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
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Robots Robert and Hubert assemble and disassemble coffee grinders. Each of them assembles a coffee grinder four times faster than the other disassembles it. When they came to the workshop in the morning, there were already some assembled coffee grinders there.
At 9:00, Hubert started assembling and Robert disassemblin... | In the morning three-hour shift, 27 coffee mugs were added, which corresponds to $27: 3=9$ coffee mugs per hour. Since Robert disassembles four times slower than Hubert assembles, Hubert alone would assemble 9 coffee mugs in $\frac{3}{4}$ of an hour, i.e., 45 minutes. Therefore, Hubert assembles one coffee mug in $45: ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Congruent rectangles $A B C D$ and $E F G H$ are placed such that their congruent sides are parallel. Points $I, J, K, L, M$, and $N$ are the intersections of the extended sides as shown in the figure.
$, the area of rectangle $B L G N$ is $28-12=16\left(\mathrm{~cm}^{2}\right)$. The ratio of the areas of rectangles $N G J C$ and $H N C K$ is the same as the ratio of the lengths of segments $N G$ and $H N$, and that is the same as the ratio of t... | 418 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Grandpa prepared a pile of hazelnuts for his six grandchildren, asking them to divide them somehow. First, Adam came, counted half for himself, took one more nut, and left. The second, Bob, the third, Cyril, the fourth, Dan, and the fifth, Eda, all did the same. Only Franta sadly looked at the empty table; no nuts were... | We will proceed from the back:
When Eda took half of the nuts that were left for him after Dan, and one more, nothing was left. This half, therefore, consisted of one nut. After Dan, two nuts were left.
Dan also took half of the nuts that were left for him after Cyril, and one more. This half, therefore, consisted of... | 62 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bětka was playing with gears, which she arranged as indicated in the picture. When she then turned one gear, all the others started turning as well. In the end, she was satisfied with a gear system where the first gear had 32 and the second 24 teeth. When the third gear made exactly eight turns, the second gear made fi... | In all gears, the number of teeth used during rotation is the same (each tooth is counted as many times as it has been in contact with another tooth on another gear). According to the problem, we can say the following about this number.
The first gear had 32 teeth and made four full turns and part of a fifth, so more ... | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In horticulture Rose, one store ordered a total of 120 roses in red and yellow, a second store a total of 105 roses in red and white, and a third store a total of 45 roses in yellow and white. The horticulture fulfilled the order in such a way that it delivered the same number of roses of the same color to each store.
... | If we sum the quantities of roses delivered to all three stores, we get 270 pieces. In this sum, the quantities of roses of each color are included twice.
In the first store, there were red and yellow roses totaling 120 pieces. Double this quantity is 240, so the number of white roses delivered to the remaining two st... | 90 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Given an isosceles right triangle $A B S$ with the base $A B$. On the circle with center at point $S$ and passing through points $A$ and $B$, there is a point $C$ such that triangle $A B C$ is isosceles.
Determine how many points $C$ satisfy the given conditions, and construct all such points.
(K. Pazourek)
Hint. Wh... | Side $AB$ of the isosceles triangle $ABC$ can be either its base or one of its legs. Depending on this, we divide the solution into two parts.
a) Side $AB$ is the base of the isosceles triangle $ABC$. In this case, $C$ is the main vertex of the triangle $ABC$ and lies on its axis of symmetry. The axis of symmetry of t... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
An arithmetic sequence is a sequence of numbers in which the difference between each number and the one preceding it is always the same; this difference is called the common difference. (For example, 2, 8, 14, 20, 26, 32 is an arithmetic sequence with a common difference of 6.)
Bolek and Lolek each had their own arith... | The ratio of Bolko's and Lolko's difference was $5: 2$. Thus, Bolko's difference was $5k$ and Lolko's $2k$, where $k$ is an unknown number that we will soon determine from other information. The difference between Bolko's and Lolko's difference can then be expressed as $5k - 2k = 3k$.
Bolko's, respectively, Lolko's se... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Two congruent equilateral triangles $A B C$ and $B D E$ are given such that the measure of angle $A B D$ is greater than $120^{\circ}$ and less than $180^{\circ}$, and points $C, E$ lie in the same half-plane defined by the line $A D$. The intersection of $C D$ and $A E$ is denoted as $F$.
Determine the measure of ang... | The internal angle at vertex $A$, resp. $D$ in quadrilateral $A F D B$ is also an internal angle of triangle $A B E$, resp. $D B C$. Triangles $A B E$ and $D B C$ are isosceles and congruent by assumption. Therefore, the internal angles at vertices $A, E, D, C$ in these triangles are congruent. The size of these angles... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Petr glued 17 dice into a snake (as shown in the picture). He always glued the sides with the same number of dots. Then he examined the snake from all sides and correctly calculated the total number of dots on its surface. What did he get? (The sum of the dots on opposite sides of a die is 7.)
(S. Bodláková, M. Dillin... | Given that the sum of the dots on opposite sides of a die is always 7, we have a total of $17 \cdot 2$ such pairs on the "long" side of the snake and one pair on the sides. In total, we therefore have 35 pairs. Therefore, Petr must have counted $35 \cdot 7=245$ dots. | 245 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
I am thinking of a three-digit natural number less than 200. If I round its triple to the hundreds, it increases by 36. Which number am I thinking of?
(M. Dillingerová) | Let the sought number be $x$. From the problem statement, we have the condition
$$
3 x+36=* 00
$$
where $*$ is one of the digits $4,5,6$. These values are given by the condition $100 \leqq x<200$. This can be rearranged to the form
$$
3 x=* 64
$$
where $*$ is one of the digits $3,4,5$. Given that the number $* 64$ ... | 188 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine the last digit of the product of all even natural numbers less than 100 that are not multiples of ten.
(M. Volfová) | The last digit of the product is determined exclusively by the last digits of the factors. Therefore, in solving the problem, we will consider only the last digits in the calculations. According to the problem, we multiply ten sets of four factors, and in each set, the factors end with the digits 2, 4, 6, and 8. The pr... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The Nováks baked wedding cakes. They took a quarter of them to their relatives in Moravia, gave a sixth to their colleagues at work, and gave a ninth to their neighbors. If they had three more cakes left, it would be half of the original number. How many cakes did they bake? | The Nováks were given a quarter, a sixth, and a ninth of the cakes, which is 3 more than half of all:
$$
\frac{1}{4}+\frac{1}{6}+\frac{1}{9}=\frac{19}{36}
$$
Half is $\frac{18}{36}$. They distributed $\frac{1}{36}$ more. $\frac{1}{36}$ represents 3 cakes, so the total number of cakes was $36 \cdot 3=108$. They baked ... | 108 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
At the elementary school U Tří lip, where Lukáš also attends, they are holding a knowledge competition with pre-arranged tasks. Each correctly solved task is scored as many points as its order. Each unsolved or partially solved task is not scored at all. Lukáš correctly solved the first 12 tasks. If he had correctly so... | To solve the problem, it is not necessary to enumerate the sums; it is sufficient to recognize the relationships between them. Let the sum of the first twelve numbers be denoted by $s$, i.e.,
$$
1+2+\cdots+11+12=s.
$$
The sum of the 2nd to 13th numbers is 12 greater than $s$, because each of the addends is 1 greater ... | 71 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
On a fairy-tale island, dragons and cyclopes live. All dragons are red, three-headed, and two-legged. All cyclopes are brown, one-headed, and two-legged. Cyclopes have one eye in the middle of their forehead, and dragons have two eyes on each head. Together, the cyclopes and dragons have 42 eyes and 34 legs.
