problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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Seats in the theater hall are divided into three categories based on their distance from the stage. "1st places" are closest to the stage, constitute two fifths of the hall's capacity, and are sold for 220 Kč. "2nd places" constitute the next two fifths of the hall and are sold for 200 Kč. The remaining "3rd places" ar... | For the calculations, it is essential to determine in which fifth of the hall the segment $\mathrm{s}$ with free tickets ends (see image). Therefore, we will divide the solution into five parts and work with a different assumption for $\mathrm{s}$ in each. For the number of seats $\mathrm{v}$ in one fifth of the hall, ... | 360 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In two rooms of an educational center, lectures were being held. The average age of eight people present in the first room was 20 years, and the average age of twelve people in the second room was 45 years. During the lecture, one participant left, and as a result, the average age of all people in both rooms increased ... | According to the problem, the sum of the ages of eight people present in the first room was $8 \cdot 20=160$ years, and the sum of the ages of twelve people present in the second room was $12 \cdot 45=540$ years. Therefore, the average age of all the people in both rooms was $\frac{160+540}{8+12}=\frac{700}{20}=$ $=35$... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Ivan and Mirka were sharing pears from a plate. Ivan always took two pears, and Mirka took half of what was left on the plate. They proceeded in this manner: Ivan, Mirka, Ivan, Mirka, and finally Ivan, who took the last two pears.
Determine who ended up with more pears and by how many.
(M. Dillingerová)
Hint. How ma... | Ivan took three times two pears, in the end, he had 6 pears. To determine how many pears Mirka ended up with, we will trace back the changes in the number of pears step by step. $\mathrm{K}$ to do this, it is enough to realize that before each of Ivan's takings, there were two more pears on the plate, and before each o... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Arnošt cut out a $4 \times 4$ square from graph paper. Kristián cut out two holes in it, see the two black squares in the picture. Arnošt tried to cut this shape along the marked lines into two identical parts.
Find at least four different ways Arnošt could have done this. (Two cuttings are considered different if the... | One of the simplest possible divisions is indicated in the following figure:
The newly created parts are — like the original shape — symmetric with respect to the center of the bounding squa... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the image, two rows of hexagonal fields are indicated, which continue to the right without limit. Fill in each field with one positive integer so that the product of the numbers in any three adjacent fields is 2018.
Determine the number that will be in the 2019th field in the upper row.
(L. Růžičková)
:
$$
1 \cdot 1 \cdot 2018, \quad 1 \cdot 2 \cdot 1009
$$
Thus, the empty fields can only be filled with one of the n... | 1009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Matěj had six different natural numbers written on a line in his notebook. The second of them was double the first, the third was double the second, and similarly, each subsequent number was double the previous one. Matěj copied all these numbers into the following table, in random order, one number per cell.
 of triangle $B D X$. Calculate the sizes of the interior angles of this triangle.
( V. Žádník) | Let $A$ be the orthocenter of triangle $B D X$. This means that on the line connecting point $A$ with vertex $B$, there is a perpendicular to side $D X$, and similarly, on the line connecting $A$ with vertex $D$, there is a perpendicular to side $B X$. Therefore, side $D X$ is perpendicular to line $A B$, and similarly... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the word PAMPELIŠKA, the same letters are to be replaced by the same non-zero digits, and different letters by different non-zero digits. Moreover, it must hold that the product of the digits of the resulting number is the square of some natural number. Find the largest number that can be obtained by replacing the l... | In the word PAMPELIŠKA, there are 8 different letters. We are allowed to use only non-zero digits, so we choose eight digits from the nine possible ones. The product
$$
P^{2} \cdot A^{2} \cdot M \cdot E \cdot L \cdot I \cdot \check{S} \cdot K
$$
should be a square of a natural number. Since the digits $P$ and $A$ are... | 8798643217 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Three friends Pankrác, Servác, and Bonifác went on a night walk through a natural labyrinth during their vacation. At the entrance, each received a candle and they set off in different directions. All of them successfully navigated the labyrinth, but each took a different path. In the following square grid, their paths... | First, we will determine which paths the friends took. For this, we need to know which cardinal directions the individual paths lead to. The path marked by a solid line leads only north and east. The dashed path leads north, east, and west. The dotted path goes in all cardinal directions. The only path that never leads... | 625 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Jolana is preparing refreshments for her friends - open-faced sandwiches. She spreads them with potato salad and wants to add more toppings: ham, hard cheese, a slice of egg, and a strip of pickled pepper. However, she does not want any two of her sandwiches to have the exact same combination of toppings. What is the m... | For clarity, we will create a table. An asterisk means that the given open-faced sandwich contains the respective ingredient, while an empty cell means that the sandwich does not contain that ingredient.
| | ham | cheese | egg | pepper | |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1 topping | $*$ | | | ... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the picture, there is a structure made of identical cubes. It is a cube with several holes through which you can see through, and these holes have the same cross-section everywhere. How many cubes is the structure made of?
(M. Krejčová)
=900 .
$$
After simplification, we get:
$$
\begin{aligned}
12 x+264... | 53 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Florián was thinking about what bouquet he would have tied for his mom for Mother's Day. In the florist's, according to the price list, he calculated that whether he buys 5 classic gerberas or 7 mini gerberas, the bouquet, when supplemented with a decorative ribbon, would cost the same, which is 295 crowns. However, if... | From the assignment, it follows that 5 classic gerberas cost the same as 7 minigerberas, meaning that the price of a classic gerbera and the price of a minigerbera are in the ratio of $7: 5$. If we represent the price of a classic gerbera with 7 equal parts, the price of a minigerbera will correspond to 5 such parts.
... | 85 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
On the table lay six cards with digits $1,2,3,4,5$ and 6. Kamila made a three-digit number from three cards, which was greater than 500 and divisible by four. Filip made a three-digit number from the remaining three cards, which was divisible by both three and five. Kamila then added both three-digit numbers, and Filip... | Filip's number must be divisible by five, so from the offered digits, it must use 5 in the units place. Therefore, 5 cannot appear in Kamila's number. At the same time, Kamila's number must be greater than 500, so it must have the digit 6 in the hundreds place.
