problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
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Blinkov A. A:
The teams held a football tournament in a round-robin format (each team played one match against every other team, with 3 points for a win, 1 point for a draw, and 0 points for a loss). It turned out that the sole winner scored less than $50 \%$ of the maximum possible points for one participant. What is... | Let's prove that there could not have been fewer than six teams. If, for example, there were five teams in the tournament, then they played $5 \cdot 4: 2=10$ matches and scored a total of at least 20 points. Therefore, the sole winner scored more than $20: 5=4$ points. However, according to the conditions, he scored no... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9,10,11 |
| :---: | :---: | :---: |
King Hiero has 11 metal ingots that are indistinguishable in appearance; the king knows that their weights (in some order) are $1, 2, \ldots, 11$ kg. He also has a bag that will tear if more than 11 kg is placed in it.
Archimedes has learned the weights of all the ingots and wants ... | Let Archimedes first put ingots weighing 1, 2, 3, and 5 kg into the bag, and then put ingots weighing 1, 4, and 6 kg. In both cases, the bag does not tear.
We will prove that this could only happen if the 1 kg ingot was used twice. Indeed, if Archimedes used ingots weighing \( w_{1}, \ldots, w_{6} \) kg instead of ing... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Ten dogs and cats were fed 56 biscuits. Each cat received 5 biscuits, and each dog - 6. How many
were dogs and how many were cats? | How many biscuits would be needed if all the animals were dogs?
## Solution
If only dogs were being fed, 106=60 biscuits would be needed. The extra 4 biscuits are required because a dog eats one more biscuit than a cat. This means there were 4 cats, and consequently, 6 dogs.
## Answer
6 dogs and 4 cats. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Sabin A. }}$.
Ofenya bought a batch of pens at a wholesale market and offers customers either one pen for 5 rubles or three pens for 10 rubles. From each customer, Ofenya makes the same profit. What is the wholesale price of a pen? | According to the condition, the difference $10-5=5$ is the wholesale price of two pens.
## Answer
2 rubles 50 kopecks. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$[$ Arithmetic of residues (other). ]
Find the remainder of the division of $2^{100}$ by 3. | $2^{100}=4^{50} \equiv 1^{50}=1(\bmod 3)$.
## Answer
1.
In a class of 30 people, can it be that 9 of them have 3 friends (in this class), 11 have 4 friends, and 10 have 5 friends?
## Solution
In the corresponding graph, there would be 30 vertices, 9 of which would have a degree of 3, 11 a degree of 4, and 10 a de... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\left.\frac{[\text { Prime numbers and their properties }]}{[\underline{\text { Evenness and oddness }}]}\right]$
Find all such natural numbers $p$, that $p$ and $3 p^{2}+1$ are primes. | If $p$ is odd, then $3 p^{2}+1$ is even.
## Answer
$p=2$. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic operations. Numerical identities ]
Find the sum of the digits in the decimal representation of the number $4^{12} \cdot 5^{21}$.
# | Let's transform: $4^{12} \cdot 5^{21}=2^{3} \cdot 10^{21}=80 \ldots 0$.
## Answer
8. | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7 chocolates are more expensive than 8 packs of cookies. What is more expensive - 8 chocolates or 9 packs of cookies?
# | Think about what is more expensive: $7 \cdot 8$ chocolate bars or $8 \cdot 8$ packs of cookies.
## Solution
7 chocolate bars are more expensive than 8 packs of cookies. Therefore, 56 chocolate bars cost more than 64 packs of cookies and even more than 63 packs of cookies. Hence,
8 chocolate bars cost more than 9 pac... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
[Mathematical logic (miscellaneous).] $[$ Arithmetic. Mental arithmetic, etc. ]
In the room, there are 85 balloons — red and blue. It is known that: 1) at least one of the balloons is red; 2) in every arbitrarily chosen pair of balloons, at least one is blue. How many red balloons are in the room? | Think about whether there can be two red balls in the room.
## Solution
Since among any two balls, one is blue, there cannot be two red balls in the room. Therefore, there are 84 blue balloons and 1 red balloon in the room.
## Answer
1 ball. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal number system ] [ Divisibility of numbers. General properties ]
In a 100-digit number 12345678901234...7890, all digits standing at odd positions were erased; in the resulting 50-digit number, all digits standing at odd positions were erased again, and so on. The erasing continued until there was nothing lef... | What characterizes the ordinal numbers of the digits that remain after the first deletion? And after the second?
## Solution
After the first deletion, only those digits will remain whose initial numbers are even, after the second, those whose initial numbers are divisible by 4, after the third - by 8, and so on. Befo... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Mathematical logic (miscellaneous).] [ Arithmetic. Mental arithmetic, etc. ]
Rabbits are sawing a log. They made 10 cuts. How many chunks did they get?
# | Into how many parts is a log divided by the first cut? How does the number of pieces change after each subsequent cut?
## Solution
The number of chunks is always one more than the number of cuts, since the first cut divides the log into two parts, and each subsequent cut adds one more chunk.
## Answer
11 chunks. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2 [Arithmetic. Mental arithmetic, etc.]
There are nuts in the boxes. In the first box, there are 6 kg fewer nuts than in the other two together. And in the second box, there are 10 kg fewer nuts than in the other two together. How many nuts are in the third box? | Of course, we can set up a system of equations, but let's try to do without it.
## Solution
Let's combine both given conditions and obtain the following statement: "In the first and second boxes, there are 6 kg + 10 kg fewer nuts than in the first, second, and two-thirds of the third box." From this, it follows that ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{l}\text { The ratio of volumes } \\ {[\text { Arithmetic. Mental calculation, etc. ]] }\end{array}\right]$
When Gulliver arrived in Lilliput, he found that everything there was exactly 12 times shorter than in his homeland. Can you tell how many Lilliputian matchboxes would fit into Gulliver's mat... | Notice that 12 Lilliputian matchboxes should fit into a Gulliver's matchbox in width, 12 - in length, and 12 - in height.
## Solution
In a Gulliver's matchbox, 12 Lilliputian matchboxes should fit in width, 12 - in length, and 12 - in height. In total, $12 \cdot 12 \cdot 12=1728$ matchboxes.
$$
\text { Send a commen... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Yashchenko I.v.
In all entrances of the house, there is the same number of floors, and on each floor, there is the same number of apartments. The number of floors in the house is greater than the number of apartments per floor, the number of apartments per floor is greater than the number of entrances, and the number ... | $105=3 \cdot 5 \cdot 7$.
## Solution
Let $p$ denote the number of entrances in the building, $f$ the number of floors, and $k$ the number of apartments per floor. Then $p \cdot f \cdot k=105=3 \cdot 5 \cdot 7$, where the numbers $3,5,7-$ are prime. Considering that $1<p<k<f$, we get $f=7$.
## Answer
7 floors. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Andjensss A.
$N$ friends simultaneously learned $N$ pieces of news, with each person learning one piece of news. They started calling each other and exchanging news.