How many ... | Since both cyclopes and dragons are bipedal, the total number of these creatures is $17(34: 2=17)$.
If all the creatures were cyclopes, they would have a total of 17 eyes. This is 25 fewer than the actual number $(42-17=25)$.
Each dragon has 5 more eyes than any cyclops, so there are 5 dragons among the creatures $(2... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Josefina the Quail dances by the marsh, using steps of double length — short steps measure $45 \mathrm{~cm}$, long steps $60 \mathrm{~cm}$. Over time, she has worn an oval path, which she dances around repeatedly during long nights. If she repeats three long steps forward and one short step back, then the ninetieth ste... | With one large four-step (three long steps forward and one short step back), Josefa moves $135 \, \text{cm} (3 \cdot 60 - 45 = 135)$. Ninety steps consist of 22 large four-steps and two long steps $(90 = 22 \cdot 4 + 2)$. Therefore, Josefa's circuit measures $3090 \, \text{cm} (22 \cdot 135 + 2 \cdot 60 = 3090)$.
With... | 162 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Radovan is reading an interesting book. Yesterday he read 15 pages, today another 12 pages. To his surprise, he realized that the sum of the page numbers he read yesterday is the same as the sum of the page numbers he read today. Which page will he start reading tomorrow? (Radovan does not skip any pages or read any pa... | It is evident that the pages Radovan read today have higher numbers than those he read yesterday. The first page read today has a number 15 higher than the first page read yesterday. ("In between" are the remaining 14 pages he read yesterday.) The second page read today again has a number 15 higher than the second page... | 74 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A secret agent is trying to decipher an access code. So far, he has obtained the following information:
- it is a four-digit number,
- it is not divisible by seven,
- the digit in the tens place is the sum of the digit in the units place and the digit in the hundreds place,
- the number formed by the first two digits ... | We will use the given information in such an order that it helps us the most at any given moment.
According to the first piece of information, we are looking for a four-digit number, and according to the fourth piece of information, we will list the possibilities that come into question:
$$
\begin{aligned}
& 15 * 1 \... | 4583 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The organizers of the exhibition "To the Moon and Beyond" felt that they had low attendance on the first day, so they reduced the admission fee by 12 Kč. As a result, the number of visitors increased by $10\%$ on the second day, but the total daily revenue decreased by $5\%$. How many korunas did the admission cost aft... | We will organize the information from the assignment into the following table:
| | first day | second day |
| :---: | :---: | :---: |
| number of visitors | $n$ | $1.1 n$ |
| admission fee (Kč per person) | $x+12$ | $x$ |
| total daily revenue (Kč) | $n(x+12)$ | $1.1 n x$, or $0.95 n(x+12)$ |
From the last cell of t... | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Danka and Janka received two identical white cubes for their birthdays, each made up of 125 small cubes as shown in the picture. To distinguish the cubes, they decided to paint them. Danka took a brush and painted three of the six faces of her cube red. Janka painted three faces of her cube green. After some time, both... | We know that both Danka and Janka painted three sides of the cube. This can only be done in two ways:
1. Two opposite sides are painted, and then one side between them is also painted; consider, for example, the top, bottom, and one side.
2. No two opposite sides are painted, so the three painted sides share a common ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Jirka collects signatures of famous athletes and singers. He has a special notebook for this purpose and has decided that signatures will be on the front side of each page only. He numbered all these pages as $1,3,5,7,9, \ldots$, to know if any page gets lost. He wrote a total of 125 digits.
- How many pages did his n... | There are five single-digit odd numbers. There are a total of $99-9=$ $=90$ two-digit numbers, of which 45 are odd. With these numbers, Jirka could number $5+45=50$ pages and would use $5+2 \cdot 45=95$ digits. There are still $125-95=30$ digits left to use. These exactly correspond to the first ten three-digit odd num... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On a night march, the Wolf and Fox patrols each received one identical candle. They lit the candles together at the start and set off. Each member of the patrol carried the candle for the duration it took for its length to be halved. The Wolves reached the finish line at the moment when the sixth carrier was passing th... | First, we will determine how long the Wolves ran. The sixth carrier passed the candle to the seventh at the moment when it still had 3 minutes left to burn. He, therefore, carried it for 3 minutes and received it from the fifth carrier at the moment when it still had $3+3=6$ minutes left to burn. The fifth carrier, the... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The cashier at the gallery sells tickets to visitors with a number according to the order in which they came that day. The first visitor gets a ticket with the number 1, the second with the number 2, etc. However, during the day, the yellow paper on which the tickets were printed ran out, so the cashier had to continue... | Let the number of yellow tickets be $n$. The first yellow ticket had the number 1, the second had 2, and so on, with the last yellow ticket having the number $n$. The first red ticket had the number $n+1$, the second red ticket had $n+2$, and so on, with the last red ticket having the number $2n$.
Notice that the firs... | 82 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Filoména has a mobile phone with the following button layout:
| 1 | 2 | 3 |
| :--- | :--- | :--- |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
| 0 | | |
| | | |
| | | |
The nine-digit phone number of her best friend Kunhuta has the following properties:
- all digits in Kunhuta's phone number are different,
- the first four ... | First, let's find all quadruples of buttons whose centers form a square. These are the buttons with the following digits:
$$
\begin{array}{ll}
1,2,4,5 & 1,3,7,9 \\
2,3,5,6 & 2,4,6,8 \\
4,5,7,8 & 5,7,9,0 \\
5,6,8,9 &
\end{array}
$$
The quadruples in the left column, however, cannot be used because, besides the square ... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Amálka observed squirrels in the garden of the nursery, where these three trees grew: spruce, beech, and fir. The squirrels were sitting quietly on the trees, so she could count them - there were 34 of them. When 7 squirrels jumped from the spruce to the beech, there were as many on the beech as on both conifers combin... | Let the original number of squirrels on the spruce be $s$, on the beech $b$, and on the pine $j$. From the problem, we can set up a system of four equations with these three unknowns:
$$
\begin{aligned}
s+b+j & =34, \\
b+7 & =j+s-7, \\
j-5 & =s-7, \\
b+7+5 & =2 j .
\end{aligned}
$$
Notice that by adding the third and... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Before the Christmas concert, the students offered for sale 60 products from their art classes. Each customer could determine the price themselves, and the entire proceeds went to charitable causes. At the beginning of the concert, the students calculated how many crowns they had earned on average for each sold product... | Let the number of items the students sold before the concert be denoted by $v$, and the amount in crowns they received for them be denoted by $k$. According to the problem, the ratio $\frac{k}{v}$ is an integer and the following equation holds:
$$
\frac{k+2505}{v+7}=130
$$
From this equation, we express the unknown $... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Mr. Sparrow was a well-known bird breeder. He had more than 50 and fewer than 100 birds in total. Canaries made up a quarter and finches a ninth of the total number.
How many birds did Mr. Sparrow keep?
(L. Hozová)
Hint. Could he have kept, for example, ninety? | The number of birds kept by Mr. Sparrow had to be divisible by nine (according to the larks) and simultaneously by four (according to the canaries). Such numbers are 36, 72, 108, etc., i.e., multiples of 36. Of these numbers, only $72 \mathrm{v}$ is in the given range.
