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 |
|---|---|---|---|---|---|---|---|---|
In the trapezoid $K L M N$, the base $K L$ has a length of $40 \mathrm{~cm}$ and the base $M N$ has a length of $16 \mathrm{~cm}$. Point $P$ lies on the segment $K L$ such that the segment $N P$ divides the trapezoid into two parts with equal areas.
Determine the length of the segment $K P$.
(L. Hozová)
Hint. Modify... | Segment $N P$ divides trapezoid $K L M N$ into triangle $K P N$ and trapezoid $P L M N$. Trapezoid $K L M N$ has the same area as triangle $K O N$, where point $O$ on line $K L$ is the image of point $N$ with respect to central symmetry about the midpoint of segment $L M$. Similarly, trapezoid $P L M N$ has the same ar... | 28 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Pat wrote on the board the example:
$$
589+544+80=2013 .
$$
Mat wanted to correct the example so that both sides would actually be equal, and he searched for an unknown number which he then added to the first addend on the left side, subtracted from the second addend, and multiplied the third addend by. After perform... | By adding an unknown number to the first addend and subtracting the same number from the second addend on the left side, the sum of these two numbers will not change and is equal to $589+544=$ $=1133$. This intermediate sum is $2013-1133=880$ less than the number on the right side of the equation. Therefore, the produc... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A Christmas tree seller sold spruces for 220 Kč, pines for 250 Kč, and firs for 330 Kč. In the morning, he had the same number of spruces, firs, and pines. In the evening, he had sold all the trees and earned a total of 36000 Kč for them.
How many trees did the seller sell that day?
Hint. Count in threes. | A trio of fir, spruce, and pine saplings stood together
$$
220+250+330=800 \text { Kč. }
$$
The seller of such trios sold $36000: 800=45$ in a day.
In total, he sold $3 \cdot 45=135$ saplings. | 135 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Jiřka constructed two congruent equilateral triangles as shown in the figure. She further wants to construct all circles that will have their center at one of the vertices and will pass through another vertex of one of the triangles.
Construct and calculate all circles that meet Jiřka's requirements.
(K. Pazourek)
!... | Let's name the vertices as shown in the following image and note that points $A$ and $C$ are always equidistant from the remaining three points (the corresponding segments form the sides of equilateral triangles). Therefore, the circle centered at point $A$ passing through point $B$ also passes through points $C$ and $... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Determine which digit is in the 1000th place after the decimal point in the decimal expansion of the number $\frac{9}{28}$.
(M. Krejčová)
Hint. What does the decimal expansion of the given number look like? | The decimal expansion of the rational number $\frac{9}{28}$ is
$$
0.32 \overline{142857}
$$
where the repeating part, consisting of six digits, is marked with a bar.
Six goes into a thousand 166 times with a remainder of four $(1000 = 166 \cdot 6 + 4)$. There are two digits between the decimal point and the repeatin... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Kuba agreed with the shepherd that he would take care of the sheep. The shepherd promised Kuba that after a year of service, he would get twenty gold coins and one sheep. But Kuba quit just when seven months of service had passed. Even so, the shepherd fairly rewarded him and paid him five gold coins and one sheep.
Ho... | Kubovi had 5 months left until the end of the year and received 15 gold coins less than he would have received for a full year of service. This means that if he had been paid only in cash, he would have received 3 gold coins per month $(15: 5=3)$.
For 7 months, Kuba should have received 21 gold coins $(7 \cdot 3=21)$.... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A subtraction pyramid is a pyramid formed by non-negative integers, each of which is the difference between the two nearest numbers from the previous level (read from bottom to top). Here is an example of a subtraction pyramid:
## 1
2
2
5 7
4
3
5
8
A significant number is the largest number in the subtraction py... | If it is 0 at the top, there must be two identical numbers in the previous row. Each excellent pyramid must have at least three floors.
An excellent pyramid with the smallest number of floors and the smallest possible numbers looks like this (except for the order of the numbers on the third line):
## 0
1
0 ..... 2
... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In triangle $ABC$, point $D$ lies on side $AC$ and point $E$ lies on side $BC$. The measures of angles $ABD, BAE, CAE$, and $CBD$ are $30^{\circ}, 60^{\circ}, 20^{\circ}$, and $30^{\circ}$, respectively.
Determine the measure of angle $AED$.
![](https://cdn.mathpix.com/cropped/2024_04_17_6d7ca75cbeb5035abbabg-4.jpg?h... | The size of angle $ABC$ is equal to the sum of the sizes of angles $ABD$ and $CBD$, i.e., $30^{\circ}+30^{\circ}=60^{\circ}$. In triangle $ABE$, two interior angles have a size of $60^{\circ}$, so the remaining angle also has a size of $60^{\circ}$, making the triangle equilateral.
Line $BD$ is the axis of the interio... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Vendelín lives between two bus stops, at a distance of three eighths of their distance. Today, he set out from home and found that whether he ran to one or the other stop, he would arrive at the stop at the same time as the bus. The average speed of the bus is $60 \mathrm{~km} / \mathrm{h}$.
What is the average speed ... | Let the first bus stop on the route be $A$, the second $B$, and Vendelín's house $D$. If Vendelín runs to $A$, he will cover the distance $D A$ before the bus arrives. If Vendelín runs to $B$, by the time the bus reaches $A$, he will have covered the same distance $D A$ towards $B$. At this point, we realize that $D$ m... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
At the mountain cabin, our coach said: "If we continue at this comfortable pace of $4 \mathrm{~km}$ per hour, we will arrive at the station 45 minutes after our train has left."
Then he pointed to a group that was just passing us: "They are using poles, and thus achieve an average speed of $6 \mathrm{~km}$ per hour. T... | Let the time from the moment the coach motivated us at the mountain hut until the departure of the train be denoted by $t$ (in hours). Let the length of the route from the hut to the station be denoted by $s$ (in km).