Filip's number must be divisible by three, meaning its di... | 939 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The teacher gave the children a task to multiply the number written on the board by three and add the number seven to the result. Kamča solved the task correctly. Her friend Růža, however, calculated a different problem: she multiplied the given number by seven and added the number three to the result. Růža's result wa... | Let's consider what would happen if the girls added the same numbers to their different intermediate results. That is, let's consider the situation where Růža, instead of adding 3 to the sevenfold of the given number, added 7 like Kamča (or the situation where Kamča, instead of adding 7 to the triple of the given numbe... | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
On each wall of the dice, we wrote a different prime number less than 20 so that the sums of two numbers on opposite walls were always the same.
We placed the dice on the first square of the... | Prime numbers less than 20 are $2,3,5,7,11,13,17$ and 19. From these, we need to take three pairs with the same sum, which are the pairs $(19,5),(17,7)$ and $(13,11)$. The cube is placed on the plan with some number, which we will denote as $a$. It is possible to discuss all the possibilities regarding the placement of... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
On the town hall tower, there are clocks that have maintenance doors near the center of the dial. However, these doors open outward, which is impractical - for example, at exactly 12:09, the minute hand will cover the doors, making them impossible to open until exactly 12:21.
(L. Šimünek) | We denote the digits of the suitable four-digit number as follows: $v$ is in the thousands place, $x$ in the hundreds place, $y$ in the tens place, and $z$ in the units place. A natural number is divisible by five if and only if it has the digit 0 or 5 in the units place. After crossing out two digits from the original... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Karel and Vojta found out that the kitchen clock at the cottage is fast by 1.5 minutes per hour and the clock in the bedroom is slow by half a minute per hour. At noon, they set both clocks to the same and correct time. The kitchen and bedroom clocks both have a standard twelve-hour dial. Determine when:
1. the kitche... | 1. The clock will show the correct time again when it runs ahead of the actual time by 12, 24, 36, ... hours. First, this will happen when it runs ahead by 12 hours, or 720 minutes. This will be achieved in $720: 1.5=480$ hours. The kitchen clock will show the correct time again for the first time after 480 hours (whic... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In triangle $ABC$, we denote the midpoints of sides $CB$ and $CA$ by the letters $K$ and $L$. We know that the quadrilateral $ABKL$ has a perimeter of $10 \text{ cm}$ and triangle $KLC$ has a perimeter of $6 \text{ cm}$. Calculate the length of segment $KL$.
(J. Mazák) | Segment $K L$ is the midline of triangle $A B C$ parallel to $A B$, since $K$ and $L$ are the midpoints of sides $B C$ and $A C$. Therefore, $2|K L|=|A B|$ and also $|A L|=|L C|$ and $|C K|=|K B|$.
The perimeter of triangle $K L C$ is $|C K|+|K L|+|L C|=6$. The perimeter of quadrilateral $A B K L$ is
$$
|A B|+|B K|+|... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On the sides of triangle $ABC$, points $D, E, F, G$ are given, as shown in the figure. It is given that quadrilateral $DEFG$ is a rhombus and segments $AD, DE$, and $EB$ are congruent.
Determine the measure of angle $ACB$.
(I. Jančigová)
Hint. On which positions could Alík and Punt be harnessed? | If Alík was first, Punta could be second, third, fourth, or fifth:
$$
A P * * * \quad A * P * * \quad A * * P * \quad A * * * P
$$
If Alík was second, Punta could be third, fourth, or fifth:
$$
* A P * * \quad * A * P * \quad * A * * P
$$
If Alík was third, Punta could be fourth or fifth:
$$
* * A P * \quad * * A ... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Petr drew a trapezoid $ABCD$, whose base $AB$ was twice as long as the base $CD$, and sides $AD, DC, CB$ were congruent. Then he drew a square that had one side in common with the shorter base of the trapezoid. The new vertex of the square, which was closer to $B$ than to $A$, he labeled $N$.
What could be the measure... | According to the assumptions in the problem, trapezoid $A B C D$ can be divided into three equilateral triangles. Therefore, the interior angle at vertex $B$ is $60^{\circ}$ and at vertex $C$ is $120^{\circ}$.
 collected by Adam, Borek, and Cenda will be denoted by $a, b$, and $c$ respectively. According to the problem statement, the following equations hold:
$$
\frac{a+b}{2}=c+10, \quad \frac{a+c}{2}=b-3 .
$$
We want to determine the difference between $\frac{b+c}{2}$ and $a$. To d... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Stars on the picture represent 16 consecutive natural multiples of the number three. The numbers in the boxes have the same sum. Determine the smallest of these 16 numbers.
(L. Šimünek)
$$
* * * * * * * * * * * * * * * * * *
$$ | Let's compare the last, i.e., the largest numbers of both frames: the largest number of the second frame lies 10 places to the right of the largest number of the first frame, so it is 30 greater. By the same comparison, we find that the second largest numbers in the frames also differ by 30, as do the third largest, th... | 150 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A group of children organized a tic-tac-toe tournament. Everyone played with everyone else, and a total of 136 games were played. Out of these, exactly 66 games were of the type girl-girl or boy-boy. How many boys and how many girls were in the group?
(Bednářová) | Let's consider a group of $n$ players. Each one plays a match with every other, and we are interested in the total number of matches. Each player will play $\mathrm{s} n-1$ opponents, and there are $n$ players, so in total we have $n \cdot(n-1)$. Now we must realize that we have counted each match twice. Therefore, we ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The average age of the Kebulov family, consisting of a father, a mother, and several children, is 18 years. When the father, who is 38 years old, is not included, the average age of the family is 14 years. How many children do the Kebulovs have?
(L. Hozová) | Let the number of members in this family be $n$. The sum of the ages of all members is equal to the product of the average age of the family and the number of members, that is, $18 \cdot n$. The family without the father has $n-1$ members, and the sum of the ages of these members is $14 \cdot(n-1)$. We know that this s... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
How many six-digit natural numbers exist that have the digit 1 in the thousands place, the digit 2 in the hundreds place, and the digit 3 in the tens place, and are divisible by 45 without a remainder?
