problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | 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 |
|---|---|---|---|---|---|---|---|---|
Determine the number of ways to write the numbers $1,2,3,4,5,6,7,8$ and 9 in the individual cells of the triangle in the picture so that the sum in each four-cell triangle is 23 and so that the given number is written in one of the cells in the direction of each arrow.
(E. Novotná)
 of Mr. Rychlý as $v_{R}$ and Mr. Louda as $v_{L}$. The time (in hours) from their departure until their meeting is denoted as $x$. Up to the meeting, Mr. Rychlý walked $x \cdot v_{R}(\mathrm{~km})$ from the cabin, and Mr. Louda walked $x \cdot v_{L}(\mathrm{~km})$ ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Petr and Pavel were picking apples and pears in the orchard. On Monday, Petr ate 2 more pears than Pavel and 2 fewer apples than Pavel. On Tuesday, Petr ate 4 fewer pears than on Monday. Pavel ate 3 more pears than Petr and 3 fewer apples than Petr on Tuesday. Pavel ate 12 apples over the two days and the same number o... | Let $x, y$ be the corresponding numbers of pears and apples that Pavel ate on Monday. According to the problem, we will patiently construct the following table:
| | Monday | Tuesday |
| :---: | :---: | :---: |
| Pavel pears | $x$ | $x+1$ |
| Pavel apples | $y$ | $12-y$ |
| Petr pears | $x+2$ | $x-2$ |
| Petr apples |... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
On July 1, 2010, Mrs. Hovorkova had a credit of 157.0 Kč on her mobile phone. The credit is gradually debited for calls, with 4.5 Kč deducted for each started minute. Mrs. Hovorkova does not send text messages and does not use any other paid services. She tops up her credit as needed, always by an amount of 400 Kč. On ... | The amount in crowns that Mrs. Hovorkova called during the period from July 1 to December 31 is a whole number because the initial and final credit balances and the top-up amounts are always whole numbers. For each minute started, 4.5 Kč is charged, and to reach a whole number, an even number of minutes must have been ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Kluci found an old minefield plan, see the picture. The numbers are on the fields where there are no mines, and they indicate the number of adjacent mined fields. Determine how many mines there are in the field and where they are. (Fields are adjacent if they share a vertex or a side.)
| 1 | | 2 | | 2 |
| :--- | :--... | We can start unambiguously filling in the plan from the cell with the number $3 \mathrm{v}$ in the first column or from the cell with the number $2 \mathrm{v}$ in the upper right corner. In both cases, all unmarked adjacent cells must contain mines (marked as 丸):
| 1 | | 2 | $\star$ | 2 |
| :---: | :---: | :---: | :-... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Consecutive natural numbers are added and subtracted according to the following guide:
$$
1+2-3-4+5+6-7-8+9+10-11-12+\ldots,
$$
that is, two positive and two negative addends always repeat.
Determine the value of such an expression whose last term is 2015.
(L. Hozová) | The sums of pairs of adjacent numbers with opposite signs are either -1 or 1. These values also alternate regularly. In the considered expression, several adjacent numbers thus cancel each other out:
$$
\begin{aligned}
& 1+2-3-4+5+6-7-8+9+10-11-12+13+\ldots \\
= & {[1+(2-3)]+[(-4+5)+(6-7)]+[(-8+9)+(10-11)]+[(-12+13)+\... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Ludvík noticed for a certain division example that if he doubles the dividend and increases the divisor by 12, he gets his favorite number as the result. He would get the same number if he reduced the original dividend by 42 and halved the original divisor.
Determine Ludvík's favorite number.
(M. Petrová) | If we denote the actors in the original example as $a: b$, then Ludvík's observation can be written as:
$$
2 a:(b+12)=(a-42): \frac{b}{2} .
$$
This is a double expression of Ludvík's favorite number, which we will denote as $\ell$ for subsequent modifications.
From the expression on the left, we get
$$
2 a=b \ell+1... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In front of Honza sat three veiled princesses, one of whom was Goldilocks. Honza's task was to find out which one she was.
The princess in the first chair said: "Goldilocks is not sitting in the third chair."
The princess in the second chair said: "I am not Goldilocks."
The princess in the third chair said: "I am Go... | We can distinguish three cases depending on where Blondie could have been sitting:
1. If Blondie was sitting in the first chair, then
- the princess in the first chair would be telling the truth,
- the princess in the second chair would be telling the truth,
- the princess in the third chair would be lying.
2. If Bl... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The commander summoned the other defenders of the castle and decided how to divide their reward:
"The first will take one gold coin and a seventh of the remainder, the second will take two gold coins and a seventh of the new remainder, and so on. That is, the $n$-th defender will take $n$ gold coins and in addition a ... | Let the number of defenders be $p$ and consider from the back. The last $p$-th defender took $p$ gold coins, and thus all the gold coins were taken. The second-to-last $(p-1)$-th defender took $p-1$ gold coins and a seventh of the current remainder, which we denote as $z$. Since both defenders received the same amount,... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Given a triangle $ABC$. On side $AB$ lies point $X$ and on side $BC$ lies point $Y$ such that $CX$ is a median, $AY$ is an altitude, and $XY$ is a midline of triangle $ABC$. Calculate the area of the shaded triangle in the figure, given that the area of triangle $ABC$ is $24 \, \text{cm}^2$.