At a comfortable pace, we would walk $\frac{s}{4}$ hours and arrive three-quarters of an hour after t... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In Mr. Goat's garden, several cherry trees were in bloom. On each cherry tree, three sparrows were sitting, and one more was sitting on the fence. Mr. Goat's dog scared them away and the sparrows flew off. After a while, they all returned and settled on the cherry trees. The cherry tree under which the dog was sleeping... | Let $x$ be the number of cherries in the garden and $y$ be the number of sparrows. If we know that all the sparrows perched on the cherries in groups of three and one remained on the fence, we can set up the following equation:
$$
y=3 x+1
$$
After the sparrows returned to the cherries, they occupied all the cherries ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Given a triangle $A B C$ such that the foot $P$ of the perpendicular from point $C$ to line $A B$ lies inside segment $A B$. From point $P$, perpendiculars $p, q$ are drawn to lines $A C$ and $B C$ (in that order). Let $S$ be the intersection of line $B C$ and line $q$, and $T$ be the intersection of line $A C$ and lin... | From the position of point $P$, it follows that the internal angles $\alpha=|\nless B A C|$ and $\beta=|\nless A B C|$ of the given triangle are acute:
From the right-angled triangles $A P T... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Gardener Mr. Malina was selling strawberries. In the last nine crates, he had 28, 51, 135, 67, 123, 29, 56, 38, and 79 strawberry plants, respectively. He sold the crates whole, never removing any plants from the crates. The gardener wanted to sell the crates to three customers so that nothing was left and each of thes... | The total number of seedlings was 606. Therefore, each of the three customers was supposed to receive 202 seedlings.
The number of seedlings in individual boxes can be divided into three groups with a sum of 202 in several ways:
$$
\begin{aligned}
& \text { a) } 135+67, \quad 123+79, \quad 56+51+38+29+28 \text {, } \... | 202 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the image, there is a number line with marked numbers 10 and 30 and other unnamed points representing whole numbers. Janek marked his favorite number and four other numbers on this line, about which we know that:
- one is half of Janek's number,
- one is 6 greater than Janek's number,
- one is 10 less than Janek's ... | Between the numbers 10 and 30, there are 10 segments. Thus, one segment has a length of 2, and the numbers marked by dots are in sequence $6, 8, 16, 22$, and 32.
From the problem statement, ... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In Zverimex, they were selling fish from one aquarium. Ondra wanted half of all the fish, but to avoid cutting any fish, he received half a fish more than he requested. Matěj wished for half of the remaining fish, but like Ondra, he received half a fish more than he requested. Finally, Petr wanted half of the remaining... | Let's consider it from the end:
Petr got half a fish more than half of all the fish that remained after Matej. Since the aquarium was then empty, that half a fish more was exactly half of what remained after Matej. Therefore, after Matej's purchase, one fish remained in the aquarium.
Matej got half a fish more than h... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Jiřina has a four-digit number written on a piece of paper. If she swaps the digits in the hundreds and units places and adds this new number to the original number, she gets a result of 3332. However, if she swaps the digits in the thousands and tens places and adds this number to the original, she gets a result of 78... | Let the digits of Jiří's number be denoted by $a, b, c, d$. Then the information from the problem can be written as follows:
$$
\begin{array}{cccc}
a & b & c & d \\
a & d & c & b \\
\hline 3 & 3 & 3 & 2
\end{array} \quad \quad \begin{array}{llll}
a & b & c & d \\
c & b & a & d \\
7 & 8 & 8 & 6
\end{array}
$$
From the... | 1468 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The encryption games involved 168 players in 50 teams, which had two to five members. The most were four-member teams, there were 20 three-member teams, and the games were attended by at least one five-member team. How many two-member, four-member, and five-member teams were there? (M. Mach) | There were 20 three-member teams, which represents 60 players. The remaining 108 players from the total number need to be divided into 30 teams of two, four, and five players. Four-member teams were the most, i.e., at least 21, which represents at least 84 players. The remaining 24 players need to be divided into 9 tea... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In front of our school, there is a flower bed. One fifth of all the flowers are tulips, two ninths are daffodils, four fifteenths are hyacinths, and the rest are pansies. How many flowers are there in total in the bed if there are no more than 60 and no fewer than 30 of any type? (M. Petrová)
Hint. Could there be, for... | The total number of flowers must be divisible by five, because one fifth of them are tulips. For a similar reason, the total number of all flowers must be divisible by nine (due to daffodils) and fifteen (due to hyacinths). Therefore, the total number of flowers in the bed is some common multiple of the numbers 5, 9, a... | 180 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The teacher wrote the following numbers on the board:
$$
1,4,7,10,13,16,19,22,25,28,31,34,37,40,43
$$
Any two adjacent numbers differ by the same value, in this case, 3. Then she erased all the numbers from the board except for 1, 19, and 43. She further added several integers between these three numbers such that an... | One way to fill in the numbers is, of course, the one the teacher erased (in this case, the difference between consecutive numbers is 3). Another possible, and probably the simplest, way is to fill in all natural numbers from 1 to 43 (in this case, the difference is 1).
Each filling according to the task is completely... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a sports area, there is a rectangular plot $A B C D$ with the longer side $A B$. The diagonals $A C$ and $B D$ form an angle of $60^{\circ}$. Runners train on the large circuit $A C B D A$ or on the small track $A D A$. Mojmír ran ten times around the large circuit, and Vojta ran fifteen times around the small track... | The intersection of the diagonals is denoted by $S$. We need to decide whether the given angle of $60^{\circ}$ refers to angle $A S B$ or $A S D$. Since for the sides of the rectangle, $|A B|>|A D|$, angle $A S B$ must be obtuse and angle $A S D$ must be acute. Therefore, the measure $60^{\circ}$ belongs to angle $A S ... | 100 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
We have a square grid with 77 nodal points. Two of them are marked as $A$ and $B$ as shown in the figure. Let $C$ be one of the remaining nodal points. Find all possible positions of point $C$ such that triangle $ABC$ has an area of 6 square units.
(E. Novotná)
$. If 17 points were exchanged only for stickers, two points would be unused (the remainder of $17: 5$ is 2), which d... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Eliška was placing cakes into boxes, from which she then built a pyramid as shown in the picture. Each box in a higher row contained as many cakes as the two adjacent boxes below it combined. In the three boxes marked with stars, there were three, five, and six cakes.