Each call lasts 1 hour. Any number of news items can be shared in one call.
What is the minimum number of hours required for everyone to learn all the ... | a) The news known to one of the friends will be known to no more than two (including the first) after 1 hour, no more than four after the second hour, ..., and no more than 32 after the 5th hour. Therefore, it will take no less than 6 hours.
We will show that 6 hours are sufficient. The conversations can proceed accor... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A number is guessed from 1 to 144. You are allowed to select one subset of the set of numbers from 1 to 144 and ask whether the guessed number belongs to it. For an answer of "yes," you have to pay 2 rubles, and for an answer of "no" - 1 ruble. What is the smallest amount of money needed to surely guess the number?
# | Let $a_{1}=2, a_{2}=3, a_{i}=a_{i-1}+a_{i-2}$ for $i \geq 2$. Then $a_{10}=144$. We will prove by induction that among no fewer than $a_{i}$ numbers, the guessed number cannot be found by paying less than $i+1$ rubles.
For $i=1$ and $i=2$, this is true.
Suppose there are no fewer than $a_{i}$ numbers. Then either the... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## Similar triangles (other). $\quad[$ [Ratio of areas of triangles with a common base or common height] $]$ Difficult [The area of a figure is equal to the sum of the areas of the figures into which it is divided Cauchy's inequality
The diagonals of a convex quadrilateral $ABCD$ intersect at point $E$. It is known th... | Prove that $B C \| A D$.
## Solution
According to problem 35162, $B C \| A D$.
Let $B C = x$. From the similarity of triangles $B E C$ and $D E A$, it follows that $S_{B E C} = \frac{B E}{E D} \cdot S_{D C E} = \frac{x}{3}, S_{D E A} = \frac{D E}{B E} \cdot S_{A B E} = \frac{3}{x}$.
By the condition $S_{A B C D} \l... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Chessboards and chess pieces $]$ [ Examples and counterexamples. Constructions ] [ Evenness and oddness
Authors: Pechkovsky A.N., Itenberg I.
Given an infinite grid paper with a cell side equal to one. The distance between two cells is defined as the length of the shortest path of a rook from one cell to another (t... | An example of coloring a square grid in four colors, such that any two cells at a distance of 6 are colored differently, is shown in the figure on the left.
Another example. Introduce a coor... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.i.
A set of five-digit numbers $\left\{N_{1}, N_{k}\right\}$ is such that any five-digit number, all digits of which are in non-decreasing order, coincides in at least one digit with at least one of the numbers $N_{1}, N_{k}$. Find the smallest possible value of $k$. | A set with the specified properties cannot consist of a single number. Indeed, for each $N=\overline{a b c d e}$, there is a number $G=\overline{g g g 9 g}$ that differs from $N$ in all digits, where $g$ is a non-zero digit different from $a, b, c, d, e$. We will show that the numbers $N_{1}=13579$ and $N_{2}=12468$ fo... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Blinkov A.d:
a) In a football tournament in a round-robin format, 75 teams participated. A team received 3 points for a win, 1 point for a draw, and 0 points for a loss. It is known that every two teams scored a different number of points. Find the smallest possible difference in points between the teams that took fir... | b) The minimum gap between the first and last place cannot be less than $n-1$. We will prove that for $n>3$ it is possible to construct a tournament scheme to achieve a gap of $n-1$ (if $n=2$ or 3, the minimum gap is obviously 3 points).
By induction, we will construct a table for $n$ teams, where the results are all ... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The equation with integer coefficients $x^{4}+a x^{3}+b x^{2}+c x+d=0$ has four positive roots, counting multiplicities.
Find the smallest possible value of the coefficient $b$ under these conditions. | Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the roots of the equation (some of them may coincide). By Vieta's theorem, $b=x_{1} x_{2}+x_{1} x_{3}$ $+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}$
$d=x_{1} x_{2} x_{3} x_{4}$, hence $b$ and $d$ are positive. Notice that
$$
\frac{b}{\sqrt{d}}=\left(\sqrt{\frac{x_{1} x_{2}}{x_... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$[\underline{\text { equations in integers }}]$
Solve the equation $\underbrace{\sqrt{n+\sqrt{n+\ldots \sqrt{n}}}}_{1964 \text { times }}=m$ in integers
# | Let's solve the more general equation $A_{y}(x)=z$ in integers, where $A_{y}(x)=\underbrace{\sqrt{x+\sqrt{x+\ldots \sqrt{x}}}}_{y \text { times }}$. Clearly, $x+A_{y-1}(x)=z^{2}$, or $A_{y-1}(x)=z^{2}-x$.
Thus, if the number $A_{y}(x)=z$ is an integer, then the number $A_{y-1}(x)=\left(z^{2}-x\right)$ is also an integ... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Khachaturyan A.V. }}$
13 children sat at a round table and agreed that boys would lie to girls, but tell the truth to each other, and girls, on the contrary, would lie to boys, but tell the truth to each other. One of the children said to their right neighbor: "The majority of us are boys." That ch... | It is clear that there were both boys and girls at the table. A group of boys sitting next to each other is followed by a group of girls, then boys again, then girls, and so on (a group can consist of one person). Groups of boys and girls alternate, so their number is even. The incorrect statements were made at the tra... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
16 cards with integers from 1 to 16 are laid face down in a \(4 \times 4\) table so that cards with consecutive numbers are adjacent (touching by a side). What is the minimum number of cards that need to be flipped simultaneously to definitely determine the location of all numbers (regardless of how the... | Evaluation. Number the cells as shown in Figure 1.
| 1 | 2 | 3 | 4 |
| :--- | :--- | :--- | :--- |
| 2 | 1 | 4 | 3 |
| 5 | 6 | 7 | 8 |
| 6 | 5 | 8 | 7 |
Fig. 1
Fig. 2
![](https://cdn.mat... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Find the maximum value of the expression
$$
x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}
$$ | Answer: The maximum value of the expression is 1. Since the expression $\sqrt{1-x^{2}}$ is defined, $|x| \leq 1$, which means there exists an angle $\varphi$ such that $x=\cos \varphi, \sqrt{1-x^{2}}=\sin \varphi$. Similarly, $y=\cos \psi, \sqrt{1-y^{2}}=\sin \psi$.
Therefore,
$$
x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=\co... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Tasks with inequalities. Case analysis Examples and counterexamples. Constructions
Author: Shapovosov A.B.
The plan of the shah's palace is a $6 \times 6$ square, divided into rooms of size $1 \times 1$. There is a door in the middle of each wall between the rooms. The shah told his architect: "Knock down some walls... | Consider an arbitrary path from the bottom-left corner of the palace to the top-right corner. Since it is necessary to "climb" 5 horizontal levels and "shift to the right" 5 vertical levels, one would have to pass through at least 10 doors, visiting at least 11 rooms (including the starting and ending rooms).