(M. Dillingerová)
 | First, we will determine how many of each type of delicacy can be prepared from all the plums:
- Four cups of plums are enough for $4 \cdot \frac{1}{2}=2$ whole trays of fruit slices.
- Four cups of plums are enough for $4 \cdot 4=16$ small cakes.
- Four cups of plums are enough for $4 \cdot 16=64$ small pastries.
In... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In a greyhound race, 36 dogs participated. The number of dogs that finished before Punta was four times smaller than the number of those who finished after him.
What was Punta's position?
(L. Hozová) | All the dogs without Punto was $36-1=35$. One fifth of this number is $35: 5=7$; before Punto, 7 dogs ran, behind Punto, $7 \cdot 4=28$ dogs ran. Punto ran eighth.
Evaluation. 3 points for dividing 35 dogs without Punto into fifths; 3 points for dividing the dogs before/behind Punto and placing Punto.
# | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Anička set out from the hotel for a walk, walked $5 \mathrm{~km}$ north, then $2 \mathrm{~km}$ east, $3 \mathrm{~km}$ south, and finally $4 \mathrm{~km}$ west. Thus, she arrived at a pond where she took a bath. Vojta set out from the camp, walked $3 \mathrm{~km}$ south, $4 \mathrm{~km}$ west, and $1 \mathrm{~km}$ north... | According to the task, we will sequentially illustrate Anička's route, Vojta's route, and their mutual relationship (each marked tile represents $1 \mathrm{~km}$):
Now we see that Anička has... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a given group of numbers, one number is equal to the average of all, the largest number is 7 greater than the average, the smallest is 7 less than the average, and most of the numbers in the group have below-average values.
What is the smallest number of numbers that can be in the group?
(K. Pazourek)
Hint. What ... | Let's denote the average of the numbers in the group as $p$. The smallest number in the group is $p-7$, and the largest is $p+7$. The average of these three numbers is $p$, so the average of the remaining numbers in the group must also be $p$.
Therefore, some of the remaining numbers must be less than $p$, and some mu... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A natural number $N$ is called bombastic if it does not contain any zero in its representation and if no smaller natural number has the same product of digits as the number $N$.
Karel first became interested in bombastic prime numbers and claimed that there are not many of them. List all two-digit bombastic prime numb... | All two-digit prime numbers are listed in the first row of the following table. In the second row, the digit products of the individual numbers are listed. In the third row, the smallest natural numbers with the corresponding digit products are listed (these numbers can be determined by comparing the factorizations wit... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Pan Stříbrný uspořádal výstavu. Vystavoval 120 prstenů, které ležely na stolech podél stěn sálu a tvořily tak jednu velkou kružnici. Prohlídka začínala u vchodových dveří v označeném směru. Odtud každý třetí prsten v řadě byl zlatý, každý čtvrtý prsten byl starožitný a každý desátý prsten měl diamant. Prsten, který nem... | Every 3rd ring was golden, every 4th was antique, and every 10th had a diamond. Thus,
- the number of golden rings was $120: 3=40$,
- the number of antique rings was $120: 4=30$,
- the number of rings with a diamond was $120: 10=12$.
When counting rings with multiple properties, we first determine the regularity with... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
On the meadow, there were 45 sheep and several shepherds. After half of the shepherds and a third of the sheep left, the remaining shepherds and sheep together had 126 legs. All the sheep and shepherds had the usual number of legs.
How many shepherds were originally on the meadow?
(L. Hozová)
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 |
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 |
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 |
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 |
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 |
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 |
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
$$
a^{\prime}=\frac{n-1}{n} a, \quad b... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Nela and Jane choose positive integer $k$ and then play a game with a $9 \times 9$ table. Nela selects in every of her moves one empty unit square and she writes 0 to it. Jane writes 1 to some empty (unit) square in every her move. Furthermore $k$ Jane's moves follows each Nela's move and Nela starts. If sum of num... |
Solution. Let us show at first that Jane wins for $k=3$. Let us assume $3 \times 3$ squares $A_{1}, A_{2}$ and $A_{3}$ (see the picture). We will call the $3 \times 3$ square covered if just one 1 is in each its row and column. If Jane covers squares $A_{1}, A_{2}$ and $A_{3}$ without writing to other squares, she win... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Let $\left(a_{n}\right)_{n=1}^{\infty}$ be an infinite sequence such that for all positive integers $n$ we have
$$
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 |
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 |
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 |
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}
$$
have to be in $A$. This is contradiction because $0<b-a<a+c$ and ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Determine all triples $(x, y, z)$ of positive numbers satisfying the system of equations
$$
\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 |
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 |
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 .
$$
Apparently, the numbers $17,19,23,29$, and 31 can never be erased and can never be a part of any change. In any step, ther... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find all integers greater than 1 by which a cancellation can be made in some fraction of the form
$$
\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}}}
$$
Choosin... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
A sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers satisfies $a_{1}>5$ and $a_{n+1}=5+6+\cdots+a_{n}$ for all positive integers $n$. Determine all prime numbers $p$ such that, regardless of the value of $a_{1}$, this sequence must contain a multiple of $p$.