The sum of all the given numbers is
$$
4 a ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Maruška received a magic pouch from her grandmother, which always doubled the amount of gold coins it contained at midnight. On Monday at noon, Maruška put some gold coins into the empty pouch. On Tuesday and Wednesday, she took out 40 gold coins each day and did not put anything in. On Thursday, she took out 40 gold c... | 1. On Thursday, after taking out 40 gold coins, the purse was empty, so there were those 40 gold coins before taking them out.
On Wednesday, after taking out (before the nightly doubling), there were $40: 2=20$ gold coins in the purse, so before taking them out, there were $20+40=60$ gold coins.
On Tuesday, after tak... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A grid with nine cells, as shown in the image, is filled with nine consecutive natural numbers. These numbers are arranged in ascending order from left to right and from top to bottom (i.e., the smallest number is in the top left, and the largest is in the bottom right). The thick broken line divides the grid into two ... | On each line, the number in its rightmost field is 2 greater than the number in its leftmost field. The sum of the numbers in the rightmost column is therefore $3 \cdot 2=6$ greater than the sum of the numbers in the leftmost column. The number in the lower field of the middle column is 3 greater than the number in its... | 94 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Mr. Dolphin and Mr. Shark were skilled fishermen. Once, they caught 70 fish together. Five ninths of the fish caught by Mr. Dolphin were trouts. Two seventeenths of the fish caught by Mr. Shark were carps.
How many fish did Mr. Dolphin catch?
(L. Hozová) | The number of fish caught by Mr. Dolphin was a multiple of 9 and did not exceed 70. Therefore, Mr. Dolphin could have caught
$$
9, \quad 18, \quad 27, \quad 36, \quad 45, \quad 54, \quad 63
$$
fish. The number of fish caught by Mr. Shark was a multiple of 17 and did not exceed 70. Therefore, Mr. Shark could have caug... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A right-angled triangle has an area of $36 \mathrm{~m}^{2}$. Inside it, a square is placed such that two sides of the square are parts of two sides of the triangle, and one vertex of the square is at one-third of the longest side.
Determine the area of this square.
(E. Novotná) | The only possible placement of the square within the triangle is as shown in the following image. Additional division of the sides into thirds and connecting the corresponding points with dashed lines divides the given triangle into 9 congruent smaller triangles:
![](https://cdn.mathpix.com/cropped/2024_04_17_768244f2... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Monika is thinking of a four-digit number that has the following properties:
- the product of the two outer digits is 40,
- the product of the two inner digits is 18,
- the difference between the two outer digits is the same as the difference between the two inner digits,
- the difference between the thought number an... | The product of the two extreme digits is 40, and this is only possible as $40=5 \cdot 8$. The difference between these digits is $8-5=3$. The product of the two inner digits is 18, which is possible either as $18=2 \cdot 9$ or as $18=3 \cdot 6$. In the first case, the difference is $9-2=7$, which is different from the ... | 8635 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Vojta bought 24 identical square tiles. The side of each tile measured $40 \mathrm{~cm}$. Vojta wanted to create a rectangular platform in front of the cottage with the smallest possible perimeter. How many meters did the perimeter of the paved rectangle measure, given that no tiles were left over? Vojta did not cut or... | Let's go through all the possibilities; for each, we will provide the dimensions of such a rectangle in "tiles" and further the perimeter in "tiles," resp. in "sides of tiles":
- dimensions: $1 \times 24$, perimeter: $2 \cdot(1+24)=50$,
- dimensions: $2 \times 12$, perimeter: $2 \cdot(2+12)=28$,
- dimensions: $3 \time... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The thief Rumcajs is teaching Cipísek to write numbers. They started writing from one and wrote consecutive natural numbers. Cipísek didn't find it very entertaining and pleaded for them to stop. Rumcajs eventually gave in and promised that he would stop writing as soon as Cipísek had a total of 35 zeros in the written... | By writing down single-digit numbers, we obtain no zeros.
Two-digit numbers containing a zero are:
$$
10,20, \ldots, 90 \quad 9 \text{ zeros}
$$
Three-digit numbers containing a zero are:
| 100 | 2 zeros | (total 11 zeros) |
| :--- | :--- | :--- |
| $101,102, \ldots, 109$ | 9 zeros | (total 20 zeros) |
| $110,120, ... | 204 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On a wire, 9 swallows are sitting at regular intervals from each other. The distance between the first and the last swallow is $720 \mathrm{~cm}$.
- What is the distance between neighboring swallows?
- How many swallows would be sitting on the wire if three more swallows sat between each pair of the already sitting sw... | Among nine swallows, there are only 8 gaps.
1. The distance between neighboring swallows is $720: 8=90(\mathrm{~cm})$.
2. If 3 new swallows perched in each of the eight gaps, a total of $8 \cdot 3=24$ swallows would perch. There would then be $9+24=33$ swallows on the wire.
Evaluation. 1 point for the observation abo... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Káta wants to give presents to her friends and is thinking: if I bought each of them a hair clip for 28 Kč, I would still have 29 Kč left, but if it were a teddy bear for 42 Kč, I would be short of 13 Kč. How many friends does Káta have and how much money does she have for gifts?
(M. Volfová) | The difference in the price of a gift for one friend is $42-28=14$ (CZK). The difference in the total price that Káta would pay for gifts for all her friends is $29+13=42$ (CZK). If we multiply the difference in the price of a gift for one friend by the number of friends, we get the difference in the total price. There... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Ondra spent $\frac{2}{3}$ of his money on a trip and then gave $\frac{2}{3}$ of the remainder to a school for children from Tibet. He then spent $\frac{2}{3}$ of the new remainder on a small gift for his mother. From the remaining money, he lost $\frac{4}{5}$ due to a hole in his pocket, and when he gave half of what w... | The number of Ondra's crowns before the trip is denoted as $x$.
- Ondra spent $\frac{2}{3}$ of his money on the trip, leaving him with $\frac{1}{3} x$ crowns.
- He donated $\frac{2}{3}$ of the remaining money to a school in Tibet, leaving him with $\frac{1}{3} \cdot \frac{1}{3} x=\frac{1}{9} x$ crowns.