Mr. Sparrow kept 72 birds.
Note. Alternatively, ... | 72 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Magda cut out two identical isosceles triangles, each with a perimeter of $100 \mathrm{~cm}$. First, she formed a quadrilateral by placing the triangles together along their legs. Then, she formed a quadrilateral by placing the triangles together along their bases. In the first case, the quadrilateral she obtained had ... | The perimeter of the first quadrilateral consists of two sides and two bases, while the perimeter of the second quadrilateral consists of four sides. Since the first quadrilateral had a perimeter $4 \mathrm{~cm}$ shorter than the second, the base must have been $2 \mathrm{~cm}$ shorter than the side.
The perimeter of ... | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Seven dwarfs were born on the same day in seven consecutive years. The sum of the ages of the three youngest dwarfs was 42 years. When one dwarf left with Snow White to fetch water, the remaining dwarfs discovered that their average age was the same as the average age of all seven.
How old was the dwarf who went with ... | The sum of the ages of the first three dwarfs was 42 years, so on average they were 14 years old $(42: 3=14)$. The ages of these, or the remaining four dwarfs, were $13,14,15$, and $16,17,18$, and 19 years, respectively.
The sum of the ages of all seven dwarfs was 112 years, so on average they were 16 years old $(112:... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Pat and Mat were practicing arithmetic. In a square grid oriented according to the cardinal directions, they assigned the following arithmetic operations to each move:
- when moving north (S) they added seven,
- when moving east (V) they subtracted four,
- when moving south (J) they divided by two,
- when moving west ... | The task can be illustrated as a so-called arithmetic snake:
We will gradually fill in all intermediate results from the back:
 | If the edges of the smaller cubes are $n$ times smaller than the edge of the original cube, then the original cube was divided into $n \cdot n \cdot n=n^{3}$ smaller cubes. The surface area of the cube is $6 \cdot 12^{2}\left(\mathrm{~cm}^{2}\right)$. The sum of the surface areas of all the smaller cubes is
$$
n^{3} \... | 512 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Martina came up with a method for generating a numerical sequence. She started with the number 52. From it, she derived the next term in the sequence as follows: $2^{2}+2 \cdot 5=4+10=14$. Then she continued in the same way and from the number 14, she got $4^{2}+2 \cdot 1=16+2=18$. Each time, she would take a number, d... | It is evident that if a number appears for the second time in the sequence, the segment bounded by these two numbers will repeat indefinitely. Therefore, we will list the numbers in the sequence until they start repeating.
- 1st number: 52,
- 2nd number: 14,
- 3rd number: 18,
- 4th number: $8^{2}+2 \cdot 1=66$,
- 5th ... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the circle $k$ with center $S$ and radius $52 \, \text{mm}$, two perpendicular chords $A B$ and $C D$ are given. Their intersection $X$ is $25 \, \text{mm}$ away from the center $S$. How long is the chord $C D$, if the length of the chord $A B$ is $96 \, \text{mm}$?
(L. Hozová) | The midpoints of chords $AB$ and $CD$ are denoted as $E$ and $F$, see the figure. Triangles $AES$, $EXS$, and $SFD$ are right-angled, $|AE|=\frac{1}{2}|AB|=48 \text{ mm}$, $|SX|=25 \text{ mm}$, and $|SA|=|SD|=$ $=52 \text{ mm}$.
Hint. How many sheep we... | After one third of the sheep left, two thirds of the original number remained on the meadow, which is 30 sheep $\left(\frac{2}{3} \cdot 45=30\right)$. These have a total of 120 legs $(30 \cdot 4=120)$.
The shepherds who remained on the meadow had a total of 6 legs $(126-120=6)$, so 3 shepherds remained $(6: 2=3)$. Sin... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Marta is playing a game where she guesses a five-digit number consisting of different digits. The course of the first three rounds is as follows:
| 1st round | 2 | 6 | 1 | 3 | 8 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 2nd round | 4 | 1 | 9 | 6 | 2 |
| 3rd round | 8 | 1 | 0 | 2 | 5 |
The color of the cell reve... | In the guessed number, the digits from the gray fields, i.e., $0,3,5,6,9$, do not appear. There remain exactly five digits - $1,2,4,7,8$ - whose order we will try to determine step by step according to the remaining fields. In each step, we consider all previous conclusions:
- The digit 1 can only be in the second pos... | 71284 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Nikola had one three-digit and one two-digit number. Each of these numbers was positive and composed of mutually different digits. The difference between Nikola's numbers was 976.
What was their sum?
(L. Hozová)
Hint. Which three-digit numbers are possible? | The largest three-digit number with distinct digits is 987. If this were one of Niko's numbers, the other would have to be 11 (to make the difference 976). This is a two-digit number, but it is not composed of distinct digits, so it could not have been Niko's number.
The next smaller three-digit number with distinct d... | 996 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Jindra collects dice, all of the same size. Yesterday he found a box and started arranging the dice in it. He managed to fill the square bottom with one layer of dice. He then filled five more layers in the same way, but ran out of dice halfway through the next layer. Today, Jindra received 18 more dice from his grandm... | The 18 new cubes from grandma represented half of all the cubes in an incomplete (as well as in any other) layer. Thus, one layer contained 36 cubes $(2 \cdot 18=36)$. The layers are indeed square, consisting of 6 rows of 6 cubes each $(6 \cdot 6=36)$.
Yesterday, Jindra had six full layers and half of the seventh laye... | 234 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
I am Ivan and I am three times younger than my father. I have brothers Vincent and Jakub, who are 11 and 9 years old. My age is equal to five times the third of the age of the younger of the brothers. A peculiarity of our family is that we were all born on April 12th, so today we are celebrating our birthdays.
How man... | The younger of the brothers is 9 years old, so I am 15 years old $\left(\frac{9}{3} \cdot 5=15\right)$. I am three times younger than my father, so my father is 45 years old $(3 \cdot 15=45)$. The sum of the ages of the three of us brothers is 35 years $(11+9+15=35)$.
The father's age is 10 years older than the sum of... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Dwarves go to the stream to fetch water. Each dwarf's bucket has a different volume: 3, 4, 5, 6, 7, 8, and 9 liters. The dwarves do not lend their buckets to each other and always bring them back full of water.
- Kejchal brings more water in his bucket than Štístko.
- Dřímal would have to go for water three times to b... | From the second condition, it follows that Drímal's jug has a volume of 3 liters and Stydlín's 9 liters (since $3 \cdot 3=9$, and if Drímal's jug were different, Stydlín's jug would have to be at least a twelve-liter one).
Now, from the fourth condition, it follows that Šmudl's jug is 3 liters larger than Štístko's. T... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Teacher Smolná prepared a test for her class in three versions to prevent students from copying. $V$ In each version, she specified three edges of a cuboid and asked the students to calculate its volume. However, she did not solve the problems in advance and did not realize that the result was the same in all three ver... | Let's factorize the lengths of all edges into prime factors:
$$
\begin{gathered}
12=2 \cdot 2 \cdot 3,18=2 \cdot 3 \cdot 3,20=2 \cdot 2 \cdot 5, \\
24=2 \cdot 2 \cdot 2 \cdot 3,30=2 \cdot 3 \cdot 5,33=3 \cdot 11,70=2 \cdot 5 \cdot 7
\end{gathered}
$$
In these products, the factors 7 and 11 are present, so the resulti... | 154 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
One interior angle in a triangle measures $50^{\circ}$. What is the angle between the bisectors of the remaining two interior angles?