(L. Simuinek) | A number is divisible by 45 if and only if it is divisible by both 5 and 9. Therefore, the digit in the units place must be 0 or 5, and the sum of its digits must be a multiple of nine.
First, we determine the number of the desired numbers that have the digit 0 in the units place. We denote these numbers as $\overline... | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The seventh grades of our school competed in collecting caps from PET bottles. Class A collected half of what classes B and C collected together, class B collected a third of what classes A and C collected together, and class C collected 150 caps.
Determine how many caps these three classes collected in total.
(M. Vo... | Class A collected half of what classes B and C collected together; this means that class A collected a third of the total number of caps. Class B collected a third of what classes A and C collected together; this means that class B collected a quarter of the total number of caps. Therefore, classes A and B together col... | 360 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The square-shaped square is being reconstructed with a side length of 20 meters. If the entire square were to be paved with cheaper light-colored paving stones, the material costs would be 100,000 Kč. If the entire square were to be covered with more expensive dark paving stones, the material costs would be 300,000 Kč.... | a) Light paving is three times cheaper than dark paving. The costs for the light and dark parts of the square are the same, so the light area is three times larger than the dark area. The dark area thus forms one quarter of the square, and the cost for it is 300000 : $4=75000$ (Kč). The costs for the light part are the... | 150000 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Honzík received his pocket money and wants to buy something nice with it. If he bought four rolls, he would have 5 Kč left. If he wanted to buy five rolls, he would be short of 6 Kč. If he bought two rolls and three pretzels, he would spend his entire pocket money without any remainder.
How much does one pretzel cost?... | Honzík's pocket money can be expressed in three ways, namely as
- the sum of the price of 4 cakes plus 5 Kč,
- the sum of the price of 5 cakes minus 6 Kč,
- the sum of the prices of 2 cakes and 3 pretzels.
From the first two expressions, it follows that one cake costs $5+6=11$ Kč. From this, we also find that Honzík'... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Four squirrels ate a total of 2020 nuts, each at least 103 nuts. The first squirrel ate more nuts than any of the other three squirrels. The second and third squirrels ate a total of 1277 nuts.
How many nuts did the first squirrel eat?
(L. Hozová) | We will indicate the possibilities for how many hazelnuts the second and third squirrels could have eaten, i.e., the number 1277 will be decomposed into two addends, each of which is at least 103:
$$
1277=1174+103=\ldots=640+637=639+638 .
$$
From this, it follows that the first squirrel ate at least 640 hazelnuts (sh... | 640 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the homework for calculating the value of the expression
$$
2-3+4-5+6-7+8-9+10=
$$
Radek forgot to write two parentheses, so he got a result that was 18 greater than what he would have obtained if he had written the parentheses. Add the parentheses in two ways and write down what number Radek got and what number h... | Radek calculated
$$
2-3+4-5+6-7+8-9+10=6 \text {. }
$$
He got a result 18 more than it should have been. The result should have been $6-18=-12$. To get a smaller number, it is necessary to subtract a larger value, i.e., place the parentheses after the minus sign. Two required solutions:
$$
\begin{aligned}
& 2-3+(4-5... | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
During the first eleven days, 700 people responded to the survey question. Each of them chose exactly one of the three offered options. The frequency ratio of the individual responses was $4: 7: 14$. On the twelfth day, several more people participated in the survey, changing the frequency ratio of the responses to $6:... | 700 people are to be divided in the ratio $4: 7: 14$ (a total of 28 parts, with 25 people per part), which in the given ratio is 112 : 196 : 392. The new ratio $6: 9: 16$ has 31 parts, and the new number must be divisible by 31. It holds that $700: 31=22$ (remainder 18). The nearest multiple of 31 greater than 700 is t... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Martin has written a five-digit number with five different digits on a piece of paper, with the following properties:
- by crossing out the second digit from the left (i.e., the digit in the thousands place), he gets a number that is divisible by two,
- by crossing out the third digit from the left, he gets a number t... | Let's denote the digits of Martin's number as $a, b, c, d, e$, and the number formed by them as $\overline{a b c d e}$. Now, let's analyze each of the five conditions:
1. The number $\overline{a c d e}$ is divisible by two, so the digit $e$ is $0, 2, 4, 6$, or 8,
2. The number $\overline{a b d e}$ is divisible by thre... | 98604 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The price of the book "New Riddles" was reduced by $62.5\%$. Matěj found out that both prices (before and after the reduction) are two-digit numbers and can be expressed with the same digits, just in a different order. By how many Kč was the book discounted?
(M. Volfová) | The original price of the book in Kč will be written in the form $10a + b$, where $a$ and $b$ are unknown non-zero digits. After the discount, the price of the book was $10b + a$. The price reduction was $62.5\%$, which means it was reduced to $37.5\%$, so
$$
\frac{37.5}{100} \cdot (10a + b) = 10b + a
$$
Given the eq... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a fairy tale valley, three-headed and six-headed dragons lived. Together they had 117 heads and 108 legs. Each dragon has 4 legs. Find out how many three-headed and how many six-headed dragons lived there. | Since every dragon (both three-headed and six-headed) has 4 legs, we can find the total number of dragons in the fairy tale valley: $108: 4=27$ dragons. If every dragon in the valley had three heads, they would have a total of $3 \cdot 27=81$ heads. This means that the remaining $117-81=36$ heads are the extra heads th... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
We built a structure similar to the one in the picture, but instead of four floors, it had twenty-five. Then we painted all the outer walls of the building. The vertical walls were red and the horizontal walls were blue. (The building stands on the ground, so we did not paint the base.)
a) How many times more red pain... | It is important to realize that each colored wall of this building can be seen from one of the following views: from the front, from the right, from the back, from the left, and from above (orientation according to
 | Karel was supposed to add the numbers
$$
10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95 \text {. }
$$
Their sum is 945. However, Karel calculated 1035, which is 90 more. Karel counted the number 45 three times. | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The mole started to dig a new tunnel. First, the tunnel led 5 meters north, then $23 \mathrm{dm}$ west, $150 \mathrm{~cm}$ south, $37 \mathrm{dm}$ west, $620 \mathrm{~cm}$ south, $53 \mathrm{~cm}$ east, and $27 \mathrm{dm}$ north. How many centimeters does he still need to dig to get back to the start of the tunnel?