- The gift for ... | 270 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Šárka declared:
"I am the youngest of three sisters. Líba is three years older, and Eliška is eight years older. Our mom loves to hear that the average age of all of us (including her) is 21. When I was born, mom was already 29."
How many years ago was Šárka born?
(M. Volfová) | If we denote Śárka's age in years by $x$, then Líbě is $x+3$, Eliška is $x+8$, and the mother is $x+29$ years old. The average age of all of them is 21 years, i.e.,
$$
(x+(x+3)+(x+8)+(x+29)): 4=21,
$$
after rearrangement
$$
\begin{aligned}
4 x+40 & =84, \\
x & =11 .
\end{aligned}
$$
Śárka was born 11 years ago. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The image represents a top view of a three-layer pyramid composed of 14 identical cubes. Each cube is assigned a natural number such that the numbers corresponding to the cubes in the bottom layer are all different, and the number on each subsequent cube is the sum of the numbers from the four adjacent cubes in the low... | The number belonging to the topmost cube is determined by four numbers from the second layer, and these are entirely determined by the numbers in the first layer. Each corner cube in the first layer is adjacent to one cube from the second layer, each non-corner cube on the edge of the first layer is adjacent to two cub... | 64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Let's consider a four-digit natural number with the following property: if we swap its first two-digit number with the second, we get a four-digit number that is 99 less.
How many such numbers are there in total, and how many of them are divisible by 9? (K. Pazourek) | Let the original four-digit number be $\overline{a b c d}=1000 a+100 b+10 c+d$. According to the problem, $\overline{a b c d}=\overline{c d a b}+99$, where $b$ and $d$ are integers from 0 to 9, $a$ and $c$ are integers from 1 to 9, and $a \geq c$. By expanding and rearranging the above equation, we get:
$$
\begin{alig... | 89 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Construct a square $A B C D$ with a side length of $6 \text{~cm}$ and mark the intersection of its diagonals as $S$. Construct point $K$ such that together with points $S, B, C$ it forms a square $B K C S$. Construct point $L$ such that together with points $S, A, D$ it forms a square $A S D L$. Construct the segment $... | Construction:
- Square $A B C D$ with side length $6 \mathrm{~cm}$,
- point $S$ as the intersection of segments $A C$ and $B D$,
- point $K$ as the intersection of the perpendicular line $\mathrm{k} S B$ at point $B$ and the perpendicular line to $S C$ at point $C$,
- point $L$ as the intersection of the perpendicular... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the summer camp, there are 50 children. One sixth of the girls and one eighth of the boys cannot swim. 43 children can swim. How many girls are in the camp? | Let the number of girls in the camp be $x$, then the number of boys is $50-x$. From the problem statement, we have that
$$
\frac{x}{6}+\frac{50-x}{8}=7 .
$$
The solution to this equation is $x=18$.
There are 18 girls in the camp.
[2 points for setting up the equation, 6 points for the complete solution] | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
From a cube with a surface area of $384 \mathrm{~cm}^{2}$, we have cut out a cuboid with a square base as shown in the figure. The volume of the resulting octagonal prism is equal to three quarters of the volume of the original cube. Calculate the surface area of the prism.
 What percentage of the remaining matches would the team need to win to achieve their intended goal?
b) If the team won all the remaining matches, what percentage of th... | a) The team won $\frac{55}{100}$ of a third of the matches, and to win $\frac{3}{4}$ of all matches, they need to win another $p / 100$ of $\frac{2}{3}$ of the matches:
$$
\frac{55}{100} \cdot \frac{1}{3}+\frac{p}{100} \cdot \frac{2}{3}=\frac{3}{4}
$$
From this, we get
$$
p=85
$$
Thus, the team would need to win $8... | 85 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Knight Milivoj was preparing to go to Veselín for a tournament. The tournament takes place on Wednesday. Since the journey from Knight's home, Rytířov, to Veselín would take two days, he set off on Monday. The road leads through two other towns, Kostín and Zubín. On the first day of travel, he covered 25 miles and spen... | Let's first visualize the situation from the assignment.
The journey to the tournament:
The journey from the tournament:
Hint. How old are all the grandchildren together? | The grandchildren together are 35 years old $(5 \cdot 7=35)$. Grandpa, grandma, and the grandchildren together are 182 years old $(7 \cdot 26=182)$. Therefore, grandpa and grandma together are 147 years old $(182-35=147)$.
The average age of grandma and grandpa is 73 and a half years $(147: 2=73.5)$, and grandma is on... | 73 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Two congruent equilateral triangles $A B C$ and $B D E$ are given such that points $A, B$, $D$ lie on the same line and points $C, E$ lie in the same half-plane defined by that line. The intersection of $C D$ and $A E$ is marked as $F$.
Determine the measure of angle $A F D$.
(I. Jančigová)
The length of segment $S X$, as well as $S ... | 82 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
From the right pocket of my pants, I moved 4 five-crown coins to the left pocket, and from the left pocket, I moved 16 two-crown coins to the right pocket. Now I have 13 crowns less in the left pocket than in the right. In which pocket did I have more crowns at the beginning, and by how much?
(Bednářová) | We solve the example from the end, using the $\mathrm{k}$ table:
state at the end
left pocket (L) right pocket $(\mathrm{P}) \quad$ difference
move 16 two-crowns from $\mathrm{P}$ to $\mathrm{L}$
32
13
13
move 4 five-crowns from $\mathrm{L}$ to $\mathrm{P}$
12
owe 19
51
1
Conclusion: I had 11 crowns more in... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find the smallest number written only with the digits 0 and 1, which is divisible without remainder by the product of the six smallest natural numbers.
(Bednářová) | First, we determine the product of the six smallest natural numbers: $1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6=2^{4} \cdot 3^{2}$. $\cdot 5=10 \cdot 2^{3} \cdot 3^{2}$. The sought number $n$ can thus be written as $n=10 \cdot a$, where the number $a$ is the smallest number written only with the digits 0 and 1 that is ... | 111111110000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the city of records and curiosities, they built a pyramid from cubes. In the top layer, there is one cube, and the number of cubes in each layer increases by two as you move downward (the top layers of the structure are shown in the image).
 whether the structure can be divided into such triplets without a r... | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Pankrác, Servác, and Bonifác bought a boat. Pankrác paid $60\%$ of the boat's price, Servác paid $40\%$ of the remaining price, and Bonifác paid the missing amount, which was 30 gold coins.