# | We claim that the only prime number of which the sequence must contain a multiple is $p=2$. To prove this, we begin by noting that
$$
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 |
. A finite set of integers is called bad if its elements add up to 2010. A finite set of integers is a Benelux-set if none of its subsets is bad. Determine the smallest integer $n$ such that the set $\{502,503,504, \ldots, 2009\}$ can be partitioned into $n$ Benelux-sets.
(A partition of a set $S$ into $n$ subsets is ... | As $502+1508=2010$, the set $S=\{502,503, \ldots, 2009\}$ is not a Benelux-set, so $n=1$ does not work. We will prove that $n=2$ does work, i.e. that $S$ can be partitioned into 2 Benelux-sets.
Define the following subsets of $S$ :
$$
\begin{aligned}
A & =\{502,503, \ldots, 670\}, \\
B & =\{671,672, \ldots, 1005\}, \... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 4. Let $n \geqslant 3$ be an integer. For each pair of prime numbers $p$ and $q$ such that $p<q \leqslant n$, Morgane has written the sum $p+q$ on the board. She then notes $\mathcal{P}(n)$ as the product of all these sums. For example, $\mathcal{P}(5)=(2+3) \times(2+5) \times(3+5)=280$.
Find all values of $n... | Solution to Exercise 4 Let $n$ be a potential solution, and let $r$ be the largest prime number such that $r \leqslant n$. Since $r$ divides $n$!, it divides $\mathcal{P}(n)$, so there exist two prime numbers $p$ and $q$ such that $p<q \leqslant n$ and $r$ divides $p+q$. By the maximality of $r$, we know that $p+q<2 r$... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 1. Let $x$ and $y$ be two real numbers. We define
$$
M=\max \{x y+1, x y-x-y+3,-2 x y+x+y+2\}
$$
Prove that $M \geqslant 2$, and determine the cases of equality. | Solution to Exercise 1 Among the three numbers $x y+1, x y-x-y+3$ and $-2 x y+x+y+2$, we denote $K$ as the smallest, $L$ as the second smallest, and $M$ as the largest. Therefore, $K \leqslant L \leqslant M$ and
$$
3 M \geqslant K+L+M=(x y+1)+(x y-x-y+3)+(-2 x y+x+y+2)=6
$$
which means that $M \geqslant 2$.
Furtherm... | 2 | Algebra | proof | Yes | Yes | olympiads | false |
Exercise 6. Find the largest integer $n \geqslant 3$ for which there exists a set $\mathcal{S}$ of $n$ points in the plane with the following property: every triangle (even degenerate) whose vertices belong to $\mathcal{S}$ is isosceles but not equilateral. | Solution to Exercise 6 We say that a set $\mathcal{S}$ having the required property is cute. We will prove that any cute set contains at most 6 elements. Furthermore, we will denote by $\omega_{A, B}$ the circle centered at $A$ passing through $B$, and by $\ell_{A, B}$ the perpendicular bisector of the segment $[A B]$.... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Exercise 6. For any integer $k \geqslant 0$, we denote $F_{k}$ as the $k^{\text{th}}$ Fibonacci number, defined by $F_{0}=0$, $F_{1}=1$, and $F_{k}=F_{k-2}+F_{k-1}$ when $k \geqslant 2$. Let $n \geqslant 2$ be an integer, and let $S$ be a set of integers with the following property:
For every integer $k$ such that $2 ... | Solution to Exercise 6 First, let $m=\lceil n / 2\rceil$, and let $S$ be the set $\left\{F_{2 \ell}: 0 \leqslant \ell \leqslant m\right\}$. For any integer $k$ such that $2 \leqslant k \leqslant n$, we choose $x=F_{k}$ and $y=F_{0}=0$ if $k$ is even, or $x=F_{k+1}$ and $y=F_{k-1}$ if $k$ is odd. In both cases, $x$ and ... | +1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 1. Morgane and Bosphore are playing the following game. Morgane has written the integers from 1 to 8 on the vertices of a regular octagon: each integer is written on one of the eight vertices of the octagon. Bosphore then chooses a vertex and calculates the sum of the numbers written on that vertex and its two... | Solution to Exercise 1 First, if Morgane distributes the integers from 1 to 8 as follows, we indeed observe that Bosphore will have to give her at least 12 candies; next to each vertex, we have indicated the sum $s$ that Bosphore would calculate if he chose this vertex.