(... | After converting all data to centimeters, we can draw the mole's tunnel. The mole started digging the tunnel at point $S$ and ended at point $K$ (using the usual orientation of cardinal directions).
Hint. How much did they earn from... | Mr. Krbec with the cat Kokeš collected a total of $210 \cdot 25=5250$ groschen for children's tickets. They collected $5950-5250=700$ for adult tickets. Therefore, they sold $700: 50=14$ of these tickets. | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Mom cooked homemade currant syrup and poured it into bottles. The bottles were of two types: small with a volume of $500 \mathrm{ml}$ and large with a volume of $750 \mathrm{ml}$. In the end, she had 12 empty small bottles left, and all other bottles were completely filled. Then Mom realized that she could have poured ... | Mom has 12 unfilled bottles, each with a volume of $500 \mathrm{ml}$. They could hold $6000 \mathrm{ml}$ of syrup $(12 \cdot 500=6000)$.
The same amount would fit into 8 large bottles $(6000: 750=8)$. If Mom were to pour the syrup the second way, she would be left with 8 empty large bottles.
Note. One large bottle ha... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
I am thinking of a five-digit number composed of even digits. If I swap the digit in the third position $\mathrm{s}$ with any other, the number decreases. Furthermore, I will reveal that the first digit is double the last, and the second digit is double the second to last.
What number am I thinking of?
(M. Mach)
Hin... | The number is composed of even digits, i.e., digits $0,2,4,6,8$, not necessarily all of them.
The property of digit swapping means that the digit in the third position is smaller than any of the preceding digits and simultaneously larger than any of the following ones.
The first two positions are occupied by even dig... | 88644 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Kája was supposed to multiply two two-digit numbers. Due to inattention, he switched the digits in one of the factors and got a product that was 4248 less than the correct result.
What should Kája have correctly obtained?
(L. Hozová) | Let $x$ and $y$ be Kája's two-digit numbers, and suppose he swapped the digits in the first number. If the digits of the number $x$ are denoted by $a$ and $b$, then we have
$$
(10 a+b) y-(10 b+a) y=4248 .
$$
After simplification, we get $9(a-b) y=4248$ or $(a-b) y=472$. This implies that the number $y$ is a two-digit... | 5369 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Below is a part of a square grid consisting of 4 rows and 2023 columns.
Determine the number of squares, all of whose vertices are grid points of the square grid.
(K. Pazourek)
 | The smallest square $\mathrm{s}$ with vertices at grid points has dimensions $1 \times 1$, and the largest has dimensions $4 \times 4$. Within these constraints, we will find additional cases, which we will distinguish as follows.
- Squares with sides parallel to the grid:
Hint. How few and how many candies can they have together at minimum and maximum? | Vojta has fewer candies than Míša, so he can have
$$
0,1,2,3 \text {, or } 4
$$
candies. Vendelín has five more candies than Vojta, so he can have
$$
5,6,7,8, \text { or } 9
$$
candies. All three together have twice as many as Vendelín, so they can have
$$
10,12,14,16 \text {, or } 18
$$
candies.
Among these num... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Šárka enjoys celebrating, so in addition to birthdays, she came up with antibirthdays: the date of the antibirthday is created by swapping the day and month numbers in the birth date. She was born on 8.11., so her antibirthday is on 11.8. Her mother cannot celebrate her antibirthday: she was born on 23.7., her antibirt... | Those born on January 1st have their anti-birthday on the same day as their birthday. Those born between January 2nd and January 12th have their anti-birthday on a different day than their birthday. Those born between January 13th and January 31st cannot celebrate their anti-birthday because there are only 12 months in... | 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the new clubroom, there were only chairs and a table. Each chair had four legs, the table was three-legged. Scouts came into the clubroom. Each of them sat on their own chair, two chairs remained unoccupied, and the number of legs in the room was 101.
Determine how many chairs were in the clubroom.
(L. Hozová)
Hi... | A table and two unoccupied chairs have a total of $3+8=11$ legs. The occupied chairs thus account for $101-11=90$ legs. Each such chair had one scout with two legs sitting on it. Therefore, there were $90: 6=15$ occupied chairs in the club room, and $15+2=17$ chairs in total. | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Two guards are overseeing the order in a room whose shape and dimensions are shown in the diagram. Each two adjacent walls are perpendicular to each other, and the dimensions are given in meters. The guards stand close to the wall at the locations marked with squares. Together, they can oversee the entire room, but onl... | The entire room can be divided into squares with a side of 10 meters as shown in the image. The parts of the room visible to each guard are marked with two types of hatching. The part that both guards can see is thus doubly hatched:
 | For 5 sticks, there are 5 hats. For 1 stick, there is therefore 1 hat, and thus 5 sticks and 1 hat have the same value as 6 sticks.
Furthermore, we know that for 4 sticks, there are 6 coats. For 2 sticks, there are therefore 3 coats, and thus 6 sticks have the same value as 9 coats.
In total, we can see that 5 sticks... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Anička has saved 290 coins in her piggy bank, consisting of one-crown and two-crown coins. When she uses a quarter of all the two-crown coins, she collects the same amount as when she uses a third of all the one-crown coins.
What is the total amount Anička has saved?
(L. Růžičková) | Let's start from the requirement that a quarter of the two-crown coins gives the same amount as a third of the one-crown coins. From this, it follows that the number of two-crown coins is a multiple of four and the number of one-crown coins is a multiple of three, and that these numbers are in the ratio $4: 6$. Thus, t... | 406 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Grandpa keeps geese, pigs, goats, and chickens - a total of 40 animals. For every goat, there are 3 geese. If there were 8 fewer chickens, there would be as many as geese and pigs combined. If Grandpa traded a quarter of the geese for chickens at a ratio of 3 chickens for 1 goose, he would have a total of 46 animals.