How many gold coins did the boat, which the boys bought, cost?
(L. Hozová) | Let the cost of the boat be denoted by $z$. Pankrác paid $\frac{6}{10} z$, leaving
$$
\left(1-\frac{6}{10}\right) z=\frac{4}{10} z
$$
Servác paid $\frac{4}{10} \cdot \frac{4}{10} z=\frac{16}{100} z$, leaving
$$
\left(\frac{4}{10}-\frac{16}{100}\right) z=\frac{24}{100} z=\frac{6}{25} z
$$
The remaining amount was pa... | 125 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Karolína wrote all three-digit numbers formed by the digits 1, 2, and 3, in which no digit was repeated and 2 was in the tens place. Nikola wrote all three-digit numbers formed by the digits 4, 5, and 6, in which no digit was repeated. Kuba chose one number from Karolína and one number from Nikola so that the sum of th... | Karolína wrote the numbers
$$
123, \quad 321 .
$$
Nikola wrote the numbers
$$
456, \quad 465, \quad 546, \quad 564, \quad 645, \quad 654 .
$$
Both of Karolína's numbers are odd. For an even sum, Kuba had to choose an odd number from Nikola. The even sums are given by the following cases:
$$
123+465, \quad 123+645,... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In 1966, there were 30 more women than men living in the village of Bezdíkov. To the present day, the number of women living in the village has decreased fourfold, and the number of men living in the village has decreased by 196. Now, there are 10 more women than men in Bezdíkov.
How many women and men in total live i... | The difference between the original number of women and the current number of men is $30+196=226$, which is 10 more than the difference between the original and current number of women. This means that three quarters of the original number of women are equal to $226-10=216$. The remaining one quarter, corresponding to ... | 134 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In rectangle $A B C D$ with side $A D$ of length $5 \mathrm{~cm}$, point $P$ is located such that triangle $A P D$ is equilateral. Ray $A P$ intersects side $C D$ at point $E$, and segment $C E$ measures $5 \mathrm{~cm}$.
How long is segment $A E$ and what is the measure of angle $A E B$?
(L. Hozová) | Triangle $A P D$ is equilateral, so the lengths of all its sides are $5 \mathrm{~cm}$ and the measures of all its internal angles are $60^{\circ}$.
The internal angles of rectangle $A B C D$ are right angles, so the measures of angles $P D E$ and $P A B$ are $30^{\circ}$. Angle $P E D$ is the third angle of triangle $... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In all nine fields of the diagram, natural numbers are to be filled in such a way that the following conditions are met:
- each of the numbers $2, 4, 6$, and 8 is used at least once,
- four of the fields in the inner square contain the products of the numbers in the adjacent fields of the outer square,
- in the circle... | For a sufficiently large number in one of the corner fields of the outer square, the corresponding products in the inner square can be greater than any arbitrarily chosen number. Therefore, the sum in the circle can also be arbitrarily large.
We will determine the smallest sum that can be in the circle. It certainly c... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
From point $A$ to point $C$, there is an educational trail passing through point $B$, and there is also a red tourist trail, see the image. In addition, a shortcut not shown in the image can be used, which is 1500 meters long, starting at $A$ and ending on the educational trail. Vojtěch found that
1. the trip from $A$... | The shortcut could have led to the educational trail either in the section between $A$ and $B$, or in the section between $B$ and $C$.
The shortcut was $1500 \mathrm{~m}$ long, and with its ... | 4500 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Katka thought of a five-digit natural number. On the first line of her notebook, she wrote the sum of the thought number and half of the thought number. On the second line, she wrote the sum of the thought number and one fifth of the thought number. On the third line, she wrote the sum of the thought number and one nin... | If we denote the five-digit number Katka thought of by $x$, then the numbers written on the first three lines were $x+\frac{1}{2} x=\frac{3}{2} x, x+\frac{1}{5} x=\frac{6}{5} x$ and $x+\frac{1}{9} x=\frac{10}{9} x$. The sum on the fourth line was equal to
$$
\left(\frac{3}{2}+\frac{6}{5}+\frac{10}{9}\right) x=\frac{34... | 11250 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Mice have built an underground house consisting of chambers and tunnels:
- each tunnel leads from one chamber to another (i.e., none are dead ends),
- from each chamber, exactly three tunnels lead to three different chambers,
- from each chamber, one can reach any other chamber via tunnels,
- the house has exactly one... | We will denote cells with circles and tunnels with lines. We start with a critical tunnel, whose collapse divides the house into two separate parts. If we denote the cells at the ends of this tunnel as $A$ and $B$, then each cell belongs to exactly one of the following two groups:
a) cell $A$ and all cells that can be... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
At the tournament, teams Akouska, Bovenska, Colska, and Demecka met. Each team played against each other exactly once. The winning team received three points, the losing team received no points, and in the event of a draw, each of the drawing teams received one point. After all six matches were played, Akouska had 7 po... | The team names are abbreviated to A, B, C, and D. Each team played three matches against the remaining teams, and could earn 0, 1, or 3 points per match. The distribution of points for teams A, B, and D is therefore uniquely determined, while for team C there are two possibilities:
$$
7=3+3+1, \quad 4=3+1+0, \quad 3=3... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Pat wrote a strange example on the board:
$$
550+460+359+340=2012.
$$
Mat wanted to correct it, so he searched for an unknown number that he could add to each of the five given numbers to make the example numerically correct. What was this number?
(L. Hozová)
Hint. How many numbers will Mat add to the left and how ... | The sum of four numbers on the left side of the equation is $550+460+359+340=1709$; this is a number $2012-1709$ = 303 smaller than the number on the right side of the equation. To the left side, we add
Mato's number four times, and to the right side only once. The difference of 303 between both sides must therefore b... | 101 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The worm consists of a white head and several segments, see the image.
