CZK. Mrs. Teacher of 6.B suggested that he buy tickets for the children of both classes at onc... | If when buying tickets for children from 6.A, thanks to the mentioned discount, 4 tickets were saved, there must have been at least $4 \cdot 5=20$, but fewer than $5 \cdot 5=25$ children from this class. With a number of children from 20 to 24, 4 fewer tickets, i.e., 16 to 20, would always have to be paid. The amount p... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Prokop constructed a triangle $ABC$, where the interior angle at vertex $A$ was greater than $60^{\circ}$ and the interior angle at vertex $B$ was less than $60^{\circ}$. Jirka drew a point $D$ in the half-plane defined by the line $AB$ and point $C$, such that triangle $ABD$ was equilateral. Then, the boys found that ... | The sizes of the internal angles in triangle $ABC$ are denoted as $\alpha, \beta, \gamma$ respectively. In the equilateral triangle $ABD$, all internal angles have a size of $60^{\circ}$.
The... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Ferda and David meet in the elevator every day. One morning, they found that if they multiplied their current ages, they would get 238. If they did the same four years later, the product would be 378. Determine the sum of the current ages of Ferda and David.
(M. Petrová)
Hint. Could Ferda (or David) be 8, 15, or 47 y... | The number 238 can be factored into the product of two numbers in the following ways:
$$
238=1 \cdot 238=2 \cdot 119=7 \cdot 34=14 \cdot 17 .
$$
Among these pairs are the current ages of Ferdy and David. After adding 4 to each of them, we should get a product of 378. Let's consider all possibilities:
$$
\begin{align... | 31 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The king gave the mason Václav the task of building a wall 25 cm thick, 50 m long, and 2 m high. If Václav worked without breaks and at a constant pace, he would build the wall in 26 hours. However, according to the valid royal regulations, Václav must adhere to the following conditions:
- During the work, he must tak... | Václav has to take exactly six half-hour breaks. This will cost him $6 \cdot 1 / 2=3$ hours. He can schedule the breaks so that at the start of work and after each break, he can work for a full hour at an increased pace (for example, with an even distribution). This will gain him $7 \cdot 1 / 4$ hours, i.e., 1 hour and... | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.