(L. Hozová)
Hint. You don't need to know the sizes of the remaining interior angles to solve the problem. | Consider a triangle $ABC$ with an angle of $50^{\circ}$ at vertex $A$; the unknown angles at vertices $B$ and $C$ are denoted by $\beta$ and $\gamma$. The intersection of the angle bisectors is denoted by $O$, the angle $BOC$ is denoted by $\omega$, and the angle adjacent to it is denoted by $\psi$.
Hint. Express one of ... | Let's denote the three numbers from the problem as $a, b, c$, where $a<b<c$. The first condition means
$$
\frac{\frac{a+b}{2}+\frac{b+c}{2}}{2}=\frac{a+b+c}{3}
$$
We can simplify this equation as follows:
$$
\begin{aligned}
\frac{a+2 b+c}{4} & =\frac{a+b+c}{3}, \\
3 a+6 b+3 c & =4 a+4 b+4 c, \\
2 b & =a+c \\
b & =\f... | 6066 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For the sequence of numbers starting
$$
1,3,4,7,11,18, \ldots
$$
it holds that each number starting from the third is the sum of the two preceding ones.
What is the last digit of the 2023rd number in this sequence?
Hint. Use the sequence formed by the last digits to help yourself. | The last digit of each number corresponds to the remainder of that number when divided by ten. Therefore, it is sufficient to deal with the sequence of corresponding remainders:
$$
1,3,4,7,1,8, \ldots
$$
which means the sequence where each number starting from the third is the remainder of the sum of the previous two... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Petra had written down the natural numbers from 1 to 9. She added two of these numbers, erased them, and wrote down the resulting sum instead of the erased addends. She then had eight numbers, which she managed to divide into two groups with the same product.
Determine the largest possible value of this product.
(E. ... | For comparing the products of the resulting groups of numbers, prime factors will be useful. To have the products of the numbers in both groups be the same, the total number of each individual prime factor must be even. If the number of some prime factors is odd, then dividing them into groups with the same product is ... | 504 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
There are 33 students in the class. Before Christmas, they went to the forest with the gamekeeper to fill the feeders. The girls unpacked the hay packages. The boys split into two groups: some took 4 bags of carrots and 3 bags of nuts, and the others took one bag of apples and one bag of nuts. The ratio of the number o... | Let the number of girls be denoted by $d$, the number of boys with carrots by $x$, and the number of boys with apples by $y$. Then the number of bags of nuts is $3x + y$, the number of bags of apples is $y$, and the number of bags of carrots is $4x$. The equality of the ratios of the quantities given in the problem is
... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
On a clean board, we wrote a three-digit natural number with yellow chalk, composed of mutually different non-zero digits. Then we wrote on the board with white chalk all other three-digit numbers that can be obtained by changing the order of the digits of the yellow number. The arithmetic mean of all the numbers on th... | The digits of the sought number will be denoted by $a, b, c$, assuming
$$
0<a<b<c \text {. }
$$
The numbers on the board, arranged from the smallest to the largest, are
$$
\overline{a b c}, \overline{a c b}, \overline{b a c}, \overline{b c a}, \overline{c a b}, \overline{c b a}
$$
The arithmetic mean of the first t... | 361 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all seven-digit numbers that contain each of the digits 0 to 6 exactly once and for which the first and last two-digit numbers are divisible by 2, the first and last three-digit numbers are divisible by 3, the first and last four-digit numbers are divisible by 4, the first and last five-digit numbers are divisible... | The last digit must be even, since the last two-digit number is divisible by two. For the last five-digit number to be divisible by five, the last digit must be 0 or 5. Of these, the even one is 0. The first five-digit number is also divisible by five, so for the fifth position, the digit 5 remains. So far, we have the... | 3216540 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all four-digit numbers that have exactly five four-digit and exactly nine one-digit divisors.
( $S$. Bednářová) | The sought numbers should have divisors $1,2,3,4,5,6,7,8,9$, so they must be divisible by their least common multiple:
$$
5 \cdot 7 \cdot 8 \cdot 9=2520 \text {. }
$$
We will consider the multiples of the number 2520 and determine their four-digit divisors:
- The number 2520 is a four-digit number, but it has only t... | 5040 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a certain polygon, the ratio of the sum of the magnitudes of its interior angles to the sum of the magnitudes of $\mathrm{k}$ supplementary angles is $3: 5$. (To explain: a supplementary angle complements a given angle to a full angle.)
How many vertices does that polygon have?
(I. Jančigová) | The sum of the sizes of the interior angles in every triangle is $180^{\circ}$, in every quadrilateral $360^{\circ}$, etc. Generally, it is true that every $n$-sided polygon can be composed of $n-2$ triangles, so the sum of its interior angles is $(n-2) \cdot 180^{\circ}$.
, and the other always lies (liars). Researchers there encountered several groups of natives and asked each member of the groups how many honest ones were in their group.
- From one four-member group, they received the s... | If the first group consisted only of honest people, the researchers would receive the numbers 4, 4, 4, 4 as answers. If the group had three, two, or one honest person, the researchers could not receive four identical numbers as answers. If the group consisted only of liars, the researchers could receive any set of four... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Snow White and the Seven Dwarfs collected cones for a campfire. Snow White said that the total number of cones is divisible by two. The first dwarf stated that it is a number divisible by three, the second dwarf said it is a number divisible by four, the third dwarf said it is a number divisible by five, the fourth dwa... | The number of cones is a number less than 350, which is not divisible by exactly one pair of consecutive numbers from 2, 3, 4, 5, 6, 7, 8, 9, and by all others yes. None of the pairs $(2,3)$, $(3,4)$, $(4,5)$, $(5,6)$, and $(6,7)$ can be it, because there would always be some other number among the remaining ones by wh... | 180 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In an acute triangle $K L M$, $V$ is the intersection of its altitudes and $X$ is the foot of the altitude on side $K L$. The bisector of angle $X V L$ is parallel to side $L M$ and angle $M K L$ has a size of $70^{\circ}$.
What are the sizes of angles $K L M$ and $K M L$?
(L. Hozová)
Hint. Visualize the situation d... | In a given triangle, we denote $Y, Z$ as the feet of the other altitudes and $N, O$ as the intersections of the angle bisector of $\angle XVL$ with the sides $KL, KM$, see the figure. In the figure, we also mark the congruent angles:
,
- he added the new numbers and wrote down ... | After dividing the original number, Roman had two at most two-digit numbers. Their sum can therefore be a one-digit, two-digit, or three-digit number. Their difference can be 0, a one-digit, or a two-digit number. Moreover, the sum is always greater than the difference.