When the worm is born, it has a head and one white segment. Each day, the worm gains a new segment in one of the follow... | On the first day of its life, a worm has only one white segment (and a head):
On the second day, it can grow a new segment only in the first of the mentioned ways; all worms that are two days... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Calculate $3 \cdot 15+20: 4+1$.
Then add parentheses to the expression so that the result is:
1. the largest possible integer,
2. the smallest possible integer.
(M. Volfová)
Hint. Determine where parentheses can be placed throughout the expression. | Given example without any parentheses yields:
$$
3 \cdot 15+20: 4+1=45+5+1=51 .
$$
Now we will place parentheses and compare the results step by step. We aim to exhaust all possibilities, while ignoring parentheses that are unnecessary, for example:
$$
[3 \cdot(15+20)]: 4+1=3 \cdot(15+20): 4+1
$$
and so on. First, ... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Seven dwarfs stood around the perimeter of their garden, one in each corner, and stretched a rope around the entire garden. Snow White came out from Smudla and walked along the rope. First, she walked four meters to the east, where she met Prof. From there, she continued two meters to the north until she reached Rejpal... | Let's draw the entire garden on a square grid, where one square represents one square meter.
Now we can easily determine the area of the shape. First, we count all the whole squares, which a... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Grandma told her grandchildren: "Today I am 60 years and 50 months and 40 weeks and 30 days old."
When was the last birthday Grandma had?
(L. Hozová)
Hint. How many whole years are 50 months? | 50 months are four years and 2 months. Thus, the grandmother was at least 64 years old. The remaining 2 months, 40 weeks, and 30 days represent roughly one additional year. We need to accurately determine whether it is more or less, so we will convert the given data into days.
2 months can have 59 days (January and Fe... | 65 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Once, a king called all his pages and lined them up. He gave the first page a certain number of ducats, the second page two ducats less, the third page again two ducats less, and so on. When he reached the last page, he gave him the appropriate number of ducats, turned around, and proceeded in the same manner back to t... | The last page boy received ducats only once, while the other page boys received them twice.
However, the second-to-last page boy received double what the last one did: first 2 ducats more than the last one, then 2 ducats less.
Everyone except the last page boy received the same amount: first, each received 2 ducats m... | 16 | Number Theory | 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.)
Among three-digit numbers, determine whether there a... | Both odd and even digits are five in number. For comic numbers, any of the odd digits $1,3,5,7,9$ can be in the first position. For each of these five possibilities, any of the even digits $0,2,4,6,8$ can be in the second position, which gives $5 \cdot 5=25$ possibilities. For each of these 25 possibilities, any of the... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bětka wrote down thirty consecutive integers and added all these numbers together. Then she erased the second, fifth, eighth, and every subsequent third number. She added all the numbers that remained after erasing and found that the new sum was 265 less than the original sum. Determine the number that Bětka erased fir... | Let the original thirty numbers be denoted as follows:
$$
a, a+1, a+2, a+3, a+4, a+5, \ldots, a+27, a+28, a+29 .
$$
The erased numbers were ten, specifically
$$
a+1, a+4, a+7, a+10, a+13, a+16, a+19, a+22, a+25, a+28 .
$$
The sum of these numbers is $10a + 145$, which corresponds to the given difference of 265. Fro... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
During a gallery tour, a group of boys and girls met. During the tour, no one disturbed or left. After the tour, 15 girls left, and thus there were twice as many boys as girls left in the gallery. Subsequently, 45 boys left, and five times as many girls as boys remained in the gallery.
How many girls were in the galle... | Let $d$ and $ch$ denote the number of girls and boys during the tour. The relationships from the problem can be written as
$$
ch=2(d-15), \quad d-15=5(ch-45),
$$
The number of girls after the tour, expressed from the first equation, is $d-15=\frac{1}{2} ch$. Together with the second equation, we get an equation with ... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Iveta gradually listed natural numbers composed of the digits 1, 3, 5, 7. She used no other digits, proceeded in ascending order from the smallest number, and did not omit any number. She wrote the numbers immediately after each other, thus forming one extraordinarily long number:
$$
1357111315173133 \ldots
$$
Which ... | From the given digits, Iveta created 4 one-digit numbers. All these numbers occupy 4 places in Iveta's long number.
$Z$ from the given digits, Iveta created $4^{2}=16$ two-digit numbers. All these numbers occupy $2 \cdot 16=32$ places in Iveta's long number. The last digit of the last two-digit number is at the 36th p... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In five bags, there are 52 balls in total. In no two bags is the number of balls the same, and some bags may be empty. All the balls from any (non-empty) bag can be moved to the other four bags so that they will have the same number of balls.
a) Find some distribution of balls into the bags that has all the mentioned ... | After moving the balls, one bag is empty and the other four contain the same number. In total, there are 52 balls, so after moving, there are 13 balls in each non-empty bag $(52: 4=13)$. Therefore, originally, there could not have been more than 13 balls in any bag.
a) Possible original numbers of balls in the bags co... | 12 | Logic and Puzzles | proof | Yes | Yes | olympiads | false |
Eva inscribed a circle in a given triangle. Then she drew three segments that touched the inscribed circle and formed three smaller triangles within the original triangle, see the figure. The perimeters of these three smaller triangles were $12 \mathrm{~cm}, 14 \mathrm{~cm}$, and $16 \mathrm{~cm}$.
Determine the perim... | Let's denote the vertices of the original triangle as $A, B, C$, the points of tangency of the inscribed circle as $D, E, F$, and its center as $S$. The endpoints of the added segments are $G, H, I, J, K, L$, and their points of tangency with the circle are $M, N, O$.
 is solved by all children who like mathematics, but only one-tenth of those who do not like mathematics.