^{2}+(y-z x)^{2}+(z-x y)^{2}
$$
can take. | Solution to Exercise 2 The quantity we wish to maximize can be rewritten as
$$
\begin{aligned}
\mathcal{S} & =\left(x^{2}+y^{2}+z^{2}\right)+\left((x y)^{2}+(y z)^{2}+(z x)^{2}\right)-6 x y z \\
& =(x+y+z)^{2}-2(x y+y z+z x)+(x y+y z+z x)^{2}-2 x y z(x+y+z)-6 x y z \\
& =1-2(x y+y z+z x)+(x y+y z+z x)^{2}-8 x y z \\
&... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 6. Let $n \geqslant 1$ be an integer. Morgane initially has $n$ piles, each containing one coin. She then allows herself operations of the following form: she chooses two piles, takes as many coins from the first pile as from the second, and forms a new pile with the coins she has taken.
Determine, as a funct... | Solution to Exercise 6 If $n$ is a power of 2, Morgane can manage to end up with only one pile by proceeding as follows: she merges the $n$ piles of size 1 into $n / 2$ piles of size 2, then merges these $n / 2$ piles into $n / 4$ piles of size 4, and so on.
Conversely, when Morgane extracts $k$ pieces from two piles ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 3. Consider 2017 lines in the plane, which intersect pairwise at distinct points. Let $E$ be the set of these intersection points.
We want to assign a color to each point in $E$ such that any two points on the same line, whose segment connecting them contains no other point of $E$, are of different colors.
H... | Solution to Exercise 3 The minimum $m$ sought is $m=3$ and, with what follows, it will be quite evident that the result remains true for $n \geq 3$ lines.
First, we note that, in the obtained configuration, there is at least one non-subdivided region that is a triangle. Indeed, three non-concurrent lines and two never... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 4. We want to color the three-element subsets of $\{1,2,3,4,5,6,7\}$ such that if two of these subsets have no element in common, then they must be of different colors. What is the minimum number of colors needed to achieve this goal? | ## Solution to Exercise 4
Consider the sequence of sets $\{1,2,3\},\{4,5,6\},\{1,2,7\},\{3,4,6\},\{1,5,7\},\{2,3,6\},\{4,5,7\}, \{1,2,3\}$.
Each set must have a different color from the next, so there must be at least two colors. If there were exactly two colors, then the colors would have to alternate, which is impo... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 1. Determine the maximum value of $\sqrt{x}+\sqrt{2 y+2}+\sqrt{3 z+6}$ when $x, y, z$ are strictly positive real numbers satisfying $x+y+z=3$. | Solution to Exercise 1 According to the Cauchy-Schwarz inequality, we have
$$
\sqrt{x} \sqrt{1}+\sqrt{y+1} \sqrt{2}+\sqrt{z+2} \sqrt{3} \leqslant \sqrt{x+y+1+z+2} \sqrt{1+2+3}=6
$$
Equality is achieved when $x=y=z=1$. Therefore, the maximum value is 6. | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Exercise 3. Let $x, y$ and $z$ be three real numbers such that $x^{2}+y^{2}+z^{2}=1$. Find the minimum and maximum possible values of the real number $x y+y z-z x$. | Solution to Exercise 3 According to the identity
$$
(x-y+z)^{2}=x^{2}+y^{2}+z^{2}-2(x y+y z-z x)=1-2(x y+y z-z x)
$$
the goal here is to find the extreme values that the number $(x-y+z)^{2}$ can take.
First, since $(x-y+z)^{2} \geqslant 0$, with equality when $x=y=1 / \sqrt{2}$ and $z=0$, we have
$$
x y+y z-z x=\fr... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 1. Determine all real numbers $x, y, z$ satisfying the following system of equations: $x=\sqrt{2 y+3}, y=\sqrt{2 z+3}, z=\sqrt{2 x+3}$. | Solution to Exercise 1 It is clear that the numbers $x, y, z$ must be strictly positive. The first two equations give $x^{2}=2 y+3$ and $y^{2}=2 z+3$. By subtracting them, we get $x^{2}-y^{2}=2(y-z)$. We deduce that if $x \leqslant y$ then $y \leqslant z$, and similarly if $y \leqslant z$ then $z \leqslant x$. Therefor... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 2. The real numbers $a, b, c$ are distinct and non-zero, and we assume that there exist two real numbers $x$ and $y$ such that $a^{3} + a x + y = 0, b^{3} + b x + y = 0$ and $c^{3} + c x + y = 0$.
Prove that $a + b + c = 0$. | ## Solution.
On a
$$
\left\{\begin{array}{l}
a^{3}+a x+y=0 \\
b^{3}+b x+y=0 \\
c^{3}+c x+y=0
\end{array}\right.
$$
Subtract the first and the third equation: $\left(a^{3}-c^{3}\right)+(a-c) x=0$. Since $a^{3}-c^{3}=(a-c)\left(a^{2}+\right.$ $\left.a c+c^{2}\right)$, we have $(a-c)\left(a^{2}+a c+c^{2}+x\right)=0$. G... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
Exercise 1. The integers $1,2, \ldots, 2018$ are written on the board. Then 2017 operations are performed as follows: choose two numbers $a$ and $b$, erase them, and write $a+b+2ab$ in their place. At the end, only one integer remains on the board.