11 rooms... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $A$ be the sum of the digits of the number $4444^{4444}$, and $B$ be the sum of the digits of the number $A$. Find the sum of the digits of the number $B$.
# | The sum of the digits is comparable to $4444^{4444} \equiv (-2)^{4444} = (2^6)^{740} \cdot 2^4 \equiv 1^{740} \cdot 7 \pmod{9}$. Moreover, $4444^{4444} < 10^{4 \cdot 5000}$, meaning this number has no more than 20000 digits. Therefore, $A \leq 180000, B \leq 45$, and the sum of the digits of the number $B$ is no more t... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $\Phi$.
On the board, $N \geq 9$ different non-negative numbers, each less than one, are written. It turns out that for any eight different numbers on the board, there is a ninth, different from them, such that the sum of these nine numbers is an integer. For which $N$ is this possible? | Clearly, when $N=9$, the required is possible - it is sufficient to write 9 different positive numbers with a unit sum on the board. We will show that when $N>9$, the required is impossible.
Assume the contrary; let $S$ be the sum of all numbers on the board. Choose any numbers $\alpha_{1}, \alpha_{2}, \ldots, \alpha_... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Solve the equation $2 \sin \pi x / 2 - 2 \cos \pi x = x^{5} + 10 x - 54$.
# | Transferring all terms to one side, we get the equation $x^{5}+10 x-54-2 \sin \pi x / 2+2 \cos \pi x=0$. Consider the function
$f(x)=x^{5}+10 x-54-2 \sin \pi x / 2+2 \cos \pi x$. Note that $f^{\prime}(x)=5 x^{4}+10-\pi \cos \pi x / 2-2 \sin \pi x>5 x^{4}+10-3 \pi>0$. Therefore, $f(x)$ is an increasing function, so the... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Numerical tables and their properties ]
Each cell of a $7 \times 8$ table (7 rows and 8 columns) is painted in one of three colors: red, yellow, or green. In each row, the number of red cells is not less than the number of yellow cells, and not less than the number of green cells, while in each column, the number of... | 1) In each row of the table, there are no fewer red cells than yellow ones, so in the entire table, there are no fewer red cells than yellow ones.
In each column of the table, there are no fewer yellow cells than red ones, so in the entire table, there are no fewer yellow cells than red ones.
Thus, in the table, ther... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$Among the actors of Karabas Barabas theater, a chess tournament was held. Each participant played exactly one game with each of the others. One solido was given for a win, half a solido for a draw, and nothing for a loss. It turned out that among any three participants, there would be a ... | Example. Let's denote the participants with letters A, B, V, G, D. Suppose A won against B, B won against V, V won against G, G won against D, D won against A, and all other matches ended in a draw. The condition of the problem is satisfied.
Evaluation. From the condition, it follows that for this tournament, two stat... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal numeral system ] [Arithmetic. Mental calculation, etc.]
Given the number: $123456789101112 \ldots$. What digit is in the 2000th place? | Let's find out from which point the decimal representation of the given number will start to include three-digit numbers: 2000 - $9 \cdot 1$ $90 \cdot 2=1811$.
$1811: 3$ = 603 (remainder 2), which means that the 2000th position is the second digit of the 604th three-digit number. This number is 703, so the required di... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The bathtub fills with cold water in 6 minutes 40 seconds, and with hot water in 8 minutes. In addition, if the plug is removed from a full bathtub, the water will drain in 13 minutes 20 seconds. How long will it take to fill the bathtub completely, given that both taps are open, but the bathtub is not plugged?
# | Let's replace the time in seconds with time in minutes: 6 minutes 40 seconds will be replaced by $62 / 3$, or $20 / 3$, and 13 minutes 20 seconds by $40 / 3$. In one minute, the cold water will fill $3 / 20$ of the bathtub, the hot water will fill $-1 / 8$ of the bathtub, and $3 / 40$ of the bathtub will drain. Therefo... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the books of Novgorod scribes of the $X V$ century, such measures of liquid are mentioned: barrel, nozzle, and bucket. From these same books, it became known that a barrel and 20 buckets of kvass are equal to three barrels of kvass, and 19 barrels, a nozzle, and 15.5 buckets are equal to twenty barrels and eight buc... | From the first condition, it follows that the barrel's capacity is 10 buckets, and from the second condition, that 7.5 buckets and one nozzle can fit into the barrel. Therefore, one nozzle holds 2.5 buckets, or a quarter of the barrel, meaning that 4 nozzles can fit into the barrel.
## Answer
They can. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7 wolves eat 7 sheep in 7 days. How many days will it take for 9 wolves to eat 9 sheep?
# | The number of wolves has increased by the same factor as the number of sheep, so the time of consumption will not change.
## Answer
In 7 days. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\left[\right.$ Arithmetic. Mental calculation, etc.] $\left[\begin{array}{l}\text { Work problems. }\end{array}\right]$
Three diggers dug three holes in three hours. How many holes will six diggers dig in five hours? | First, calculate how many pits 6 diggers will dig in 3 hours.
## Solution
Six diggers in 3 hours will dig $2 \cdot 3=6$ pits. And in 5 hours, they will dig $5 / 3 \cdot 6=10$ pits.
## Answer
10 pits. | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Several gnomes, loading their luggage onto a pony, set off on a long journey. They were spotted by trolls, who counted 36 legs and 15 heads in the caravan. How many gnomes and how many ponies were there?
# | There were 15 creatures in the caravan. If all of them were gnomes, they would have $15 \cdot 2=30$ legs; but in reality, there are 6 more legs, which means there were $6:(4-2)=3$ ponies and $15-3=12$ gnomes.
## Answer
12 gnomes and 3 ponies. | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Text problems (miscellaneous) ] [Arithmetic. Mental calculation, etc.]
In response to a question about the age of his children, the mathematician answered:
- My wife and I have three children. When our first child was born, the total age of the family members was 45 years, a year ago, when the third child was born ... | A year ago, the total age of the children was 11 years, which means the parents' total age was 59 years. And on the day of the first child's birth, this sum was 45. Therefore, the time between these two events is (59 - 45) : $2=7$ years. Thus, the first child was 7 years old a year ago, and the second was 4.
## Answer... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The graphs of the functions $y=k x+b$ and $y=b x+k$ intersect. Find the abscissa of the point of intersection.
# | First method. The desired abscissa is the solution to the equation $k x + b = b x + k$. This equation can be simplified to: $(k - b) x = k - b$. Since the given graphs intersect (do not coincide), $k \neq b$, therefore $x = 1$.
Second method. Notice that $x = 1$ is a solution to the problem: when $x = 1$, both given l... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc. ] [ Problems on percentages and ratios ]
Author: Raskina I.V.