The number 171 could only be obtained by Roman i... | 18810 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Grandpa Vendelin went with his two grandsons Cyril and Metoděj to buy fishing rods. The price of the rods caught the attention of both Cyril and Grandpa. It was a four-digit number, where the first digit was one greater than the third digit, but one less than the last digit. The sum of all four digits was 6. Cyril also... | The number representing the price of rods is a four-digit number, where the first digit is one greater than the third digit and one less than the fourth. The three mentioned digits are therefore distinct, and the sum of all four digits is 6. The only numbers that satisfy the above conditions are 2013 and 1302. However,... | 61 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The warehouse received cement in bags weighing $25 \mathrm{~kg}$ and $40 \mathrm{~kg}$. There were twice as many smaller bags as larger ones. The warehouse keeper reported the total number of bags to the manager but did not specify how many of each. The manager assumed all bags weighed $25 \mathrm{~kg}$. Therefore, he ... | Thieves stole $60 \cdot 40=2400 \text{ kg}$. Before the theft, $15 \text{ kg}$ was missing from each large bag according to the documentation. The total undocumented weight equals the stolen weight, so the number of large bags delivered was
$$
2400: 15=160 \text{.}
$$
There were twice as many small bags, i.e., 320; i... | 12000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A quarter of the students in the class are non-swimmers. Half of the non-swimmers signed up for the swimming course. Four non-swimmers did not sign up for the course. How many students in the class can swim and how many students are there in total in the class? | Four non-swimmers, who did not sign up for the course, make up half of the non-swimmers, i.e., there are a total of 8 non-swimmers. Non-swimmers represent a quarter of the students in the class, i.e., there are 24 swimmers (three times the number of non-swimmers) in the class. In total, there are 32 children in the cla... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Vítek has written down two numbers, 541 and 293. From the six digits used, he first needs to strike out two such that the sum of the two resulting numbers is the largest possible. Then, he needs to strike out two digits from the original six digits such that the difference between the two resulting numbers is the small... | First, we will be crossing out digits so that the sum is as large as possible. We can either cross out two digits from the first number, or we can cross out two digits from the second, or we can cross out one digit from each number. In each case, we cross out digits so that the resulting sum is as large as possible. We... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Four friends Adam, Mojmír, and twins Petr and Pavel earned a total of 52 smileys in math class, each getting at least 1. The twins together have 33, but the most successful was Mojmír. How many did Adam get?
(M. Volfová) | All smileys amount to 52, while the twins got 33 and Adam at least one. For Mojmír, there are at most $52-33-1=18$ smileys left. For him to have the most of all, each of the twins can have at most 17 smileys. This means that Petr got exactly 17 and Pavel 16, or vice versa. If one had less than 16, the other would have ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Mr. Tik and Mr. Tak sold alarm clocks in the stores Before the Corner and Behind the Corner. Mr. Tik claimed that Before the Corner sold 30 more alarm clocks than Behind the Corner, while Mr. Tak claimed that Before the Corner sold three times as many alarm clocks as Behind the Corner. In the end, it turned out that bo... | From this information, it follows that if the number of alarm clocks sold in the store Behind the Corner represents one part, then the number of alarm clocks sold in the store In Front of the Corner represents three of these parts. From Tik's information, it then follows that two of these parts correspond to 30 alarm c... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Mrs. Siroka was expecting guests in the evening. First, she prepared 25 sandwiches for them. Then she calculated that each guest could take two, but three would not be enough for everyone. She thought that if she made another 10 sandwiches, each guest could take three, but not four. This still seemed too little to her.... | First, let's work with the part of the problem where 25 sandwiches are considered. According to this, Mrs. Siroka expected no more than 12 guests, because $25: 2=12$, remainder 1, which means that 12 people could take two sandwiches each, but only one would be left. Here we also find out that Mrs. Siroka expected more ... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Let $p, q$ and $r$ be positive real numbers such that the equation
$$
\lfloor p n\rfloor+\lfloor q n\rfloor+\lfloor r n\rfloor=n
$$
is satisfied for infinitely many positive integers $n$.
(a) Prove that $p, q$ and $r$ are rational.
(b) Determine the number of positive integers $c$ such that there exist ... |
Solution. We will first prove that $(p+q+r=1 \wedge p,, q,, r \in \mathbb{Q})$ is an equivalent statement for the above.
(a) From
$$
n=\lfloor p n\rfloor+\lfloor q n\rfloor+\lfloor r n\rfloor \leq p n+q n+r n
$$
for some positive integer $n$ we infer $p+q+r \geq 1$.
Let us write $p+q+r=1+t$ for some $t \geq 0$. Th... | 15 | Number Theory | proof | Yes | Yes | olympiads | false |
1. Some objects are in each of four rooms. Let $n \geqslant 2$ be an integer. We move one $n$-th of objects from the first room to the second one. Then we move one $n$-th of (the new number of) objects from the second room to the third one. Then we move similarly objects from the third room to the fourth one and from ... |
Solution. Let us compute backwards. Firstly we find the number of the objects in two rooms before the move. Let $a$ and $b$ be number of the objects in the rooms $A$ and $B$ before the move. This number after the move we denote by $a^{\prime}$ and $b^{\prime}$. By the conditions
$$
a^{\prime}=\frac{n-1}{n} a, \quad b... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Nela and Jane choose positive integer $k$ and then play a game with a $9 \times 9$ table. Nela selects in every of her moves one empty unit square and she writes 0 to it. Jane writes 1 to some empty (unit) square in every her move. Furthermore $k$ Jane's moves follows each Nela's move and Nela starts. If sum of num... |
Solution. Let us show at first that Jane wins for $k=3$. Let us assume $3 \times 3$ squares $A_{1}, A_{2}$ and $A_{3}$ (see the picture). We will call the $3 \times 3$ square covered if just one 1 is in each its row and column. If Jane covers squares $A_{1}, A_{2}$ and $A_{3}$ without writing to other squares, she win... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In how many ways can you partition the set $\{1,2, \ldots, 12\}$ into six mutually disjoint two-element sets in such a way that the two elements in any set are coprime?
(Martin Panák)
|
Solution. No two even numbers can be in the same set (pair). Let us call partitions of $\{1,2, \ldots, 12\}$ with this property, that is one even and one odd number in each pair, even-odd partitions. The only further limitations are, that 6 nor 12 cannot be paired with 3 or 9 , and 10 cannot be paired with 5 .
That m... | 252 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Let $\left(a_{n}\right)_{n=1}^{\infty}$ be an infinite sequence such that for all positive integers $n$ we have
$$
a_{n+1}=\frac{a_{n}^{2}}{a_{n}^{2}-4 a_{n}+6}
$$
a) Find all values $a_{1}$ for which the sequence is constant.
b) Let $a_{1}=5$. Find $\left\lfloor a_{2018}\right\rfloor$.
(Vojtech Bálint)
|
Solution. a) Assume $\left(a_{n}\right)_{n=1}^{\infty}$ is constant. Then $a_{2}=a_{1}$, hence
$$
a_{1}=\frac{a_{1}^{2}}{a_{1}^{2}-4 a_{1}+6}
$$
which rewrites as $a_{1}\left(a_{1}-2\right)\left(a_{1}-3\right)=0$. It follows that $a_{1} \in\{0,2,3\}$. On the other hand, once $a_{2}=a_{1}$, the sequence is clearly co... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Let $n$ be a positive integer. Tom and Jerry play a game on a board consisting of a row of 2018 cells. Initially, Jerry places a piece at some cell. In each subsequent step, Tom says an integer from the interval $[1, n]$ and Jerry moves the piece by the said number of cells, by his choice either to the left or to t... |
Another solution. We describe another strategy for Tom when $n=1010$. This strategy forces a win in 3 steps. As before, label the cells 1,2, . , 2018 left to right and denote the current position of the piece by $k$. If $k \in\{1009,1010\}$, Tom says 1010 and wins immediately. If $k \leqslant 504$, Tom says $1009-k$. ... | 1010 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let $A B C$ be a triangle with $\angle B A C=90^{\circ}$. Points $D, E$ on its hypotenuse $B C$ satisfy $C D=C A, B E=B A$. Let $F$ be such a point inside $\triangle A B C$ that $D E F$ is an isosceles right triangle with hypotenuse DE. Find $\angle B F C$. (Patrik Bak)
|
Solution. We show that $F$ is the incenter of $\triangle A B C$. Simple angle-chasing in $\triangle B F C$ then yields $\angle B F C=180^{\circ}-\frac{1}{2} B-\frac{1}{2} C=135^{\circ}$.