What are the possible values that the units digit of this integer can ... | Solution to Exercise 1 Since there are 1009 odd numbers between 1 and 2018, we know that the sum of the integers initially written on the board is odd. Furthermore, one of these integers is congruent to $2(\bmod 5)$. An immediate induction then shows that, after each operation, the sum of the integers written on the bo... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 3. Maena and Théodore are playing a game. They play on a square grid consisting of $99 \times 99$ cells. Two cells are considered adjacent if they share a vertex or a side.
Initially, Maéna numbers the cells of the grid from 1 to $99^{2}$, in an arbitrary manner. Théodore then places a token on one of the cel... | Solution to Exercise 3 First, Théodore can always manage to perform at least three moves. To do this, he just needs to select a $2 \times 2$ square inside the $99 \times 99$ square, then traverse its four cells, which are necessarily adjacent since they share a vertex.
Conversely, here is how Maena can proceed to prev... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 5. Martin is trying to fill each cell of a rectangular grid with 8 rows and $\mathrm{n}$ columns with one of the four letters $\mathrm{P}, \mathrm{O}, \mathrm{F}$, and $\mathrm{M}$ such that for any pair of distinct rows, there is at most one column where the intersections of the two rows are cells with the sa... | Solution to Exercise 5 In this problem, we are looking for the largest integer satisfying a certain property. Suppose we want to show that the largest integer sought is the integer c. To show that c is indeed the largest integer, we will on the one hand show that if an integer n satisfies the property, then n ≤ c, and ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 12. Martin is trying to fill each cell of a rectangular grid with 8 rows and $n$ columns with one of the four letters $\mathrm{P}, \mathrm{O}, \mathrm{F}$, and $\mathrm{M}$ such that for any pair of distinct rows, there is at most one column where the intersections of the two rows are cells with the same lette... | Solution to Exercise 12 In this problem, we are looking for the largest integer satisfying a certain property. Suppose we want to show that the largest integer sought is the integer c. To show that c is indeed the largest integer, we will on the one hand show that if an integer n satisfies the property, then n ≤ c, and... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 1. Let $x$ be a strictly positive real number. Show that $x^{3}+\frac{1}{x} \geqslant 2 x$ and find the cases of equality. | Solution to Exercise 1 Given that $x$ is a strictly positive real number, the same is true for $x^{3}$ and $\frac{1}{x}$. The arithmetic-geometric mean inequality then indicates that
$$
x^{3}+\frac{1}{x} \geqslant 2 \sqrt{\frac{x^{3}}{x}}=2 \sqrt{x^{2}}=2 x
$$
To determine the case of equality, one must look at the e... | 1 | Inequalities | proof | Yes | Yes | olympiads | false |
Exercise 6. Determine all sequences $\left(a_{n}\right)_{n \geqslant 1}$ of real numbers such that $a_{i}=a_{i+2020}$ for all integers $i \geqslant 1$, and such that
$$
a_{j}+2 a_{j+2} \geqslant a_{j+1}^{2}+a_{j+1}+1
$$
for all integers $\mathrm{j} \geqslant 1$. | Solution to Exercise 6: Here, the terms $a_{j+1}^{2}$ and 1 in the inequality seem a bit troublesome, as they are not of degree 1 like the other terms in the inequality. We would like to reduce it to a homogeneous inequality, that is, an inequality that remains true if we multiply the different terms by the same factor... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Exercise 18. We say that a set $A$ of polynomials with real coefficients is beautiful if, whenever $P$ and $Q$ are two distinct elements of $A$, there exist positive integers $a_{1}>\cdots>a_{2020}$ such that
$$
P Q=\sum_{i=1}^{2020} i X^{a_{i}}
$$
What is the maximal cardinality of a beautiful set? | Solution to Exercise 18 Let $A$ be a magnificent set. Suppose that $A$ has 3 distinct elements, which we denote as $P, Q, R$.
Since $\mathrm{P}^{2}=\frac{(\mathrm{PQ}) \times(\mathrm{PR})}{\mathrm{QR}}, \mathrm{P}^{2}$ is the product of two polynomials with integer coefficients, divided by a polynomial with integer co... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 8. In a tournament organized between 6 teams, each team plays against each other team exactly once. When a team wins, it gets 3 points, and the losing team receives 0 points. If the game is a draw, both teams receive 1 point. Determine the values of \( a \) for which it is possible that the final scores of the... | Solution to Exercise 8: In each match, between two and three points are awarded in total. Given that there are 15 matches in total, there must have been between 30 and 45 points distributed among the teams. Therefore, we must have:
$$
\begin{aligned}
& 30 \leqslant a+(a+1)+(a+2)+\cdots+(a+5) \leqslant 45 \\
& 30 \leqs... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 16. Consider a circular necklace with 2021 beads. Each bead can be colored either white or green. A coloring of the necklace is said to be chic if, among any 21 consecutive beads, there is always at least one green bead. Show that the number of chic colorings of the necklace is odd. | Solution to Exercise 16 Let's consider a slightly different problem, which involves calculating, for a given integer $\mathrm{N}$, the parity of the number of lines of $\mathrm{N}$ pearls where each pearl is either white or green, and such that among 21 consecutive pearls there is always one green. Let $a_{N}$ be this ... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false |
Exercise 3. Let E be a finite set of strictly positive real numbers such that, for any strictly positive real number $x$, E contains as many elements strictly greater than $x$ as elements strictly less than $\frac{1}{x}$. Determine the product of all elements of $E$. | ## Solution.