Children went to the forest to pick mushrooms. If Anya gives half of her mushrooms to Vitya, all the children will have the same number of mushrooms, and if instead Anya gives all her mushrooms to Sasha, Sasha will ha... | Let Anya give half of her mushrooms to Vitya. Now all the kids have the same number of mushrooms (this means Vitya had no mushrooms of his own). For Sasha to get all of Anya's mushrooms, he needs to take the mushrooms from Vitya and Anya. Then he will have the mushrooms of three kids - Vitya's, Anya's, and his own. The... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The sum of the three smallest distinct divisors of some number $A$ is 8. How many zeros can the number $A$ end with?
# | The number 8 can be represented as the sum of three different natural numbers in two ways: $8=1+2+5=$ $1+3+4$. The numbers 1, 3, and 4 cannot be the three smallest divisors of the number $A$: if $A$ is divisible by 4, then it is also divisible by 2. Therefore, the three smallest divisors of $A$ are 1, 2, and 5. Thus, $... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Proovvoov v.V.
A chess king has traversed the entire $8 \times 8$ board, visiting each square exactly once and returning to the starting square on the last move.
Prove that he made an even number of diagonal moves. | At each non-diagonal move, the color of the field the king stands on changes; at a diagonal move - it does not change. Since the king has toured the entire board and returned, the color of the field changed from white to black as many times as from black to white, which means the king made an even number of non-diagona... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false |
Booin D.A.
The whole family drank a full cup of coffee with milk, and Katya drank a quarter of all the milk and a sixth of all the coffee. How many people are in the family? | Let $n$ be the number of cups (the number of people in the family), and $x$ be the amount of milk consumed (in cups). Then the amount of coffee consumed is $n-x$. Katya drank one cup of coffee with milk, which consisted of one quarter of all the milk $(x / 4)$ and one sixth of all the coffee $((n-x) / 6)$. We get
$$
\... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Inclusion-Exclusion Principle] [Arithmetic. Mental calculation, etc.]
In a group of 50 children, some know all the letters except "r", which they simply skip when writing, while others know all the letters except "k", which they also skip. One day, the teacher asked 10 students to write the word "кот" (cat), 18 other... | The word "крот" was not written correctly by anyone, because no one can write both the letter "p" and the letter "к" at the same time. The word "рот" or "кот" should have been written by $10+18=28$ people. Note that only
| the words "рот", "кот", and "от" were written. The first two words were written 15 times each, s... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Tiling with domino bones and tiles $]$ [ Text problems (other). $]$
A square $8 \times 8$ was cut into squares $2 \times 2$ and rectangles $1 \times 4$. As a result, the total length of the cuts was 54.
How many figures of each type were obtained? | In an $8 \times 8-64$ square, there are 64 cells, and each of the resulting figures has 4 cells. Therefore, there are 16 figures in total.
Let's find the sum of the perimeters of all the resulting figures. Since each cut boundary is part of the perimeter of two figures, we add to the perimeter of the square twice the ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A chess player played 20 games in a tournament and scored 12.5 points. How many more games did he win than he lost?
# | If a chess player had an equal number of wins and losses, he would have scored 10 points (the same as if he had drawn all 20 games). Since in reality he scored 2.5 points more, five draws need to be replaced by five wins, meaning the chess player has 5 more wins than losses.
## Answer
By 5 games. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
The sequence $a_{n}$ is defined by the condition: $a_{n+1}=a_{n}-a_{n-1}$. Find $a_{100}$, if $a_{1}=3, a_{2}=7$. | Note that $a_{n+3}=a_{n+2}-a_{n+1}=-a_{n}$. Therefore, $a_{n+6}=a_{n}$, which means the terms of the sequence repeat with a period of 6. Since 100 divided by 6 leaves a remainder of 4, we have $a_{100}=a_{4}=-a_{1}$.
## Answer
$-3$. | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}{[\text { Prime numbers and their properties }} \\ {[\underline{\text { Arithmetic of residues (other) })]}}\end{array}\right.$
a) $p, p+10, p+14$ - prime numbers. Find $p$.
b) $p, 2 p+1,4 p+1$ - prime numbers. Find $p$. | Consider the remainders when dividing by 3. One of these numbers is divisible by 3.
## Answer
$p=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Prove that for any $n$
a) $7^{2 n}-4^{2 n}$ is divisible by 33;
b) $3^{6 n}-2^{6 n}$ is divisible by 35. | a) $7^{2 n}-4^{2 n}$ is divisible by $7^{2}-4^{2}=33$.
b) First method. $3^{6 n}-2^{6 n}$ is divisible by $3^{6}-2^{6}=665=19 \cdot 35$.
Second method. $3^{6 n}-2^{6 n}=(-2)^{6 n}-2^{6 n}=0(\bmod 5), 27^{2 n}-8^{2 n}=(-1)^{2 n}-1^{2 n}=0(\bmod 7)$.
$a \equiv 68(\bmod 1967), \quad a \equiv 69(\bmod 1968)$. Find the re... | 5 | Number Theory | proof | Yes | Yes | olympiads | false |
9,10 Can integers be written in the cells of a $4 \times 4$ table so that the sum of all the numbers in the table is positive, while the sum of the numbers in each $3 \times 3$ square is negative? | The central square of size $2 \times 2$ is contained in each square of size $3 \times 3$. Place the number -9 in one of the cells of the central square, and fill the rest of the cells in this table with ones. Then the sum of all numbers in the table is $15+(-9)=6$, and the sum of the numbers inside any $3 \times 3$ squ... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Multiplied several natural numbers and got 224, and the smallest number was exactly half of the largest.
How many numbers were multiplied?
# | $224=2^{5} \cdot 7$. Consider the two numbers mentioned in the condition: the smallest and the largest. If one of them is divisible by 7, then the other must also be divisible by 7. But 224 is not divisible by $7^{2}$, so both of these numbers must be powers of two. From the condition, it also follows that these are tw... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
When it is noon in Bratsk - it is 6 a.m. in Gusev, and 2 p.m. in Komsomolsk-on-Amur. When it is noon in Zlatoust - it is 6 p.m. in Elizovo, and 9 a.m. in Gusev. What time is it in Komsomolsk-on-Amur when it is noon in Elizovo? | When it is noon in Elizovo, it is 3 o'clock in Gusev (from the second condition). And when it is 3 o'clock in Gusev, it is 11 o'clock in Komsomolsk-on-Amur (from the first condition).
## Otвет
11 o'clock in the morning. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Text problems (other).]
In a seven-story building, house spirits live. The elevator travels between the first and the last floors, stopping at each floor. On each floor, starting from the first, one house spirit entered the elevator, but no one exited. When the thousandth house spirit entered the elevator, the eleva... | Let's find out how many house elves ended up in the elevator for a trip from the first to the seventh floor and back, until the elevator returned to the first floor. One house elf entered on the first and seventh floors, and two house elves entered on all other floors. Thus, 12 house elves enter the elevator in one tri... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[The Fundamental Theorem of Arithmetic. Prime Factorization]
## Authors: Raskina I.V., Fedumin L.E.
A hunter told a friend that he saw a wolf with a meter-long tail in the forest. That friend told another friend that a wolf with a two-meter-long tail had been seen in the forest. Passing on the news further, ordinary ... | Each storyteller either doubled or tripled the length of the tail, so 864 is the product of a certain number of twos and threes. Since
$864=2^{5} \cdot 3^{3}$, the wolf's tail was "lengthened" by five ordinary people and three creative individuals.