Since $B A=B E$, triangle $B A E$ is isosceles and $\angle B A E=90^{\circ}-\frac{1}{2} \angle B$. Similarly, $\angle D A C=90^{\ci... | 135 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Find the maximal value of $a^{2}+b^{2}+c^{2}$ for real numbers $a, b, c$ such that $a+b$, $b+c, c+a$ all lie in the interval $[0,1]$.
(Ján Mazák)
|
Solution. By symmetry of the problem, we can WLOG suppose $a \geqslant b \geqslant c$.
Then both $b-c$ and $b+c$ are non-negative (by our assumptions), hence their product will also be non-negative and $b^{2} \geqslant c^{2}$. Analogously, both $1-b-a$ and $1-b+a$ are non-negative and $(1-b)^{2} \geqslant a^{2}$.
Th... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A number $n$ is a product of three (not necessarily distinct) prime numbers. Adding 1 to each of them, after multiplication we get a larger product $n+963$. Determine the original product $n$.
(Pavel Novotný)
|
Solution. We look for $n=p \cdot q \cdot r$, with primes $p \leqslant q \leqslant r$ satisfying
$$
(p+1)(q+1)(r+1)=p q r+963 .
$$
If $p=2$, the right-hand side of (1) is odd, hence the factors $q+1, r+1$ on the left must be odd too. This implies that $p=q=r=2$, which contradicts to (1). Thus we have proved that $p \... | 2013 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Determine the number of all coverings of a chessboard $3 \times 10$ by (nonoverlapping) pieces $2 \times 1$ which can be placed both horizontally and vertically.
(Stanislava Sojáková)
|
Solution. Let us solve a more general problem of determining the number $a_{n}$ of all coverings of a chessboard $3 \times 2 n$ by pieces $2 \times 1$, for a given natural $n .{ }^{2}$ We will attack the problem by a recursive method, starting with $n=1$.
The value $a_{1}=3$ (for the chessboard $3 \times 3$ ) is evid... | 571 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Let $n$ be a natural number whose all positive divisors are denoted as $d_{1}, d_{2}, \ldots, d_{k}$ in such a way that $d_{1}<d_{2}<\cdots<d_{k}$ (thus $d_{1}=1$ and $d_{k}=n$ ). Determine all the values of $n$ for which both equalities $d_{5}-d_{3}=50$ and $11 d_{5}+8 d_{7}=3 n$ hold.
(Matúš Harminc)
|
Solution. We distinguish whether $n$ is odd or even.
(i) The case of $n$ odd. Since all the $d_{i}$ 's are odd too, it follows from $11 d_{5}+8 d_{7}=3 n$ that $d_{7} \mid 11 d_{5}$ as well as $d_{5} \mid 8 d_{7}$, hence $d_{5} \mid d_{7}$. In view of $d_{7}>d_{5}$, the relations $d_{5}\left|d_{7}\right| 11 d_{5}$ im... | 2013 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Let us call by an "edge" any segment of length 1 which is common to two adjacent fields of a given chessboard $8 \times 8$. Consider all possible cuttings of the chessboard into 32 pieces $2 \times 1$ and denote by $n(e)$ the total number of such cuttings that involve the given edge $e$. Determine the last digit of... |
Solution. The total number of the vertical edges is $7 \cdot 8=56$ as well as the total number of the horizontal edges. Thus the number of all the edges under consideration is $56 \cdot 2=112$.
The number of edges, which are not involved in a given cutting, is equal to 32, because each of these edges must coincide wi... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. There are 234 visitors in a cinema auditorium. The visitors are sitting in $n$ rows, where $n \geqslant 4$, so that each visitor in the $i$-th row has exactly $j$ friends in the $j$-th row, for any $i, j \in\{1,2, \ldots, n\}, i \neq j$. Find all the possible values of $n$. (Friendship is supposed to be a symmetric... |
Solution. For any $k \in\{1,2, \ldots, n\}$ denote by $p_{k}$ the number of visitors in the $k$-th row. The stated condition on given $i$ and $j$ implies that the number of friendly pairs $(A, B)$, where $A$ and $B$ are from the $i$-th row and from $j$-th row respectively, is equal to the product $j p_{i}$. Interchang... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find all real numbers $s$ for which the equation
$$
4 x^{4}-20 x^{3}+s x^{2}+22 x-2=0
$$
has four distinct real roots and the product of two of these roots is -2 .
|
Solution. Assume that $s$ is a number as above, and denote the four roots of the equation by $x_{1}, x_{2}, x_{3}$ and $x_{4}$ in such a way that
$$
x_{1} x_{2}=-2 .
$$
From the factorization
$$
4 x^{4}-20 x^{3}+s x^{2}+22 x-2=4\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)
$$
we ... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let natural numbers $p, q(p<q)$ be given. Find the least natural number $m$ with the following property: the sum of all fractions whose denominators (in lowest terms) are equal to $m$ and whose values lie in the open interval $(p, q)$ is at least $56\left(q^{2}-p^{2}\right)$.
|
Solution. We show that the least $m$ is 113 (independent of $p, q$ ). Clearly $m>1$. For arbitrary natural numbers $c1$, let $S_{m}(c, d)$ denote the sum of all fractions (in their lowest terms) which lie in the open interval $(c, d)$ and whose denominator is $m$. Then we have the inequality
$$
S_{m}(c, c+1) \leqslan... | 113 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Denote by $\mathbb{N}$ the set of all natural numbers and consider all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any $x, y \in \mathbb{N}$,
$$
f(x f(y))=y f(x) .
$$
Find the least possible value of $f(2007)$.
|
Solution. Let $f$ be any function with the given property. We claim first of all that $f$ is injective. Indeed, if $f\left(y_{1}\right)=f\left(y_{2}\right)$, then for all natural $x$
$$
y_{1} f(x)=f\left(x f\left(y_{1}\right)\right)=f\left(x f\left(y_{2}\right)\right)=y_{2} f(x) \text {, }
$$
and as $f(x)$ is a natu... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. There are two touching circles, $k_{1}\left(S_{1}, r_{1}\right.$ and $k_{2}\left(S_{2}, r_{2}\right)$ in an rectangle $A B C D$ with $|A B|=9,|B C|=8$. Moreover $k_{1}$ touches $A D$ and $C D$, while $k_{2}$ touches $A B$ and $B C$.
a) Prove $r_{1}+r_{2}=5$.
b) What is the least and what is the greatest possible ... |
Solution. a) Let $M$ and $N$ be intersections of the line through $S_{1}$ parallel to $A D$. Analogously let $K$ and $L$ be intersections of the line through $S_{2}$ parallel to $A B$. Let $P$ be the intersection of $K L$ and $M N$ (see Fig. 1). The Pythagoras theorem for $S_{1} P S_{2}$ gives
$$
\begin{gathered}
\le... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Let $A, B$ be sets of positive integers such that a sum of arbitrary two different numbers from $A$ is in $B$ and a ratio of arbitrary two different numbers from $B$ (greater one to smaller one) is in $A$. Find the maximum number of elements in $A \cup B$.