Let $x_{0} \in E$ such that $x_{0}>1$. Then, by hypothesis,
- the sets $E \cap]-\infty, \frac{1}{x_{0}}[$ and $E \cap] x_{0},+\infty[$ have cardinals of the same parity,
- the sets $E \cap]-\infty, x_{0}[$ and $E \cap] \frac{1}{x_{0}},+\infty[$ have cardinals of the same parity.
Consequently, the two se... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 2. Find all real numbers $x, y$ such that
$$
x(x-2 y)+y(2 y-4)+4=0
$$ | ## Solution to Exercise 2
Let's develop the equation to obtain
$$
x^{2}-2 x y+2 y^{2}-4 y+4=0
$$
The terms $x^{2}-2 x y$ remind us of the remarkable identity $x^{2}-2 x y+y^{2}=(x-y)^{2}$. We then separate the $2 y^{2}$ into $y^{2}+y^{2}$ to write
$$
(x-y)^{2}+y^{2}-4 y+4=0
$$
But we then recognize a second remark... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 7. An integer $k>1$ is said to be superb if there exist $m, n$, and $a$, three strictly positive integers such that
$$
5^{\mathrm{m}}+63 \mathrm{n}+49=\mathrm{a}^{\mathrm{k}}
$$
Determine the smallest superb integer. | ## Solution to Exercise 7
Suppose $k=2$ is superb: there exist three strictly positive integers $m, n$, and $a$ such that $5^{m} + 63n + 49 = a^{2}$. By looking modulo 3, we have $5^{m} + 1 \equiv 2^{m} + 1 \equiv a^{2} \pmod{3}$. The powers of 2 modulo 3 alternate between 2 if $m$ is odd and 1 if $m$ is even, so $2^{... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 11. Félix wishes to color the integers from 1 to 2023 such that if $a, b$ are two distinct integers between 1 and 2023 and $a$ divides $b$, then $a$ and $b$ are of different colors. What is the minimum number of colors Félix needs? | ## Solution to Exercise 11
We can try to color the numbers greedily: 1 can be colored with a color we denote as $a$, 2 and 3 with the same color $b$ (but not color $a$), then 4 and 6 with color $c$ (we can also color 5 and 7 with color $c$), etc. It seems that an efficient coloring is to color all numbers $n$ such tha... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 3. Alice puts 100 balls in a bag. On each ball she has written the square of an integer. Bob wants to find two balls such that the difference of the numbers is a multiple of 7. How many balls does he need to draw at a minimum to be sure to achieve his goal, regardless of the numbers chosen by Alice at the begi... | ## Solution to Exercise 3
A congruence table modulo 7 shows that the square of an integer can only take the values $0, 1, 2,$ or $4$ modulo 7. Thus, if Bob draws 5 balls, he will have two that have the same remainder modulo 7, according to the pigeonhole principle. The difference in the numbers on these two balls will... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 3. Let $x, y, z$ be non-zero real numbers such that $x+y+z=0$. Suppose that
$$
\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{x}{z}+\frac{z}{y}+\frac{y}{x}+1
$$
Determine the value of $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$. | Solution to Exercise 3 Let \( a \) be the common value of \( \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \) and \( \frac{x}{z} + \frac{z}{y} + \frac{y}{x} + 1 \) so that
\[
\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = a
\]
and
\[
\frac{x}{z} + \frac{z}{y} + \frac{y}{x} + 1 = a
\]
Let's add these two equations:
\[
2a = \f... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 1. Find all integers $p$ such that $p, p+2$ and $p+4$ are all three prime?
A prime number is an integer $\geqslant 2$ that is divisible only by 1 and itself. | Solution to Exercise 1 First, we can expect there to be very few. We wish to obtain information about these prime numbers.
Note that a prime number divisible by 3 is necessarily equal to 3. We consider 3 cases based on the remainder of the division of $p$ by 3 (modulo 3):
- $p=3k: p$ is divisible by 3, so $p=3$. $\{3... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 5. In a grid of size $n \times n$, some cells are white and some are black. We assume that for any pair of columns and any pair of rows, the 4 cells formed by the intersections of these two columns and these two rows are never all the same color. Find the largest value of $n$ for which this is possible. | Solution to Exercise 5: We will show that it is not possible for \( n \geqslant 5 \), and that it is possible for \( n \leqslant 4 \).