## Answer
5 ordinary people and 3 creative individuals. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Each of the artists in the creative association "Patience and Labor" has their own work schedule. Six of them paint one picture every two days, another eight artists - one picture every three days, and the rest never paint pictures. From September 22 to 26, they painted a total of 30 pictures. How many pictures will th... | Let's see how many paintings the artists will create from September 22 to 27 inclusive. Each of the six artists, who paint one painting every two days, will create three paintings (one for each pair of days 22-23, 24-25, and 26-27), and each of the eight artists, who paint one painting every three days, will create two... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Identical Transformations ]
It is known that $a^{2}+b=b^{2}+c=c^{2}+a$. What values can the expression $a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-\right.$ $\left.a^{2}\right)$ take? | From the condition, it follows that $a^{2}-b^{2}=c-b, b^{2}-c^{2}=a-c$ and $c^{2}-a^{2}=b-a$. Therefore, $a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+$ $c\left(c^{2}-a^{2}\right)=a(c-b)+b(a-c)+c(b-a)=0$.
## Answer
0. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Text problems (other) ]
At each of the four math club sessions, 20 schoolchildren were present. Nine students attended exactly three out of these four sessions, five students - exactly two sessions, and three students only attended one session. How many schoolchildren attended all sessions? | Let's fill in the "attendance record" for this club. In total, for the specified four sessions, there will be $20 \cdot 4=80$ attendance marks. Each student who attended three sessions is marked three times in the record, so nine such students are marked a total of $9 \cdot 3=27$ times. Similarly, those who attended tw... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic operations. Numerical identities ]
In a test, each of the 25 students received one of the grades "3", "4", or "5". By how many more were there fives than threes, if the sum of all grades is 106?
# | Let $a$ students received a grade of 3, $b$ students received a grade of 4, and $c$ students received a grade of 5. From the problem, we have $a+b+c=25$ and $3a+4b+5c=106$.
Multiply both sides of the first equation by 4: $4a+4b+4c=100$. Now subtract the obtained equation from the second equation, then $c-a=6$.
## Ans... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
Prosperous Mole in autumn dug up 8 bags of grain. For each winter month, he needs either 3 bags of grain or 1 bag of grain and 3 bags of millet. Mole can trade with other moles 1 bag of grain for 2 bags of millet. But his burrow cannot hold more than 12 bags, and in winter Mole ... | A mole can exchange 3 bags of grain for 6 bags of millet, then in its burrow there will be 11 bags: 5 bags of grain and 6 bags of millet. For one month, it will spend 3 bags of grain, and for each of the other two - 1 bag of grain and 3 bags of millet. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Motion problems ]
A cyclist rode from point A to point B, where he stayed for 30 minutes, and then returned to A. On the way to B, he overtook a pedestrian, and 2 hours later met him on the return trip. The pedestrian arrived in B at the same time the cyclist returned to A. How much time did it take the pedestrian t... | The first method. The distance that the pedestrian covers in 2 hours, we will take as a unit. Then the cyclist covers this same distance in 30 minutes. From the moment of their first meeting, the pedestrian has walked one unit, while the cyclist has traveled three units (he rested for half an hour in point B). Therefor... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find the last digit of the number $2^{50}$.
# | $2^{50}=(16)^{3} \cdot 4.6$ any power of 6 ends in 6, so $2^{50}$ ends with the same digit as $6 \cdot 4=24$.
## Answer
4. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Periodicity and Non-periodicity] [
Find the last digit of the number $7^{1988}+9^{1988}$.
# | $7^{1988}+9^{1988}=49^{994}+9^{1988} \equiv(-1)^{994}+(-1)^{1988}=2(\bmod 10)$.
## Answer
2. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
| Reverse Process |
| :---: | :---: |
| | [ Arithmetic. Mental calculation, etc.] |
| | [ Word problems (miscellaneous). [ Iterations |
John had a full basket of trampolines. First, he met Anna and gave her half of his trampolines and another half trampoline. Then he met Banna and gave her half of the remaining tra... | Notice that before meeting Vanna, John had one tremponch left, as half of this amount was half a tremponch. Before meeting Banna, he had 3 tremponchs, because half of this amount was one and a half tremponchs, that is, one and a half. Similarly, we get that initially there were 7 tremponchs.
## Answer
7 tremponchs. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Fomin S.B.
A natural number $n$ is written in the decimal system. It is known that if any digit is included in this representation, then $n$ is divisible by this digit (0 does not appear in the representation). What is the maximum number of different digits that this representation can contain? | If the digit 5 is included in the number, then the number must end in 5. Thus, it is odd and, consequently, contains only odd digits. Therefore, it cannot have more than five digits. If 5 does not appear in the decimal representation of the number, then it can include all the other 8 digits. Here is an example: 1471963... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7,8,9 |
| $[\underline{\text { Modular Arithmetic (other) }}]$ | | |
Find all prime numbers that are equal to the sum of two prime numbers and the difference of two prime numbers. | The given prime number $p$ is odd, so the sum and difference involve numbers of different parity. Thus, $p=q+2=r-2$. From this, it is clear that the numbers give different remainders when divided by 3, meaning one of them is divisible by 3, and since it is prime, it equals 3.
## Answer
5.
$ of positive integers with $m<n$ fulfill the equation
$$
\frac{3}{2008}=\frac{1}{m}+\frac{1}{n} ?
$$
## Answer: 5. | Let $d$ be the greatest common divisor of $m$ and $n$, and let $m=d x$ and $n=d y$. Then the equation is equivalent to
$$
3 d x y=2008(x+y) \text {. }
$$
The numbers $x$ and $y$ are relatively prime and have no common divisors with $x+y$ and hence they are both divisors of 2008. Notice that $2008=8 \cdot 251$ and 251... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $N$ be a positive integer. Two persons play the following game. The first player writes a list of positive integers not greater than 25 , not necessarily different, such that their sum is at least 200 . The second player wins if he can select some of these numbers so that their sum $S$ satisfies the condition $200-... | If $N=11$, then the second player can simply remove numbers from the list, starting with the smallest number, until the sum of the remaining numbers is less than 212. If the last number removed was not 24 or 25 , then the sum of the remaining numbers is at least $212-23=189$. If the last number removed was 24 or 25 , t... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find all integers $n>1$ such that any prime divisor of $n^{6}-1$ is a divisor of $\left(n^{3}-1\right)\left(n^{2}-1\right)$. | Clearly $n=2$ is such an integer. We will show that there are no others.