... | Let's denote the quantities of each type of animal by their initial letters. The information from the problem can be written step by step as follows:
\[
\begin{aligned}
h + p + k + s & = 40, \\
h & = 3k, \\
s - 8 & = h + p, \\
40 - \frac{1}{4}h + \frac{3}{4}h & = 46.
\end{aligned}
\]
From the last equation, we get \(... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Last year, there were 30 more boys than girls in our scout troop. This year, the number of children in the troop increased by $10 \%$, while the number of boys increased by $5 \%$ and the number of girls increased by $20 \%$.
How many children do we have in the troop this year?
(L. Hozová) | Both the number of children and their current year's increments are expressed by natural numbers. Given that $10 \% = 10 / 100 = 1 / 10$, the original total number of children must have been a multiple of 10. Similarly, the original number of boys was a multiple of 20, since $5 \% = 5 / 100 = 1 / 20$, and the original ... | 99 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Triangle $A B C$ is divided by segments as shown in the figure. Segments $D E$ and $A B$ are parallel. Triangles $C D H, C H I, C I E, F I H$ have the same area, which is $8 \mathrm{dm}^{2}$.
Determine the area of quadrilateral $A F H D$.
(E. Semerádová)
 | The smallest of the ten written numbers is denoted by $p$. The numbers on the cards were then:
$$
p, p+1, p+2, p+3, \ldots, p+9
$$
The sum of the numbers on all ten cards was $10 p+45$. The equation:
First, let's assume that the card with the number $p$ was lost. Then the following would apply:
$$
\begin{aligned}
(... | 223 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Vojta wanted to add several three-digit natural numbers on a calculator. On the first attempt, he got the result 2224. To check, he added the numbers again and got 2198. He calculated once more, and this time the sum was 2204. It turned out that the last three-digit number was cursed - each time Vojta did not press one... | A cursed number will be denoted as $\overline{A B C}$. However, Vojta added two-digit numbers $\overline{A B}, \overline{A C}, \overline{B C}$, which we will generally denote as ??. We will denote as **** the sum of the numbers that Vojta could add without making a mistake. The schematic addition is expressed as follow... | 2324 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The wolves were pressing apple cider. They had it in two containers of the same volume, with almost the same amount in both. If they poured 1 liter from the first to the second, they would have the same amount in both, but neither container would be full. So they decided to pour 9 liters from the second to the first. T... | Let the number of liters in the first barrel before pouring be $x$, and in the second barrel be $y$. After pouring one liter, there would be $x-1$ liters in the first barrel and $y+1$ liters in the second barrel, and it would hold that
$$
x-1=y+1 \text {. }
$$
After pouring 9 liters, there would be $x+9$ liters in th... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Mr. Rychlý and Mr. Louda set out on the same hiking tour at the same time, but Mr. Rychlý started from the mountain hut and Mr. Louda from the bus at the bottom of the town, heading up to the hut. They met on the trail at 10 o'clock. Mr. Rychlý was in a hurry and reached his destination by 12 o'clock. On the other hand... | Let's denote the speed (in $\mathrm{km} / \mathrm{h}$) of Mr. Rychlý as $v_{R}$ and Mr. Louda as $v_{L}$. The time (in hours) from their departure until their meeting is denoted as $x$. Up to the meeting, Mr. Rychlý walked $x \cdot v_{R}(\mathrm{~km})$ from the cabin, and Mr. Louda walked $x \cdot v_{L}(\mathrm{~km})$ ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Petr and Pavel were picking apples and pears in the orchard. On Monday, Petr ate 2 more pears than Pavel and 2 fewer apples than Pavel. On Tuesday, Petr ate 4 fewer pears than on Monday. Pavel ate 3 more pears than Petr and 3 fewer apples than Petr on Tuesday. Pavel ate 12 apples over the two days and the same number o... | Let $x, y$ be the corresponding numbers of pears and apples that Pavel ate on Monday. According to the problem, we will patiently construct the following table:
| | Monday | Tuesday |
| :---: | :---: | :---: |
| Pavel pears | $x$ | $x+1$ |
| Pavel apples | $y$ | $12-y$ |
| Petr pears | $x+2$ | $x-2$ |
| Petr apples |... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
On July 1, 2010, Mrs. Hovorkova had a credit of 157.0 Kč on her mobile phone. The credit is gradually debited for calls, with 4.5 Kč deducted for each started minute. Mrs. Hovorkova does not send text messages and does not use any other paid services. She tops up her credit as needed, always by an amount of 400 Kč. On ... | The amount in crowns that Mrs. Hovorkova called during the period from July 1 to December 31 is a whole number because the initial and final credit balances and the top-up amounts are always whole numbers. For each minute started, 4.5 Kč is charged, and to reach a whole number, an even number of minutes must have been ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In rectangle $K L M N$, the distance from the intersection of the diagonals to line $K L$ is $2 \mathrm{~cm}$ less than its distance from line $L M$. The perimeter of the rectangle is $56 \mathrm{~cm}$. What is the area of rectangle $K L M N$?
(L. Hozová) | The intersection of the diagonals in rectangle $K L M N$ is denoted as $S$. If the distance from point $S$ to side $K L$ is $2 \mathrm{~cm}$ less than the distance to side $L M$, this means that side $K L$ is $2+2=4(\mathrm{~cm})$ longer than side $L M$.
 | Since $84: 12=7$, 7 students had a half-price ticket, i.e., a ticket for 25 Kč. Since $84: 35=2$ (remainder 14), 2 students had free entry. In total, the tickets cost
$$
7 \cdot 25+2 \cdot 0+(84-7-2) \cdot 50=175+0+75 \cdot 50=3925(\mathrm{~K} ̌) .
$$ | 3925 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Kluci found an old minefield plan, see the picture. The numbers are on the fields where there are no mines, and they indicate the number of adjacent mined fields. Determine how many mines there are in the field and where they are. (Fields are adjacent if they share a vertex or a side.)
| 1 | | 2 | | 2 |
| :--- | :--... | We can start unambiguously filling in the plan from the cell with the number $3 \mathrm{v}$ in the first column or from the cell with the number $2 \mathrm{v}$ in the upper right corner. In both cases, all unmarked adjacent cells must contain mines (marked as 丸):
| 1 | | 2 | $\star$ | 2 |
| :---: | :---: | :---: | :-... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Consecutive natural numbers are added and subtracted according to the following guide:
$$
1+2-3-4+5+6-7-8+9+10-11-12+\ldots,
$$
that is, two positive and two negative addends always repeat.