Let's say $x \%$ of children like mathematics, and thus solve MO. Then $(46-x) \%$ of children do not like mathematics and also solve MO. The total number of children who do not like ... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A table of numbers has 4 columns and 99 rows and is created in the following way: starting from the second row, the quartet of numbers on each row is determined by the numbers in the previous row, specifically as the sum of the first and second numbers, the difference of the first and second numbers, the sum of the thi... | The entire table is determined by the numbers in the first row. The first few rows of the table look like this:
| 1 | $a$ | $b$ | $c$ | $d$ |
| :---: | :---: | :---: | :---: | :---: |
| 2 | $a+b$ | $a-b$ | $c+d$ | $c-d$ |
| 3 | $2 a$ | $2 b$ | $2 c$ | $2 d$ |
| 4 | $2(a+b)$ | $2(a-b)$ | $2(c+d)$ | $2(c-d)$ |
| 5 | $4 ... | 4096 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
To a toy store, new stuffed animals have been delivered: dragonflies, ostriches, and crabs. Each dragonfly has 6 legs and 4 wings, each ostrich has 2 legs and 2 wings, and each crab has 8 legs and 2 claws. In total, these delivered toys have 118 legs, 22 wings, and 22 claws. How many heads do they have in total?
(M. P... | For clarity, let's record the data about each toy in a table:
| | legs | wings | claws | heads |
| :---: | :---: | :---: | :---: | :---: |
| dragonfly | 6 | 4 | 0 | 1 |
| ostrich | 2 | 2 | 0 | 1 |
| crab | 8 | 0 | 2 | 1 |
It is clear that only crabs have claws. Since there are 22 claws in total and each crab has 2 c... | 18 | Logic and Puzzles | 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. We have submerged the entire structure in paint. How many cubes have at least one side painted?
(M. Krejčová)
 How many children are in 5.A?
b) How many girls are in 5.A? | All girls and one third of 18 boys, i.e., 6 boys, attend the dance club, which is half of the students in 5.A. 12 boys, who do not attend the dance club, make up the other half of the class.
a) There are thus 24 children in class $V$.
b) $24-18=6$. There are 6 girls.
[ 6 boys ... 1 point, 12 boys are half ... 1 point,... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Mom is sewing towels from fabric that is $120 \mathrm{~cm}$ wide. The finished towel has dimensions of $60 \mathrm{~cm} \times 38 \mathrm{~cm}$. When cutting the fabric, $2 \mathrm{~cm}$ must be added to each edge for finishing. How many centimeters of fabric does Mom need to buy at a minimum to sew 10 towels? | First, we need to realize what piece of fabric mom needs to cut out for one dishcloth. It will be a rectangle with dimensions $42 \mathrm{~cm}$ and $64 \mathrm{~cm}$. This means that we cannot fit even three dishcloths "in width" or two dishcloths "in height" on the fabric (let alone if we rotate one of them).
Let's s... | 254 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Matěj and his friends went caroling. In addition to apples, walnuts, and gingerbread, each of the boys also received an orange. Jarda got one orange, Milan too. Radek, Patrik, Michal, and Dušan each got two oranges. Matěj got as many as four oranges, which was the most of all the boys. The other boys each got three ora... | Boys whose names we know received a total of $1+1+2+2+2+2+4=14$ oranges. For the boys whose names we do not know, there are $23-14=9$ oranges left. Since each of these boys received three oranges, there must have been $9: 3=3$ of them. We know the names of seven boys, and three more boys whose names we do not know, so ... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Snail Josephine landed on a table right in the middle of a square grid consisting of 81 squares, see the image. She decided that she would not crawl away directly, but in the following way: first one square to the south, then one to the east, then two to the north, then two to the west, and again one to the south, one ... | Let's draw the entire path of the snail Josefína on this square grid.
Before Josefína slid off the square grid, she was on the gray-marked square. In total, she crawled over twenty squares of... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A correspondence mathematical competition takes place in three rounds, whose difficulty increases. Only those solvers who were successful in the first round advance to the second round, and only successful solvers of the second round advance to the third round. The winner is anyone who is a successful solver of the las... | Let's denote the number of all solvers in the first round as $x$. The number of successful solvers in the first round (and thus the number of all solvers in the second round) is $14\%$ of $x$, which is $0.14x$. The number of successful solvers in the second round (and thus the number of all solvers in the third round) ... | 5000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
We placed a cube in the center of a potter's wheel, with each of its faces inscribed with a natural number. Just before we spun the wheel, we could see three faces of the cube and thus only three numbers from our position. Their sum was 42. After rotating the potter's wheel by $90^{\circ}$, we observed from the same po... | Let the numbers we see before spinning the circle be denoted as $a, b, c$, where $c$ is the number on the top face. After a $90^{\circ}$ rotation, we lose the face with the number $a$ from our view and a face parallel to it becomes visible. According to the problem, the sum of the visible numbers changes from 42 to 34,... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Two five-digit numbers were written on the wall. Pat added a one in front of one of these numbers, and Mat added a one after the other. This resulted in two six-digit numbers, one of which was three times larger than the other.
Which five-digit numbers were originally written on the wall?
(L. Hozová)
Hint. Which of ... | Both new numbers had the same number of digits, and the larger one was three times larger than the other. The larger number, therefore, could not start with a one - it was Mat's number.
Let the original five-digit number be denoted by $x$. Pat's modification gives the number $100000 + x$, and Mat's modification gives ... | 42857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the picture, there is a structure made of 14 identical cubes. We want to paint the structure from all sides, including the bottom. What will be the paint consumption if $10 \mathrm{ml}$ of paint is enough to paint one entire cube?
(M. Mach)
. According to the problem, $\overline{a b}$ is a prime number, as well as $\overline{a c}$ and $\overline{a d}$. We are therefore looking for three different two-digit prime numbers that start with the same digit (i.e., the... | 4731 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Friends Jarda, Premek, and Robin were playing marbles. Jarda wasn't doing well, so after the game, he had the fewest marbles of all. The boys felt sorry for him, so Robin gave Jarda half of all his marbles, and Premek gave him a third of his. Then Jarda had the most marbles, so he returned seven marbles to each of his ... | We can solve the task advantageously from the end:
- Before the second (last) exchange round, Premek and Robin had seven marbles less, while Jarda had 14 more. So Premek and Robin had $25-7=18$ marbles, while Jarda had $25+14=39$.
- Before the first exchange round, Robin had double the number of marbles (he gave half ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In an eight-digit number, each of its digits (except the last one) is greater than the digit following it. How many such numbers are there?