Consider the equality
$$
n^{6}-1=\left(n^{2}-n+1\right)(n+1)\left(n^{3}-1\right) \text {. }
$$
The integer $n^{2}-n+1=n(n-1)+1$ clearly has an odd divisor $p$. Then $p \mid n^{3}+1$. Therefore, $p$ does not divide $n^{3}-1$ and consequently $p ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A lattice point in the plane is a point whose coordinates are both integral. The centroid of four points $\left(x_{i}, y_{i}\right), i=1,2,3,4$, is the point $\left(\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}, \frac{y_{1}+y_{2}+y_{3}+y_{4}}{4}\right)$. Let $n$ be the largest natural number with the following property: There are ... | To prove $n \geq 12$, we have to show that there are 12 lattice points $\left(x_{i}, y_{i}\right)$, $i=1,2, \ldots, 12$, such that no four determine a lattice point centroid. This is guaranteed if we just choose the points such that $x_{i} \equiv 0(\bmod 4)$ for $i=1, \ldots, 6, x_{i} \equiv 1(\bmod 4)$ for
$i=7, \ldot... | 12 | Combinatorics | proof | Yes | Yes | olympiads | false |
. Let $\Gamma$ be a circle in the plane and $S$ be a point on $\Gamma$. Mario and Luigi drive around the circle $\Gamma$ with their go-karts. They both start at $S$ at the same time. They both drive for exactly 6 minutes at constant speed counterclockwise around the track. During these 6 minutes, Luigi makes exactly on... | . Without loss of generality, we assume that $\Gamma$ is the unit circle and $S=(1,0)$. Three points are marked with bananas:
(i) After 45 seconds, Luigi has passed through an arc with a subtended angle of $45^{\circ}$ and is at the point $\left(\sqrt{2} / 2, \sqrt{2} / 2\right.$ ), whereas Mario has passed through an... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
. Distinct positive integers $a, b, c, d$ satisfy
$$
\left\{\begin{array}{l}
a \mid b^{2}+c^{2}+d^{2}, \\
b \mid a^{2}+c^{2}+d^{2}, \\
c \mid a^{2}+b^{2}+d^{2}, \\
d \mid a^{2}+b^{2}+c^{2},
\end{array}\right.
$$
and none of them is larger than the product of the three others. What is the largest possible number of pr... | At first we note that the given condition is equivalent to $a, b, c, d \mid a^{2}+b^{2}+c^{2}+d^{2}$.
It is possible that three of the given numbers are primes, for example for $a=2, b=3, c=13$ and $d=26$. In this case $2^{2}+3^{2}+13^{2}+26^{2}=13 \cdot 66$ which is divisible by all four given numbers. Furthermore we... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. Find all real numbers $x, y, z$ so that
$$
\begin{aligned}
x^{2} y+y^{2} z+z^{2} & =0 \\
z^{3}+z^{2} y+z y^{3}+x^{2} y & =\frac{1}{4}\left(x^{4}+y^{4}\right)
\end{aligned}
$$ | Answer: $x=y=z=0$.
$y=0 \Longrightarrow z^{2}=0 \Longrightarrow z=0 \Longrightarrow \frac{1}{4} x^{4}=0 \Longrightarrow x=0 . x=y=z=0$ is a solution, so assume that $y \neq 0$. Then $z=0 \Longrightarrow x^{2} y=0 \Longrightarrow x=0 \Longrightarrow \frac{1}{4} y^{4}=0$, which is a contradiction. Hence $z \neq 0$. Now ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
. Let $n>2$ be a given positive integer. There are $n$ guests at Georg's bachelor party and each guest is friends with at least one other guest. Georg organizes a party game among the guests. Each guest receives a jug of water such that there are no two guests with the same amount of water in their jugs. All guests now... | Answer: 2 .
If there are guests $1,2, \ldots, n$ and guest $i$ is friends with guest $i-1$ and $i+1$ modulo $n$ (e.g. guest 1 and guest $n$ are friends). Then if guest $i$ has $i$ amount of water in their jug at the start of the game, then only guest 1 and $n$ end up with a different amount of water than they started ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
On a $16 \times 16$ torus as shown all 512 edges are colored red or blue. A coloring is good if every vertex is an endpoint of an even number of red edges. A move consists of switching the color of each of the 4 edges of an arbitrary cell. What is the largest number of good colorings such that none of them can be conve... | 
Answer: 4. Representatives of the equivalence classes are: all blue, all blue with one longitudinal red ring, all blue with one transversal red ring, all blue with one longitudinal and one ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a forest each of $n$ animals $(n \geqslant 3)$ lives in its own cave, and there is exactly one separate path between any two of these caves. Before the election for King of the Forest some of the animals make an election campaign. Each campaign-making animal visits each of the other caves exactly once,
uses only the... | Answer: b) 4 .
a) As each campaign-making animal uses exactly $n$ paths and the total number of paths is $\frac{n(n-1)}{2}$, the number of campaign-making animals cannot exceed $\frac{n-1}{2}$. Labeling the caves by integers $0,1,2, \ldots, n-1$, we can construct $\frac{n-1}{2}$ non-intersecting campaign routes as fol... | 4 | Combinatorics | proof | Yes | Yes | olympiads | false |
Twelve cards lie in a row. The cards are of three kinds: with both sides white, both sides black, or with a white and a black side. Initially, nine of the twelve cards have a black side up. The cards 1-6 are turned, and subsequently four of the twelve cards have a black side up. Now cards $4-9$ are turned, and six card... | Answer: there are 9 cards with one black and one white side and 3 cards with both sides white.
Divide the cards into four types according to the table below.
| Type | Initially up | Initially down |
| :---: | :---: | :---: |
| $A$ | black | white |
| $B$ | white | black |
| $C$ | white | white |
| $D$ | black | black... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The graph of the function $f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ (where $n>1$ ), intersects the line $y=b$ at the points $B_{1}, B_{2}, \ldots, B_{n}$ (from left to right), and the line $y=c(c \neq b)$ at the points $C_{1}, C_{2}, \ldots, C_{n}$ (from left to right). Let $P$ be a point on the line $y=c$, to ... | Let the points $B_{i}$ and $C_{i}$ have the coordinates $\left(b_{i}, b\right)$ and $\left(c_{i}, c\right)$, respectively, for $i=1,2, \ldots, n$. Then we have
$$
\cot \angle B_{1} C_{1} P+\cdots+\cot \angle B_{n} C_{n} P=\frac{1}{b-c} \sum_{i=1}^{n}\left(b_{i}-c_{i}\right)
$$
The numbers $b_{i}$ and $c_{i}$ are the ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Consider the sequence $a_{k}$ defined by $a_{1}=1, a_{2}=\frac{1}{2}$,
$$
a_{k+2}=a_{k}+\frac{1}{2} a_{k+1}+\frac{1}{4 a_{k} a_{k+1}} \quad \text { for } k \geq 1
$$
Prove that
$$
\frac{1}{a_{1} a_{3}}+\frac{1}{a_{2} a_{4}}+\frac{1}{a_{3} a_{5}}+\cdots+\frac{1}{a_{98} a_{100}}<4
$$ | Note that
$$
\frac{1}{a_{k} a_{k+2}}a_{k}+\frac{1}{2} a_{k+1}
$$
which is evident for the given sequence. Now we have
$$
\begin{aligned}
\frac{1}{a_{1} a_{3}}+\frac{1}{a_{2} a_{4}} & +\frac{1}{a_{3} a_{5}}+\cdots+\frac{1}{a_{98} a_{100}} \\
& <\frac{2}{a_{1} a_{2}}-\frac{2}{a_{2} a_{3}}+\frac{2}{a_{2} a_{3}}-\frac{2... | 4 | Algebra | proof | Yes | Yes | olympiads | false |
Let $m=30030=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ and let $M$ be the set of its positive divisors which have exactly two prime factors. Determine the minimal integer $n$ with the following property: for any choice of numbers from $M$, there exist three numbers $a, b, c$ among them satisfying $a \cdot b \cdot c=... | Taking the 10 divisors without the prime 13 shows that $n \geq 11$. Consider the following partition of the 15 divisors into five groups of three each with the property that the product of the numbers in every group equals $m$.