Determine the value of such an expression whose last term is 2015.
(L. Hozová) | The sums of pairs of adjacent numbers with opposite signs are either -1 or 1. These values also alternate regularly. In the considered expression, several adjacent numbers thus cancel each other out:
$$
\begin{aligned}
& 1+2-3-4+5+6-7-8+9+10-11-12+13+\ldots \\
= & {[1+(2-3)]+[(-4+5)+(6-7)]+[(-8+9)+(10-11)]+[(-12+13)+\... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A three-digit number has a digit sum of 16. If in this number the digits in the hundreds and tens places are swapped, the number decreases by 360. If in the original number the digits in the tens and units places are swapped, the number increases by 54.
Find that three-digit number.
(L. Hozová) | Let the original three-digit number be $\overline{a b c}=100 a+10 b+c$. According to the first piece of information from the problem, we have
$$
a+b+c=16 \text {. }
$$
According to the second piece of information, $\overline{b a c}=\overline{a b c}-360$, thus
$$
\begin{aligned}
100 b+10 a+c & =100 a+10 b+c-360, \\
3... | 628 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Ludvík noticed for a certain division example that if he doubles the dividend and increases the divisor by 12, he gets his favorite number as the result. He would get the same number if he reduced the original dividend by 42 and halved the original divisor.
Determine Ludvík's favorite number.
(M. Petrová) | If we denote the actors in the original example as $a: b$, then Ludvík's observation can be written as:
$$
2 a:(b+12)=(a-42): \frac{b}{2} .
$$
This is a double expression of Ludvík's favorite number, which we will denote as $\ell$ for subsequent modifications.
From the expression on the left, we get
$$
2 a=b \ell+1... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In front of Honza sat three veiled princesses, one of whom was Goldilocks. Honza's task was to find out which one she was.
The princess in the first chair said: "Goldilocks is not sitting in the third chair."
The princess in the second chair said: "I am not Goldilocks."
The princess in the third chair said: "I am Go... | We can distinguish three cases depending on where Blondie could have been sitting:
1. If Blondie was sitting in the first chair, then
- the princess in the first chair would be telling the truth,
- the princess in the second chair would be telling the truth,
- the princess in the third chair would be lying.
2. If Bl... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The commander summoned the other defenders of the castle and decided how to divide their reward:
"The first will take one gold coin and a seventh of the remainder, the second will take two gold coins and a seventh of the new remainder, and so on. That is, the $n$-th defender will take $n$ gold coins and in addition a ... | Let the number of defenders be $p$ and consider from the back. The last $p$-th defender took $p$ gold coins, and thus all the gold coins were taken. The second-to-last $(p-1)$-th defender took $p-1$ gold coins and a seventh of the current remainder, which we denote as $z$. Since both defenders received the same amount,... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A theater company performed thirty "Evenings of Improvisation" during the season as planned. Fan, an admirer of the main protagonist, calculated at the beginning of the season how much she would spend in total on admission if she attended every performance. After a few performances, however, the admission price unexpec... | The number of shows where the entrance fee was increased by 60 Kč is denoted by $a$. Fans spent $60a$ Kč more than they had planned for these shows. After the reduction, the entrance fee was lower than at the beginning of the season, by $85-60=25$ Kč. The number of shows with this entrance fee is denoted by $b$. Fans p... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find the smallest natural number such that its half is divisible by three, its third is divisible by four, its quarter is divisible by eleven, and its half gives a remainder of five when divided by seven.
(E. Patáková) | If half of a number is divisible by three, then the number must be divisible by both 2 and 3. For similar reasons, it must also be divisible by 3 and 4, and by 4 and 11.
The least common multiple of all these numbers is the product $3 \cdot 4 \cdot 11 = 132$; the sought number must be a multiple of 132. Half of the so... | 528 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Given a triangle $ABC$. On side $AB$ lies point $X$ and on side $BC$ lies point $Y$ such that $CX$ is a median, $AY$ is an altitude, and $XY$ is a midline of triangle $ABC$. Calculate the area of the shaded triangle in the figure, given that the area of triangle $ABC$ is $24 \, \text{cm}^2$.
(M. Dillingerová)
![](https://cd... | We will mark the beds with the same vegetables with the same letter, and beds with different vegetables with different letters. To ensure that beds with the same vegetables do not neighbor each other, they must be planted as follows:
| $A$ | | $B$ | |
| :--- | :--- | :--- | :---: |
| $B$ | $C$ | $A$ | |
Katka can ... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
From four identical squares, an ornament was created as shown in the picture. The sides of the squares are $4 \mathrm{~cm}$ long, and they are either parallel or perpendicular to each other, intersecting either at their quarters or halves. Libor wanted to color the ornament and found that the cost of the paint for $1 \... | The price of the color in each part corresponds to the number of squares that part is common to. This is the same as if all the squares were individually colored with a color of the same price.