(Martin Panák)
|
Solution. Initially we will prove that the set $A$ consists from at most two numbers. Suppose that three numbers $a<b<c$ belongs to the set $A$. Then the numbers $a+b<a+c<b+c$ are in $B$ and therefore the number
$$
\frac{b+c}{a+c}=1+\frac{b-a}{a+c}
$$
have to be in $A$. This is contradiction because $0<b-a<a+c$ and ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Find the number of finite increasing sequences of natural numbers $a_{1}, a_{2}, \ldots, a_{k}$, of all possible lengths $k$, for which $a_{1}=1, a_{i} \mid a_{i+1}$ for $i=1,2, \ldots, k-1$, and $a_{k}=969969$.
|
Solution. Clearly all members of the sequence must be divisors of its last member, 969 969. Using the decomposition into prime factors,
$$
969969=3 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \text {. }
$$
we thus see that the following holds:
Each member $a_{i}$ of the sequence $a_{1}, a_{2}, \ldots, a_{k}$ is the... | 4683 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Determine all triples $(x, y, z)$ of positive numbers satisfying the system of equations
$$
\begin{aligned}
& 2 x^{3}=2 y\left(x^{2}+1\right)-\left(z^{2}+1\right), \\
& 2 y^{4}=3 z\left(y^{2}+1\right)-2\left(x^{2}+1\right), \\
& 2 z^{5}=4 x\left(z^{2}+1\right)-3\left(y^{2}+1\right) .
\end{aligned}
$$
|
Solution. For any integer $k \geqslant 3$ and any $x \geqslant 0$ we have
$$
2 x^{k} \geqslant[(k-1) x-(k-2)]\left(x^{2}+1\right) .
$$
To see this, observe that, by AM-GM inequality,
$$
x^{k}+x^{k}+\underbrace{x+x+\ldots+x}_{(k-3) \text { times }} \geqslant(k-1) x^{3}
$$
and add it to
$$
(k-2)\left(x^{2}-2 x+1\ri... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find the largest positive integer $n$ such that
$$
\lfloor\sqrt{1}\rfloor+\lfloor\sqrt{2}\rfloor+\lfloor\sqrt{3}\rfloor+\cdots+\lfloor\sqrt{n}\rfloor
$$
is a prime $(\lfloor x\rfloor$ denotes the largest integer not exceeding $x)$.
(Patrik Bak)
|
Solution. Consider the infinite sequence $\left\{a_{n}\right\}_{n=1}^{\infty}$ defined by $a_{n}=\lfloor\sqrt{n}\rfloor$. This sequence is clearly non-decreasing and since
$$
k=\sqrt{k^{2}}6$ then the fraction $\frac{1}{6}(k-1) k(4 k+1)$ is an integer sharing a prime factor with $k$, hence the whole right-hand side i... | 47 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Find the largest possible size of a set $\mathbb{M}$ of integers with the following property: Among any three distinct numbers from $\mathbb{M}$, there exist two numbers whose sum is a power of 2 with non-negative integer exponent.
(Ján Mazák)
|
Solution. The set $\{-1,3,5,-2,6,10\}$ attests that $\mathbb{M}$ can have 6 elements: The sum of any two numbers from the triplet $(-1,3,5)$ is a power of two and the same is true for triplet $(-2,6,10)$. For the sake of contradiction, assume that some set $\mathbb{M}$ has more than 6 elements.
Clearly, $\mathbb{M}$ ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Paul is filling the cells of a rectangular table alternately with crosses and circles (he starts with a cross). When the table is filled in completely, he determines his score as $X-O$ where $X$ is the sum of squares of the numbers of crosses in all the rows and columns, and $O$ is the sum of squares of the numbers... |
Solution. Let $n=67$ and denote by $k=\frac{1}{2}\left(n^{2}+1\right)$ the total number of crosses in the table. A row containing $a$ crosses and $n-a$ circles contributes $a^{2}-(n-a)^{2}=2 n \cdot a-n^{2}$ to the total score and thus all the $n$ rows combined contribute
$$
2 n \cdot k-n \cdot n^{2}=2 n \cdot \frac{... | 134 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let $A B C$ be a triangle. The $A$-angle bisector intersects $B C$ at $D$. Let $E, F$ be the circumcenters of triangles $A B D, A C D$, respectively. Given that the circumcenter of triangle $A E F$ lies on $B C$, find all possible values of $\angle B A C$. (Patrik Bak)
|
Solution. Let $O$ be the circumcenter of triangle $A E F$ and denote $\alpha=\angle B A C$. Since $\angle B A D$ and $\angle C A D$ are acute (Fig. 2), points $E, F$ lie in the half-plane $B C A$ and the Inscribed angle theorem yields
$$
\angle B E D=2 \cdot \angle B A D=\alpha=2 \cdot \angle D A C=\angle D F C \text... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find the smallest positive integer $n$ such that for any coloring of numbers 1,2 , $3, \ldots, n$ by three colors there exist two numbers with the same color whose difference is a square of an integer. (Vojtech Bálint, Michal Rolínek, Josef Tkadlec)
|
Solution. The answer is $n=29$.
First, for the sake of contradiction, assume that numbers $1,2, \ldots, 29$ can be colored by colors $A, B, C$ such that no two numbers with the same color differ by a square. Let $f(i)$ be the color of number $i$ for $i \in\{1,2, \ldots, 29\}$.
Since 9, 16, and 25 are squares, number... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Alice and Ben play the game on a board with 72 cells around a circle. First, Ben chooses some cells and places one chip on each of them. Each round, Alice first chooses one empty cell and then Ben moves a chip from one of the adjacent cell onto the chosen one. If Ben fails to do so, the game ends; otherwise, anothe... |
Solution. We show that the smallest possible number of chips is 36 .
In the first part, we describe the strategy of Ben in which he can ensure that the game will never end. At the beginning, Ben places 36 chips on even cells of the game board and the odd cells he lets empty. Moreover, he firmly divides all 72 cells i... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Let $n$ be the sum of all ten-digit numbers, which contain every decimal digit from zero to nine. Determine the remainder of $n$ when divided by 77 .
(Pavel Novotný)
|
Solution. We determine the value of $n$ first. We count for each non-zero digit the number of its occurrences in the summands (giving $n$ altogether) in the place of units, tens, hundreds, ... This enables us to evaluate the "contribution" of each digit to the sum, and thus $n$ itself. Every digit is on each of the pl... | 28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Find the greatest real $k$ such that
$$
\frac{2\left(a^{2}+k a b+b^{2}\right)}{(k+2)(a+b)} \geqslant \sqrt{a b}
$$
holds for any positive real $a$ and $b$.
(Ján Mazák)
|
Solution. If $k=2$ then the inequality is equivalent to $\frac{1}{2}(a+b) \geqslant \sqrt{a b}$ (which holds), therefore $k \geqslant 2$. For $k>2$ we have $k+2>0$
$$
2\left(a^{2}+k a b+b^{2}\right) \geqslant(k+2)(a+b) \sqrt{a b},
$$
the division by $b^{2}$ gives
$$
2\left(\frac{a^{2}}{b^{2}}+k \frac{a}{b}+1\right)... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Find all positive integers $n$ such that $\left(2^{n}+1\right)\left(3^{n}+2\right)$ is divisible by $5^{n}$.