$$
\begin{array}{ll}
\{2 \cdot 3,5 \cdot 13,7 \cdot 11\}, & \{2 \cdot 5,3 \cdot 7,11 \cdot ... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the smallest number of circles of radius $\sqrt{2}$ that are needed to cover a rectangle
(a) of size $6 \times 3$ ?
(b) of size $5 \times 3$ ?
Answer: (a) Six circles, (b) five circles. | (a) Consider the four corners and the two midpoints of the sides of length 6 . The distance between any two of these six points is 3 or more, so one circle cannot cover two of these points, and at least six circles are needed.
On the other hand one circle will cover a $2 \times 2$ square, and it is easy to see that si... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A set of 8 problems was prepared for an examination. Each student was given 3 of them. No two students received more than one common problem. What is the largest possible number of students? | Answer: 8 .
Denote the problems by $A, B, C, D, E, F, G, H$, then 8 possible problem sets are $A B C, A D E, A F G, B D G, B F H, C D H, C E F, E G H$. Hence, there could be 8 students.
Suppose that some problem (e.g., $A$ ) was given to 4 students. Then each of these 4 students should receive 2 different "supplement... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $f$ be a real-valued function defined on the positive integers satisfying the following condition: For all $n>1$ there exists a prime divisor $p$ of $n$ such that
$$
f(n)=f\left(\frac{n}{p}\right)-f(p)
$$
Given that $f(2001)=1$, what is the value of $f(2002)$ ? | Answer: 2.
For any prime $p$ we have $f(p)=f(1)-f(p)$ and thus $f(p)=\frac{f(1)}{2}$. If $n$ is a product of two primes $p$ and $q$, then $f(n)=f(p)-f(q)$ or $f(n)=f(q)-f(p)$, so $f(n)=0$. By the same reasoning we find that if $n$ is a product of three primes, then there is a prime $p$ such that
$$
f(n)=f\left(\frac{... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Two positive integers are written on the blackboard. Initially, one of them is 2000 and the other is smaller than 2000. If the arithmetic mean $m$ of the two numbers on the blackboard is an integer, the following operation is allowed: one of the two numbers is erased and replaced by $m$. Prove that this operation canno... | Each time the operation is performed, the difference between the two numbers on the blackboard will become one half of its previous value (regardless of which number was erased). The mean value of two integers is an integer if and only if their difference is an even number. Suppose the initial numbers were $a=2000$ and... | 10 | Number Theory | proof | Yes | Yes | olympiads | false |
Let's call a positive integer "interesting" if it is a product of two (distinct or equal) prime numbers. What is the greatest number of consecutive positive integers all of which are "interesting"? | The three consecutive numbers $33=3 \cdot 11,34=2 \cdot 17$ and $35=5 \cdot 7$ are all "interesting". On the other hand, among any four consecutive numbers there is one of the form $4 k$ which is "interesting" only if $k=1$. But then we have either 3 or 5 among the four numbers, neither of which is "interesting". | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Solve the system of equations:
$$
\left\{\begin{array}{l}
x^{5}=y+y^{5} \\
y^{5}=z+z^{5} \\
z^{5}=t+t^{5} \\
t^{5}=x+x^{5} .
\end{array}\right.
$$ | Adding all four equations we get $x+y+z+t=0$. On the other hand, the numbers $x, y, z, t$ are simultaneously positive, negative or equal to zero. Thus, $x=y=z=t=0$ is the only solution. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
An equilateral triangle $A B C$ is divided into 100 congruent equilateral triangles. What is the greatest number of vertices of small triangles that can be chosen so that no two of them lie on a line that is parallel to any of the sides of the triangle $A B C$ ? | An example for 7 vertices is shown in Figure 2. Now assume we have chosen 8 vertices satisfying the conditions of the problem. Let the height of each small triangle be equal to 1 and denote by $a_{i}, b_{i}, c_{i}$ the distance of the $i$ th point from the three sides of the big triangle. For any $i=1,2, \ldots, 8$ we ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A square is divided into 16 equal squares, obtaining the set of 25 different vertices. What is the least number of vertices one must remove from this set, so that no 4 points of the remaining set are the vertices of any square with sides parallel to the sides of the initial square?
Remark. The proposed solution to thi... | The example in Figure 3a demonstrates that it suffices to remove 8 vertices to "destroy" all squares. Assume now that we have managed to do that by removing only 6 vertices. Denote the horizontal and vertical lines by $A, B, \ldots, E$ and $1,2, \ldots, 5$ respectively. Obviously, one of the removed vertices must be a ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $\mathbb{N}$ denote the set of positive integers. Let $\varphi: \mathbb{N} \rightarrow \mathbb{N}$ be a bijective function and assume that there exists a finite limit
$$
\lim _{n \rightarrow \infty} \frac{\varphi(n)}{n}=L
$$
What are the possible values of $L$ ? | In this solution we allow $L$ to be $\infty$ as well. We show that $L=1$ is the only possible value. Assume that $L>1$. Then there exists a number $N$ such that for any $n \geq N$ we have $\frac{\varphi(n)}{n}>1$ and thus $\varphi(n) \geq n+1 \geq N+1$. But then $\varphi$ cannot be bijective, since the numbers $1,2, \l... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
We say that an integer $m$ covers the number 1998 if $1,9,9,8$ appear in this order as digits of $m$. (For instance, 1998 is covered by 215993698 but not by 213326798 .) Let $k(n)$ be the number of positive integers that cover 1998 and have exactly $n$ digits $(n \geqslant 5)$, all different from 0 . What is the remain... | Answer: 1.