Each square has an area of $4 \cdot 4 = 16 \mathrm{~cm}^{2}$, there are four squares, and $1 \mathrm{~cm}^{2}$ of color costs... | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a laboratory, on a shelf stands a closed glass container in the shape of a cuboid. It contains 2.4 liters of distilled water, but the volume of the container is larger. The water reaches a height of $16 \mathrm{~cm}$. When the cuboid container is placed on another of its sides, the water level will be at a height of... | Let the dimensions of the container be denoted by $a, b, c$. For each position of the container, we will set up an equation expressing the volume of water:
$$
\begin{aligned}
a \cdot b \cdot 16 & =2400 \\
a \cdot c \cdot 10 & =2400 \\
b \cdot c \cdot 9.6 & =2400
\end{aligned}
$$
We solve the system of three equations... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Read the results of a survey conducted in Pec pod Snezkou, where 1240 people were approached:
"46% of the respondents believe in the existence of Krkonose (rounded to the nearest whole number), 31% do not believe in its existence (rounded to the nearest whole number). The rest of the respondents refused to respond to ... | The number of people who responded positively will be denoted by $x$. The number of people who responded negatively will be denoted by $y$. For these numbers, the following applies:
$$
\begin{aligned}
0.455 \cdot 1240 & \leqq x < 0.465 \cdot 1240, \\
564.2 & \leqq x < 576.6, \\
0.305 \cdot 1240 & \leqq y < 0.315 \cdot... | 565 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Mommy preserves plums in jars so that the plums from one jar are enough for either 16 cups, or 4 pies, or half a sheet of fruit slices.
In the pantry, she has 4 such jars and wants to bake one sheet of fruit slices and 6 pies. $\mathrm{On}$ how many cups will the remaining plums be enough?
(M. Petrová) | First, we will determine how many of each type of delicacy can be prepared from all the plums:
- Four cups of plums are enough for $4 \cdot \frac{1}{2}=2$ whole trays of fruit slices.
- Four cups of plums are enough for $4 \cdot 4=16$ small cakes.
- Four cups of plums are enough for $4 \cdot 16=64$ small pastries.
In... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In a greyhound race, 36 dogs participated. The number of dogs that finished before Punta was four times smaller than the number of those who finished after him.
What was Punta's position?
(L. Hozová) | All the dogs without Punto was $36-1=35$. One fifth of this number is $35: 5=7$; before Punto, 7 dogs ran, behind Punto, $7 \cdot 4=28$ dogs ran. Punto ran eighth.
Evaluation. 3 points for dividing 35 dogs without Punto into fifths; 3 points for dividing the dogs before/behind Punto and placing Punto.
# | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Anička set out from the hotel for a walk, walked $5 \mathrm{~km}$ north, then $2 \mathrm{~km}$ east, $3 \mathrm{~km}$ south, and finally $4 \mathrm{~km}$ west. Thus, she arrived at a pond where she took a bath. Vojta set out from the camp, walked $3 \mathrm{~km}$ south, $4 \mathrm{~km}$ west, and $1 \mathrm{~km}$ north... | According to the task, we will sequentially illustrate Anička's route, Vojta's route, and their mutual relationship (each marked tile represents $1 \mathrm{~km}$):
Now we see that Anička has... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a given group of numbers, one number is equal to the average of all, the largest number is 7 greater than the average, the smallest is 7 less than the average, and most of the numbers in the group have below-average values.
What is the smallest number of numbers that can be in the group?
(K. Pazourek)
Hint. What ... | Let's denote the average of the numbers in the group as $p$. The smallest number in the group is $p-7$, and the largest is $p+7$. The average of these three numbers is $p$, so the average of the remaining numbers in the group must also be $p$.
Therefore, some of the remaining numbers must be less than $p$, and some mu... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In a regular pentagon $A B C D E$, an equilateral triangle $A B M$ is contained. Determine the size of the angle $B C M$.
(L. Hozová)
Hint. What are the sizes of the interior angles of a regular pentagon? | The size of the internal angles of an equilateral triangle is $60^{\circ}$, the size of the internal angles of a regular pentagon is $108^{\circ}$. At vertex $B$, we find that the size of angle $C B M$ is $108^{\circ}-60^{\circ}=48^{\circ}$.
Segments $A B, B C$, and $B M$ are congruent, so triangle $C B M$ is isoscele... | 66 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Grandpa forgot the four-digit code of his mobile phone. He only remembered that the first digit was not zero, that in the middle were either two fours or two sevens or a four and a seven (in an unknown order), and that the number was divisible by 15. How many possibilities are there for the forgotten code? What digit c... | The middle two places can be occupied in exactly four ways:
$$
* 44 *, * 77 *, * 47 *, * 74 * .
$$
A number is divisible by 15 if and only if it is divisible by both three and five. A number is divisible by five if and only if its last digit is either 0 or 5, and a number is divisible by three if and only if the sum ... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. September 2007, a language school was founded, where seven educators taught. On 1. September 2010, a new colleague joined these seven teachers, who was exactly 25 years old. By 1. September 2012, one of the teachers left the school, leaving seven teachers again. The average age of the educators at the school was the... | The sum of the ages of all seven teachers at the school on September 1, 2007, is denoted by $c$. The sum of the ages of these seven people increased by $7 \cdot 3=21$ on September 1, 2010, so the sum of the ages of all eight teachers working at the school on that day was
$$
c+21+25=c+46.
$$
The sum of the ages of the... | 62 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
My only son was born when I was 37 years old. This was exactly 32 years after my grandfather's death, and he died at the age of 64. My grandfather was 12 years older than my grandmother, and they got married in 1947, when my grandmother was 18 years old.
In which year was my son born?
Note: Ignore any discrepancies r... | We can solve the task according to the given information from the back:
In 1947, the grandmother was 18 years old and the grandfather was $30(=18+12)$.
The grandfather died at the age of 64, which was 34 years after the wedding $(64-30=34)$, which was in 1981 ( = $1947+34)$.
The son was born 32 years after the grand... | 2013 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Kuba wrote down a four-digit number, two of whose digits were even and two were odd. If he crossed out both even digits in this number, he would get a number four times smaller than if he crossed out both odd digits in the same number.
What is the largest number with these properties that Kuba could have written down?... | After crossing out the even digits, the remaining number should be four times smaller than another two-digit number. This number must therefore be smaller than $25(4 \cdot 25=100$, and that is already a three-digit number).
After crossing out the even digits, the remaining number should be composed of odd digits. The ... | 6817 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Five friends were comparing how much scrap metal they had brought to the collection. On average, it was $55 \mathrm{~kg}$, but Ivan only brought $43 \mathrm{~kg}$.
How many kg on average did the others bring without Ivan?