(I. Jančigová) | The so-called popped numbers can be constructed by removing two out of ten available digits and arranging the remaining eight in descending order. This is the same as removing two digits from the ten-digit number 9876543210:
- If we remove 9 as the first digit, then there are nine options left for which digit to remov... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Péta constructed several plane figures from identical triangles, see the image. The perimeters of the first three are $8 \mathrm{~cm}, 11.4 \mathrm{~cm}$, and $14.7 \mathrm{~cm}$, respectively.
Determine the perimeter of the fourth figure.
(E. Semerádová)
![](https://cdn.mathpix.com/cropped/2024_04_17_d675c96fae1a3d3... | In the perimeters of the first, second, and fourth figures, two out of the three sides of the basic triangle are always included, such that each of the two sides is counted twice, and in the perimeters of different figures, different pairs of sides are included.
In the perimeter of the third figure, all sides of the b... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Petr and Karel played a series of chess games. They agreed that a player would add 3 points for a win, subtract 2 points for a loss, and no points would be awarded for a draw. Their friends wanted to know how many games Petr and Karel had played and who was currently leading, but they only found out that Petr won six t... | Karel had to win so many times that after subtracting 12 points for 6 losses, he still had 9 points left. Therefore, he must have earned $9+12=21$ points from his wins, which corresponds to 7 wins. Thus, Petr lost 7 times. From the problem, we also know that Petr drew twice and won six times, so he has a total of $6 \c... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
We will call a three-digit natural number "even-loving" if it contains two even digits and the digit 1. We will call a three-digit natural number "odd-loving" if it contains two odd digits and the digit 2. Determine how many even-loving and how many odd-loving numbers there are.
(E. Novotná) | First, we determine the number of all even-loving numbers:
- If the digit 1 is in the hundreds place, the tens place can be any even digit $(0,2, 4,6,8)$ and the units place can be the same; this gives a total of $5 \cdot 5=25$ possibilities.
- If the digit 1 is in the tens place, the hundreds place can be any non-zer... | 65 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a two-story house, which is inhabited not only on both floors but also in the basement, 35 people live above someone and 45 people live below someone. At the same time, one third of all the people living in the house live on the 1st floor.
How many people live in the house in total?
(L. Hozová) | People who live above someone are residents of the 2nd and 1st floors. People who live below someone are residents of the 1st floor and the ground floor. In the sum $35+45=80$, the residents of the 1st floor are counted twice.
If we denote the number of residents on the 1st floor by $p$, then the total number of resid... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
For how many positive numbers less than 1000 is it true that among the numbers $2,3,4,5,6,7,8$ and 9 there is exactly one that is not its divisor?
(E. Semerádová) | If a number is not divisible by 2, then it is also not divisible by 4, 6, and 8. If a number is not divisible by 3, then it is also not divisible by 6 and 9. If a number is not divisible by 4, then it is also not divisible by 8. If a number is not divisible by 6, then it is not divisible by 2 or 3. Therefore, none of t... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A four-digit palindrome is any four-digit natural number that has the same digit in the units place as in the thousands place and simultaneously has the same digit in the tens place as in the hundreds place. How many pairs of four-digit palindromes exist whose difference is 3674?
(L. Šimünek) | Let's denote the digits of the palindrome with letters. The fact that two four-digit palindromes have the required difference can be written as follows:
| $A B B A$ |
| ---: |
| $-C D D C$ |
| 3674 |
In the thousands and units columns, the same digits are subtracted. In the thousands column, we see that $A>C$. Theref... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The students of 8. A elected a representative to the school parliament from four candidates. They agreed that a majority of the votes (from those present) was needed for election, and each student, including the candidates, had one vote. In the first round, no one was elected. Anička was short of 3 votes, Petr was shor... | Let the number of students who voted be denoted as $x$.
We will divide the problem into two scenarios:
a) an even number of students voted,
b) an odd number of students voted.
First, let's solve the scenario where an even number of students voted. In this case, $\frac{1}{2} x+1$ votes were needed for election. Anič... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Šárka and Terezka received a box of chocolates, containing 35 chocolate candies. Each day, each of the girls ate at least one candy, and no candy was divided into parts. On the first day, Terezka ate $\frac{2}{5}$ of what Šárka ate on the same day. On the second day, Šárka ate $\frac{3}{4}$ of what Terezka ate on the s... | On the first day, Terezka ate $\frac{2}{5}$ of what Sárka ate. The number of candies Sárka ate that day must therefore be divisible by 5. On the second day, Sárka ate $\frac{3}{4}$ of what Terezka ate. The number of candies Terezka ate that day must therefore be divisible by 4. The number of candies eaten by the girls ... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Dwarves painted cubic dice with green and white colors so that each face was entirely painted in one of these two colors. After a while, they noticed that some of the painted dice looked exactly the same after a suitable rotation and began to sort them into groups (in the same group are identically painted dice).
What... | A cube has six faces, with each face adjacent to four other faces (they share an edge) and parallel to one face (they share no points).
We can categorize the colorings based on the number of green (or white) faces. This gives us seven possibilities, for each of which we will analyze different types of coloring.
- No ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Adámek was recalculating his collection of rainbow marbles. He found that he could divide them into equal piles in several ways. If he divided them into three piles, there would be $\mathrm{v}$ more marbles in each pile than if he divided them into four piles.
How many rainbow marbles did Adámek have?
(E. Semerádová)... | Rearranging the balls from four piles into three can be done by distributing all the balls from one pile into the remaining three.
In each of the three new piles, there were eight more balls than originally, so the pile from which the balls were distributed had 24 balls $(3 \cdot 8=24)$.
All the piles were of equal s... | 96 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the parking lot, there were cars and bicycles. If one more car arrived, there would be as many of them as there are bicycles. If five more bicycles arrived, all the bicycles would have as many wheels as all the cars.
How many cars and how many bicycles were in the parking lot?
(M. Dillingerová)
Hint. Imagine a si... | A car has four wheels, a bicycle has two; one car has the same number of wheels as two bicycles.
There was one more bicycle than cars in the parking lot. Consider the situation where there would be five more bicycles in the parking lot than originally, that is, the situation where the number of wheels would match. In ... | 6 | Algebra | 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.