To do this, we first observe that if it is possible for an \( n \), it is possible for all \( k \leqslant n \). Indeed, if we take an \( n \times n \) grid satisfying the hypotheses, a... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 7. Let $n$ be a strictly positive integer. Domitille has a rectangular grid divided into unit squares. Inside each unit square is written a strictly positive integer. She can perform the following operations as many times as she wishes:
- Choose a row and multiply each number in the row by $n$.
- Choose a col... | Solution to Exercise 7 For $\mathrm{n}=1$, if we start from a configuration with one column of two rows, with different integers on these two rows, the difference between the two integers remains invariant when one of the two operations is performed. Consequently, it cannot be canceled. In particular, Domitille cannot ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 13. Let $\mathrm{n}$ be a strictly positive integer. Domitille has a rectangular grid divided into unit squares. Inside each unit square is written a strictly positive integer. She can perform the following operations as many times as she wishes:
- Choose a row and multiply each number in the row by $n$.
- Ch... | Solution to Exercise 13 For $n=1$, if we start from a configuration with one column of two rows, with different integers on these two rows, the difference between the two integers remains invariant when one of the two operations is performed. Consequently, it cannot be canceled. In particular, Domitille cannot arrive a... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 5. Let $x, y, z$ be strictly positive real numbers such that
$$
x+\frac{y}{z}=y+\frac{z}{x}=z+\frac{x}{y}=2
$$
Determine all possible values that the number $x+y+z$ can take. | Solution to Exercise 5 In such a problem, it is necessary to examine each equation separately but also to relate them. In practice, this involves looking at the equation obtained when performing the sum or product of two or more equations. Another idea is to apply known inequalities on one side or the other of the equa... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 9. Determine the smallest integer $n \geqslant 2$ such that there exist strictly positive integers $a_{1}, \ldots, a_{n}$ such that
$$
a_{1}^{2}+\ldots+a_{n}^{2} \mid\left(a_{1}+\ldots+a_{n}\right)^{2}-1
$$ | Solution to Exercise 9 Let's analyze the problem. In this problem, we are looking for the smallest integer $\mathrm{n}$ satisfying a certain property. Suppose we want to show that the smallest integer sought is the integer $c$. To show that it is indeed the smallest integer, we must on the one hand show that if an inte... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 4. Let $A B C D$ be a rectangle with area 4. Let I be the midpoint of $[A D]$ and let $J$ be the midpoint of $[B C]$. Let $X$ be the point of intersection of (AJ) and (BI), and let $Y$ be the point of intersection of (DJ) and (CI). What is the area of the quadrilateral IXJY? | Solution to exercise 4 Let's cut the rectangle $A B C D$ into isometric right triangles, as shown below. All triangles have the same area, and the rectangle is made up of 16 triangles in total: each triangle therefore has an area of $1 / 4$. The quadrilateral IXJY is formed of 4 triangles. Its area is therefore equal t... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Exercise 10. I have socks of two different colors: 6 blue socks and 6 red socks. I take several socks at random. How many should I take, at a minimum, to be sure of having two socks of different colors? | Solution to Exercise 10: I must take, at a minimum, 7 socks to be sure of having two socks of different colors. Indeed, if I take 6 socks or fewer, it is possible that these socks are all blue. Conversely, if I take 7 socks, they cannot all be blue (so I will have taken at least one red sock), and they cannot all be re... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 11. Let $A B C D$ be a rectangle with area 4. Let I be the midpoint of $[A D]$ and let $J$ be the midpoint of $[B C]$. Let $X$ be the point of intersection of (AJ) and (BI), and let Y be the point of intersection of (DJ) and (CI). What is the area of the quadrilateral IXJY? | Solution to Exercise 11 Let's cut the rectangle $A B C D$ into isometric right triangles, as shown below. All the triangles have the same area, and the rectangle is made up of 16 triangles in total: each triangle therefore has an area of $1 / 4$. The quadrilateral IXJY is formed of 4 triangles. Its area is therefore eq... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Exercise 8. Calculate
$$
\sqrt{7+\sqrt{1+\sqrt{7+\sqrt{1+\sqrt{7+\sqrt{1+\sqrt{7+\sqrt{3+\sqrt{1}}}}}}}}}
$$
Only a numerical answer is expected here. | Solution to exercise 8 We notice the following chain of simplifications: $\sqrt{3+\sqrt{1}}=2$ then $\sqrt{7+\sqrt{3+\sqrt{1}}}=\sqrt{7+2}=3$ then $\sqrt{1+\sqrt{7+\sqrt{3+\sqrt{1}}}}=\sqrt{1+3}=2$, and so on...