(L. Hozová)
Hint. By how many kg does Ivan's contribution differ from the average? | Ivan brought $43 \mathrm{~kg}$, which is $12 \mathrm{~kg}$ less than the (arithmetic) average of all his friends. These $12 \mathrm{~kg}$ correspond to an average of $3 \mathrm{~kg}$ for each of the four remaining friends $(12: 4=3)$.
Without Ivan, the friends brought an average of $58 \mathrm{~kg}$ of iron $(55+3=58)... | 58 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Anička and Blanka each wrote a two-digit number starting with a seven. The girls chose different numbers. Then each inserted a zero between the two digits, creating a three-digit number. From this, each subtracted their original two-digit number. The result surprised them.
Determine how their results differed.
(L. Ho... | A two-digit number starting with a seven is of the form $7 *$, where the asterisk can be any digit. By inserting a zero, we get a three-digit number of the form 70*. Regardless of which digit the asterisk represents in the units place, the difference is
$$
\begin{array}{r}
70 * \\
-\quad 7 * \\
\hline 630
\end{array}
... | 630 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Honza had 100 identical preserving jars, with which he built triangular pyramids. The highest floor of the pyramid always has one jar, the second floor from the top represents an equilateral triangle, the side of which consists of two jars, and so on. An example of a three-story pyramid construction is shown in the ima... | 1. We will count the glasses by floors starting from the top. From the problem statement and the guiding images, we know that in the fifth (highest) floor, there is 1 glass, in the fourth floor, there are 3 glasses, and in the third floor, there are 6 glasses. Each subsequent (lower) floor can be imagined as adding one... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the tin, there were red and green candies. Čeněk ate $\frac{2}{5}$ of all the red candies and Zuzka ate $\frac{3}{5}$ of all the green candies. Now, the red candies make up $\frac{3}{8}$ of all the candies in the tin.
How many candies were there in the tin at least originally?
(L. Růžičková)
Hint. How many candie... | Both Čeněk and Zuzka ate several fifths of the candies of the respective color. Therefore, the original number of both red and green candies must be divisible by five. We will consider the smallest number divisible by five as the original number of red candies and try to express the number of green candies:
- If there... | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A hotelier wanted to furnish the dining room with new chairs. In the catalog, he chose a type of chair. Only when placing the order did he learn from the manufacturer that as part of a discount offer, every fourth chair was offered at half price, and that he could thus save the equivalent of seven and a half chairs com... | In the promotion, every fourth chair was at half price. If the hotelier could save the equivalent of 7.5 chairs, he ordered 15 sets of four chairs and at most three additional chairs, i.e., he ordered at least 60 and at most 63 chairs.
Compared to the original plan, he could afford 9 more chairs during the promotion, ... | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A natural number $N$ is called bombastic if it does not contain any zero in its representation and if no smaller natural number has the same product of digits as the number $N$.
Karel first became interested in bombastic prime numbers and claimed that there are not many of them. List all two-digit bombastic prime numb... | All two-digit prime numbers are listed in the first row of the following table. In the second row, the digit products of the individual numbers are listed. In the third row, the smallest natural numbers with the corresponding digit products are listed (these numbers can be determined by comparing the factorizations wit... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Jitka worked on a brigade with a fixed daily wage. In three days, she earned enough money to buy a board game and still had 490 CZK left. If she had spent five days on the brigade, she could have bought two such board games and still had 540 CZK left.
How much did the board game cost?
(K. Pazourek)
Hint. First, try ... | For a three-day pay, Jitka bought one game and had 490 Kč left, so for a six-day pay, she could buy two games and have 980 Kč left. Meanwhile, for five days, she would also earn enough for two games, but she would have only 540 Kč left. Therefore, Jitka's daily wage was $440 \mathrm{~K} c$ c $(980-540=440)$.
From the ... | 830 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Pan Stříbrný uspořádal výstavu. Vystavoval 120 prstenů, které ležely na stolech podél stěn sálu a tvořily tak jednu velkou kružnici. Prohlídka začínala u vchodových dveří v označeném směru. Odtud každý třetí prsten v řadě byl zlatý, každý čtvrtý prsten byl starožitný a každý desátý prsten měl diamant. Prsten, který nem... | Every 3rd ring was golden, every 4th was antique, and every 10th had a diamond. Thus,
- the number of golden rings was $120: 3=40$,
- the number of antique rings was $120: 4=30$,
- the number of rings with a diamond was $120: 10=12$.
When counting rings with multiple properties, we first determine the regularity with... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Body $A, B, C, D$ and $E$ are vertices of an irregular five-pointed star, see the figure. Determine the sum of the marked angles.
Note: the figure is illustrative only.
Hint. What is the su... | Using the marked angles, we can express all other angles in the given polygon. We will start this way and try to find out something about the desired sum. The marked angles and the remaining vertices of the polygon are labeled as follows:[^0]
 squares, which must be the sum of the newly formed black shapes. Therefore, a black square can have a maximum size of $5 \times 5$ squares. We will examine all possible cases according to the size of the black square, determine how many squares remain for the black re... | 168 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Ctirad programmed two cooperative drawing robots, Miky and Niki. Miky can construct squares, regular pentagons, and regular hexagons. However, in one day, he draws only congruent polygons. Niki adds all the diagonals to all of Miky's polygons.
1. On Monday, Miky constructed the same number of segments as Niki. Which p... | A square has two diagonals, a pentagon has five, and a hexagon has nine:
1. Only a pentagon has the same number of sides and diagonals. On Monday, the robots drew pentagons.
2. Miky construc... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Petra filled in the circles with the numbers $1,2,3,4,5,6,7,8$ such that each was used exactly once and the sum of the numbers on each side of the triangle was the same.
What is the largest sum she could have obtained this way? Provide an example of a possible filling.
(A. Bohiniková)
![](https://cdn.mathpix.com/cro... | Numbers at the vertices of the triangle contribute to the sums on two sides, while other numbers contribute only to one. Larger numbers at the vertices of the triangle will therefore result in larger sums.
We will place the three largest numbers at the vertices of the triangle and try to fill in the remaining numbers.... | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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