(Ján Mazák)
|
Solution. The following table shows $2^{n}+1$ and $3^{n}+2$ modulo 5 (we know, that both sequences modulo 5 have to be periodic):
| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | $\ldots$ |
| ---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $2^{n}+1$ | 3 | 5 | 9 | 17 | 33 | 65 | 129 | 257 | $\l... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Among real numbers $a, b, c$, and $d$ which satisfy
$$
a b+c d=a c+b d=4 \quad a \quad a d+b c=5
$$
find these, for which the value of $a+b+c+d$ is the least possible. Find this (the least) value as well.
(Jaromír Šimša)
|
Solution. We have
$$
\begin{aligned}
(a+b+c+d)^{2} & =a^{2}+b^{2}+c^{2}+d^{2}+2(a b+c d+a c+b d+a d+b c)= \\
& =a^{2}+b^{2}+c^{2}+d^{2}+2(4+4+5)=a^{2}+d^{2}+b^{2}+c^{2}+26 .
\end{aligned}
$$
Now $a^{2}+d^{2} \geqslant 2 a d, b^{2}+c^{2} \geqslant 2 b c$ where the equality holds iff $a=d$ and $b=c$ and from (1) we ge... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the greatest possible area of a triangle ABC with medians satisfying $t_{a} \leqslant 2$, $t_{b} \leqslant 3, t_{c} \leqslant 4$.
(Pavel Novotný)
|
Solution. Let $T$ be the centroid of $A B C$ and $K, L, M$ be the midpoints of $B C$, $C A, A B$. Medians cut $A B C$ into six smaller triangles, each with the same area: for example in the triangle $A M T$ we have $A M=\frac{1}{2} c$, its altitude through $T$ is $\frac{1}{3} v_{c}$ long, that is $S_{A M T}=\frac{1}{2... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. There are numbers 1,2,.., 33 written on a blackboard. In one step we choose two numbers on the blackboard such that one of them divides the other one, we erase the two numbers and write their integer quotient instead. We proceed in this manner until no number on the blackboard divides another one. What is the least... |
Solution. In the process, apparently only numbers from the set $M=\{1,2, \ldots, 33\}$ can be on the blackboard. Prime numbers 17, 19,23, 29 a 31 are going to stay on the blackboard, in one copy each, because they have no other divisor than the number 1 and the set $M$ does not contain any other of their multiples.
W... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find all real solutions of the system
$$
\begin{aligned}
& \sqrt{x-y^{2}}=z-1, \\
& \sqrt{y-z^{2}}=x-1, \\
& \sqrt{z-x^{2}}=y-1 .
\end{aligned}
$$
|
Solution. The square roots and their arguments have to be positive, therefore $x, y, z \geqslant 1, x \geqslant y^{2}, y \geqslant z^{2}$, and $z \geqslant x^{2}$. The last three inequalities imply $x \geqslant$ $y^{2} \geqslant y \geqslant z^{2} \geqslant z \geqslant x^{2}$, and since $x \geqslant 1$, the inequality ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. There are numbers $1,2, \ldots, 33$ written on the blackboard. In one step we choose a group of numbers on the blackboard (at least two) such that their product is a square, we erase them and write the square root of their product instead. We proceed until no group can be chosen. What is the least amount of numbers... |
Solution. The product of all the numbers written on the blackboard is
$$
S=2^{31} \cdot 3^{15} \cdot 5^{7} \cdot 7^{4} \cdot 11^{3} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 .
$$
Apparently, the numbers $17,19,23,29$, and 31 can never be erased and can never be a part of any change. In any step, ther... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Let $M$ be a set of six mutually different positive integers which sum up to 60 . We write these numbers on faces of a cube (on each face one). In a move we choose three faces with a common vertex and we increase each number on these faces by one. Find the number of all sets $M$, whose elements (numbers) can be wri... |
Solution. Let us denote the faces of the cube by $S_{1}, S_{2}, \ldots, S_{6}$, where $S_{1}$ is opposite to $S_{6}$, and $S_{2}$ opposite to $S_{5}$. Let $c_{i}$ be written on $S_{i}$. Since any vertex of the cube lies just in one of any pair of opposite faces, we increase by one the sums $c_{1}+c_{6}$, $c_{2}+c_{5}$... | 84 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Find all integers greater than 1 by which a cancellation can be made in some fraction of the form
$$
\frac{3 p-q}{5 p+2 q}
$$
where $p$ and $q$ are mutually prime integers.
|
Solution. The fraction admits cancellation by an integer $d>1$ if and only if $d$ is a common divisor of its numerator and its denominator. Let us thus assume that $d \mid 3 p-q$ and at the same time $d \mid 5 p+2 q$, where $p$ and $q$ are mutually prime integers. Adding up the appropriate multiples of the two binomia... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the least positive number $x$ with the following property: if $a, b, c, d$ are arbitrary positive numbers whose product is 1 , then
$$
a^{x}+b^{x}+c^{x}+d^{x} \geqslant \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} .
$$
|
Solution. Let $a, b, c, d$ be positive numbers whose product equals 1 . From the inequality between the arithmetic and geometric mean of the three numbers $a^{x}, b^{x}$, $c^{x}$, with arbitrary $x>0$, we obtain
$$
\frac{a^{x}+b^{x}+c^{x}}{3} \geqslant \sqrt[3]{a^{x} b^{x} c^{x}}=\sqrt[3]{\frac{1}{d^{x}}}
$$
Choosin... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Pebbles are placed on a $2021 \times 2021$ board in such a way that each square contains at most one pebble. The pebble set of a square of the board is the collection of all pebbles which are in the same row or column as this square. Determine the least number of pebbles that can be placed on the board in such a way th... | Let $N \geqslant 1$ be a positive integer. We claim that the least number of pebbles that can be placed on a $(2 N+1) \times(2 N+1)$ chessboard in such a way that no two squares of the board have the same pebble set is $3 N+1$. The problem has $N=1010$, so at least 3031 pebbles are needed.
We begin by placing $(2 N+1)... | 3031 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers satisfies $a_{1}>5$ and $a_{n+1}=5+6+\cdots+a_{n}$ for all positive integers $n$. Determine all prime numbers $p$ such that, regardless of the value of $a_{1}$, this sequence must contain a multiple of $p$.
# | We claim that the only prime number of which the sequence must contain a multiple is $p=2$. To prove this, we begin by noting that
$$
a_{n+1}=\frac{a_{n}\left(a_{n}+1\right)}{2}-10=\frac{\left(a_{n}-4\right)\left(a_{n}+5\right)}{2} .
$$
Let $p>2$ be an odd prime, and choose $a_{1} \equiv-4(\bmod p)$, so $2 a_{2} \equ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the smallest possible value of the expression
$$
\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{b+c+d}{a}\right\rfloor+\left\lfloor\frac{c+d+a}{b}\right\rfloor+\left\lfloor\frac{d+a+b}{c}\right\rfloor,
$$
in which $a, b, c$ and $d$ vary over the set of positive integers.
(Here $\lfloor x\rfloor$ den... | The answer is 9 .
Notice that $\lfloor x\rfloor>x-1$ for all $x \in \mathbb{R}$. Therefore the given expression is strictly greater than
$$
\frac{a+b+c}{d}+\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+b}{c}-4,
$$
which can be rewritten as
$$
\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\le... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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