Step by step, we deduce that any expression of the type $\sqrt{1+\sqrt{7+\sqrt{1+\sqrt{\cdots}}}}$ equals 2... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 11. The integers $1,2, \ldots, 20$ have been arranged around a circle, in some order. For each of these integers $k$, Matthieu counts how many integers are less than $k$ among the 9 integers that follow $k$ when traversing the circle clockwise; he counts $A(k)$. He also counts how many integers are less than $... | Solution to Exercise 11 From now on, to simplify the terms used, when we talk about a number "before" (or "after") another number, it means that the first number is among the 9 numbers preceding (or following) the second number, looking in the clockwise direction.
In the following, we denote $o(a)$ as the number diame... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 2. Let $\mathrm{ABC}$ be a triangle, $\mathrm{O}$ a point inside this triangle. The line parallel to (BC) passing through $O$ intersects $[C A]$ at $D$, the line parallel to $(C A)$ passing through $O$ intersects $[A B]$ at $E$, and the line parallel to $(A B)$ passing through $O$ intersects $[\mathrm{BC}]$ at... | ## Solution to Exercise 2 First Solution:
Let $|A B C|$ denote the area of triangle $A B C$. We observe that
$$
\frac{B F}{B C}=\frac{|A B F|}{|A B C|}=\frac{|A B O|}{|A B C|}
$$
because triangles $A B F$ and $A B C$ have the same height and bases $B F$ and $B C$, and $A B O$ and $A B F$ have the same base $A B$ and... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Exercise 4. We are given five numbers in ascending order, which are the lengths of the sides of a quadrilateral (not crossed, but not necessarily convex, meaning a diagonal is not necessarily inside the polygon) and one of its diagonals $D$. These numbers are 3, 5, 7, 13, and 19. What can be the length of the diagonal ... | Solution to exercise 4 We can rephrase the problem: let $a, b, c, d$ be the lengths of the sides of the quadrilateral, and $e$ the length of the diagonal that separates, on one side, the sides of length $a$ and $b$, and on the other side, the sides of length $d$ and $e$. It is necessary and sufficient that the triplets... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Exercise 5. A number has been written on the board. At each step, we add to it the largest of its digits (for example, if 142 is written, the next number will be 146). What is the maximum number of odd numbers that can be written consecutively by proceeding in this way? | Solution to Exercise 5 The answer is 5. Suppose we start with an odd number $n$. We denote $n_{i}$ as the $i$-th number written, with $n_{1}=n$. Let $c_{i}$ and $d_{i}$ be the largest digit and the units digit of $n_{i}$. If $c_{1}$ is odd, then $n_{2}=n_{1}+c_{1}$ is even, and we have only written one odd number. Note... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 4. Let ABCD be a rectangle with area 4. Let I be the midpoint of $[A D]$ and let J be the midpoint of $[B C]$. Let $X$ be the point of intersection of (AJ) and (BI), and let $Y$ be the point of intersection of (DJ) and (CI). What is the area of the quadrilateral IXJY? | Solution to Exercise 4: Let's cut the rectangle ABCD into isometric right triangles, as shown below. All triangles have the same area, and the rectangle is made up of 16 triangles in total: each triangle therefore has an area of $1 / 4$. The quadrilateral IXJY is formed of 4 triangles. Its area is therefore equal to 1.... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Exercise 11. Let $A B C D$ be a rectangle with area 4. Let I be the midpoint of $[A D]$ and let $J$ be the midpoint of $[B C]$. Let $X$ be the point of intersection of (AJ) and (BI), and let $Y$ be the point of intersection of (DJ) and (CI). What is the area of the quadrilateral IXJY? | Solution to Exercise 11 Let's cut the rectangle ABCD into isometric right triangles, as shown below. All triangles have the same area, and the rectangle is made up of 16 triangles in total: each triangle therefore has an area of $1 / 4$. The quadrilateral IXJY is formed of 4 triangles. Its area is therefore equal to 1.... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Exercise 5. In a football tournament, each team plays exactly twice against each of the others. There are no draws; a win earns two points, and a loss earns none. It turns out that only one team won the tournament with 26 points, and there are two teams tied for last with 20 points each. Determine the number of teams, ... | Solution to Exercise 5 Let $n$ be the number of teams.
Each team wins a number of matches between 10 and 13. Since there are $n(n-1)$ matches, each team wins on average $n-1$ matches. The winning team and the two last teams tied win on average 11 matches, while the other teams win on average between 11 and 12 matches.... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 6. Given a point $P$ and a circle $\mathcal{C}$ in the plane, the distance from $P$ to $\mathcal{C}$ is defined as the minimum length $PM$ between $P$ and a point $M$ on the circle $\mathcal{C}$. For example, if $P$ lies on the circle, then the distance from $P$ to $\mathcal{C}$ is zero, and if $P$ is the cent... | Solution to Exercise 6: Let $\mathcal{C}$ be a circle passing at equal distance from $A, B, C, D$. The four points cannot all be inside the circle, otherwise they would be equidistant from the center, and thus would be concyclic. Similarly, they cannot all be outside.
Determine the number of circles $\mathcal{C}$ pass... | 7 | Geometry | 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.