Let $1 \leqslant g<h<i<j \leqslant n$ be fixed integers. Consider all $n$-digit numbers $a=\overline{a_{1} a_{2} \ldots a_{n}}$ with all digits non-zero, such that $a_{g}=1, a_{h}=9, a_{i}=9$, $a_{j}=8$ and this quadruple 1998 is the leftmost one in $a$; that is,
$$
\begin{cases}a_{l} \neq 1 & \text { if }... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Prove that there exists a number $\alpha$ such that for any triangle $A B C$ the inequality
$$
\max \left(h_{A}, h_{B}, h_{C}\right) \leq \alpha \cdot \min \left(m_{A}, m_{B}, m_{C}\right)
$$
holds, where $h_{A}, h_{B}, h_{C}$ denote the lengths of the altitudes and $m_{A}, m_{B}, m_{C}$ denote the lengths of the med... | Let $h=\max \left(h_{A}, h_{B}, h_{C}\right)$ and $m=\min \left(m_{A}, m_{B}, m_{C}\right)$. If the longest height and the shortest median are drawn from the same vertex, then obviously $h \leq m$. Now let the longest height and shortest median be $A D$ and $B E$, respectively, with $|A D|=h$ and $|B E|=m$. Let $F$ be ... | 2 | Inequalities | proof | Yes | Yes | olympiads | false |
For a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of real numbers it is known that
$$
a_{n}=a_{n-1}+a_{n+2} \quad \text { for } n=2,3,4, \ldots
$$
What is the largest number of its consecutive elements that can all be positive?
Answer: 5. | The initial segment of the sequence could be $1 ; 2 ; 3 ; 1 ; 1 ;-2 ; 0$. Clearly it is enough to consider only initial segments. For each sequence the first 6 elements are $a_{1} ; a_{2}$; $a_{3} ; a_{2}-a_{1} ; a_{3}-a_{2} ; a_{2}-a_{1}-a_{3}$. As we see, $a_{1}+a_{5}+a_{6}=a_{1}+\left(a_{3}-a_{2}\right)+\left(a_{2}-... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $a, b, c, d, e, f$ be non-negative real numbers satisfying $a+b+c+d+e+f=6$. Find the maximal possible value of
$$
a b c+b c d+c d e+d e f+e f a+f a b
$$
and determine all 6-tuples $(a, b, c, d, e, f)$ for which this maximal value is achieved.
Answer: 8 . | If we set $a=b=c=2, d=e=f=0$, then the given expression is equal to 8 . We will show that this is the maximal value. Applying the inequality between arithmetic and geometric mean we obtain
$$
\begin{aligned}
8 & =\left(\frac{(a+d)+(b+e)+(c+f)}{3}\right)^{3} \geq(a+d)(b+e)(c+f) \\
& =(a b c+b c d+c d e+d e f+e f a+f a ... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Determine all positive integers $n$ such that $3^{n}+1$ is divisible by $n^{2}$.
Answer: Only $n=1$ satisfies the given condition. | First observe that if $n^{2} \mid 3^{n}+1$, then $n$ must be odd, because if $n$ is even, then $3^{n}$ is a square of an odd integer, hence $3^{n}+1 \equiv 1+1=2(\bmod 4)$, so $3^{n}+1$ cannot be divisible by $n^{2}$ which is a multiple of 4 .
Assume that for some $n>1$ we have $n^{2} \mid 3^{n}+1$. Let $p$ be the sma... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. In a club with 30 members, every member initially had a hat. One day each member sent his hat to a different member (a member could have received more than one hat). Prove that there exists a group of 10 members such that no one in the group has received a hat from another one in the group. | Let $S$ be the given group of 30 people. Consider all subsets $A \subset S$ such that no member of $A$ received a hat from a member of $A$. Among such subsets, let $T$ be a subset of maximal cardinality. The assertion of the problem is that $|T| \geq 10$.
Let $U \subset S$ consist of all people that have received a ha... | 10 | Combinatorics | proof | Yes | Yes | olympiads | false |
. For a positive integer $k$, let $d(k)$ denote the number of divisors of $k$ (e.g. $d(12)=6$ ) and let $s(k)$ denote the digit sum of $k$ (e.g. $s(12)=3$ ). A positive integer $n$ is said to be amusing if there exists a positive integer $k$ such that $d(k)=s(k)=n$. What is the smallest amusing odd integer greater than... | The answer is 9 . For every $k$ we have $s(k) \equiv k(\bmod 9)$. Calculating remainders modulo 9 we have the following table
| $m$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $m^{2}$ | 0 | 1 | 4 | 0 | 7 | 7 | 0 | 4 | 1 |
| $m^{6}$ | 0 | 1 |... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. For which $k$ do there exist $k$ pairwise distinct primes $p_{1}, p_{2}, \ldots, p_{k}$ such that
$$
p_{1}^{2}+p_{2}^{2}+\cdots+p_{k}^{2}=2010 ?
$$ | We show that it is possible only if $k=7$.
The 15 smallest prime squares are:
$$
4,9,25,49,121,169,289,361,529,841,961,1369,1681,1849,2209
$$
Since $2209>2010$ we see that $k \leq 14$.
Now we note that $p^{2} \equiv 1 \bmod 8$ if $p$ is an odd prime. We also have that $2010 \equiv 2 \bmod 8$. If all the primes are ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
If the price of a product increased from $R \$ 5.00$ to $R \$ 5.55$, what was the percentage increase? | The increase in reais was $5.55-5=0.55$; so the percentage increase was
$$
\frac{0.55}{5}=\frac{0.55 \times 20}{5 \times 20}=\frac{11}{100}=11 \%
$$ | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A factory produced an original calculator that performs two operations:
- the usual addition +
- the operation $\circledast$
We know that for any natural number $a$ we have:
$$
\text { (i) } a \circledast a=a \quad \text { and (ii) } a \circledast 0=2 a
$$
and, for any four naturals $a, b, c$ and $d$
$$
\text { (i... | To calculate $(2+3) \circledast(0+3)$, we will use property (iii), and we have:
$$
(2+3) \circledast(0+3)=(2 \circledast 0)+(3 \circledast 3)
$$
Now, by (i) we have $2 \circledast 0=2 \times 2=4$, and by (ii) we have $3 \circledast 3=3$. Therefore,
$$
(2+3) \circledast(0+3)=4+3=7
$$
Now, to calculate $1024 \circled... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the figure below, $D$ is the midpoint of side $A B, C E: D E=5: 3 \text{ and } B F: E F=1: 3$. If the area of triangle $A B C$ is $192 \text{ cm}^2$, determine the area of triangle $B D F$.
![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-03.jpg?height=481&width=547&top_left_y=616&top_left_x=826... | Solution
Denote the lengths of $B D, B F$ and $D E$ by $x, z$ and $3 y$, respectively. Given the proportions, it follows that $E F=3 z, C E=5 y$ and $A D=x$. Therefore,
$$
\begin{aligned}
& \frac{A_{B D F}}{A_{B D E}}=\frac{z}{3 z+z} \\
& \frac{A_{B D E}}{A_{C D B}}=\frac{3 y}{3 y+5 y} \\
& \frac{A_{C D B}}{A_{A B C}... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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