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6. How many numbers from 1 to 999 are written in the Roman numeral system with the same number of symbols as in the decimal system?
(P. D. Mulyenko)
Reference. To write a number in Roman numerals, you need to break it down into place value addends, write each place value addend according to the table, and then write ... | Solution. Note that, regardless of the digit place and other digits of the number, a decimal digit a is written as:
- zero Roman numerals when $a=0$,
- one Roman numeral when $a=1$ and $a=5$,
- two Roman numerals when $a$ is $2,4,6,9$,
- three Roman numerals when $a=3$ and $a=7$,
- four Roman numerals when $a=8$.
Thu... | 52 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Milla and Zhena came up with a number each and wrote down all the natural divisors of their numbers on the board. Milla wrote down 10 numbers, Zhena wrote down 9 numbers, and the largest number written on the board twice is 50. How many different numbers are written on the board? | Solution. Note that a number written twice is a common divisor of the original numbers; the largest such number is their GCD. Therefore, all numbers written twice are divisors of the number 50, that is, the numbers $1,2,5,10,25,50$. Thus, among the listed numbers, exactly 6 are repeated, and the number of different num... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. On graph paper, a polygon with a perimeter of 36 is drawn, with its sides running along the grid lines. What is the maximum area it can have? | Solution. Consider the extreme verticals and horizontals. Moving from them inward does not allow reducing the perimeter, but it decreases the area. Therefore, the rectangle has the largest area. If A and B are the lengths of its sides, then A + B = 18.
By trying different rectangles with a perimeter of 36, such as (1,... | 81 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The brothers found a treasure of gold and silver. They divided it so that each received 100 kg. The eldest received the most gold - 25 kg - and one eighth of all the silver. How much gold was in the treasure? | Solution. 1) The elder brother received 75 kg of silver, which is one-eighth of the total amount; therefore, the total mass of silver is 600 kg.
2) The others received more silver than the elder brother, i.e., each received more than 75 kg. If there are at least eight brothers, then in total they would receive more th... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same... | Solution. $A$ and $C$ meet every 7 minutes, while $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 53 = 371$ minutes. At the same time, $B$ and $D$ also meet every 7 minutes, so on the 371st minute, car $D$ will be at the same point as th... | 371 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. The park is a $10 \times 10$ grid of cells. A lamp can be placed in any cell (but no more than one lamp per cell).
a) The park is called illuminated if, no matter which cell a visitor is in, there is a $3 \times 3$ square of 9 cells that contains both the visitor and at least one lamp. What is the minimum number of... | Solution. a) 4. Divide the park into 4 quarters (squares $5 \times 5$), then there must be at least one lamp in each quarter (to illuminate, for example, the corner cells). By placing one lamp in the center of each quarter, we get an example.
b) 10.
Estimate. In each corner square $3 \times 3$ there must be at least t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Let's call the efficiency of a natural number $n$ the fraction of all natural numbers from 1 to $n$ inclusive that have a common divisor with $n$ greater than 1. For example, the efficiency of the number 6 is $\frac{2}{3}$.
a) Does there exist a number with an efficiency greater than $80\%$? If so, find the smalles... | Solution. Let's move on to studying inefficiency (1 minus efficiency). From the formula for Euler's function, it follows that it is equal to $\frac{p_{1}-1}{p_{1}} \cdot \ldots \frac{p_{k}-1}{p_{k}}$, where $p_{1}, \ldots, p_{k}$ are all distinct prime divisors of $n$. Then, by adding a new prime factor, we can increas... | 30030 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Since a syllable consists of two different letters, identical letters can only appear at the junction of syllables.
First, let's find the number of combinations of two syllables with a matching letter at the junction. Such syllables (in terms of the arrangement of vowels and consonants) are either AMMO $(3 \cdot 8 ... | Answer: $2 \cdot 264 \cdot 48-2 \cdot 24^{2}=24192$ funny words. | 24192 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Once Valera left home, walked to the cottage, painted 11 fence boards there, and returned home 2 hours after leaving. Another time, Valera and Olga went to the cottage together, painted 8 fence boards (without helping or hindering each other), and returned home together 3 hours after leaving. How many boards will Ol... | Solution. This is a more complex version of problem 5 for 5th grade.
The strange result (that the characters accomplished less work in more time when working together) is explained by the different times spent walking, since the speed of "joint" walking is equal to the lower of the two walkers' speeds. The second time... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In a rectangular grid $20210 \times 1505$, two diagonals were drawn and all cells through which they passed were painted. How many cells were painted?
(O. A. Pyayve, A. A. Tessler) | Solution. This is a more complex version of problem 3 for 5th grade.
First, let's determine how many cells one diagonal crosses. Note that $20210=215$. $94,1505=215 \cdot 7$. Therefore, the ... | 42986 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On a plane, an equilateral triangle and three circles with centers at its vertices are drawn, and the radius of each circle is less than the height of the triangle. A point on the plane is painted yellow if it lies inside exactly one of the circles; green if inside exactly two; and blue if inside all three. It turns... | Solution. The sum of the areas of the three circles is $1000+2 \cdot 100+3 \cdot 1=1203$; the sum of the areas of the three "lenses" is $100+3 \cdot 1=103$ (a "lens" is the intersection of two circles).
The area of the triangle is $S_{1}-S_{2}+S_{3}$, where
$S_{1}=1203 / 6-$ the sum of the areas of the three 60-degre... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. How many five-digit numbers are roots of the equation $x=[\sqrt{x}+1][\sqrt{x}]$? The symbol $[a]$ denotes the integer part of the number $a$, that is, the greatest integer not exceeding $a$.
(0. A. Pyayve) | Solution. Let $n=[\sqrt{x}]$, then $[\sqrt{x}+1]=[\sqrt{x}]+1=n+1$, which means $x=n(n+1)$.
All numbers of the form $x=n(n+1)$ are suitable, since for them $n<\sqrt{x}<n+1$, meaning $[\sqrt{x}]$ is indeed equal to $n$.
It remains to count the five-digit numbers of this form. Note that 99$\cdot$100 $<10000<100 \cdot 1... | 216 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Given a rectangle of size $2021 \times 4300$. Inside it, there is a billiard ball. It is launched in a straight line, forming a $45^{\circ}$ angle with the sides of the rectangle. Upon reaching a side, the ball reflects at a $45^{\circ}$ angle; if the ball hits a corner, it exits along the same line it entered. (An ... | # Solution.
We will replicate the rectangle multiple times by reflecting it relative to its sides. In adjacent columns (rows), copies of the rectangle will be oriented differently, but when shifted by an even number of columns and an even number of rows, the orientation will match the initial one. Now, we can consider... | 294 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In the country, there are 100 cities, and several non-stop air routes are in operation between them, such that one can travel from any city to any other, possibly with layovers. For each pair of cities, the minimum number of flights required to travel from one to the other was calculated. The transportation difficul... | Solution. First, we prove that the maximum transportation difficulty occurs when the cities are connected "in a chain."
Indeed, we can consider the graph as a tree (otherwise, we can remove some edges to leave a tree - the difficulty will increase). Choose the longest path in the tree. Suppose there are vertices not i... | 8332500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Marina needs to buy a notebook, a pen, a ruler, a pencil, and an eraser to participate in the Olympiad. If she buys a notebook, a pencil, and an eraser, she will spend 47 tugriks. If she buys a notebook, a ruler, and a pen, she will spend 58 tugriks. How much money will she need for the entire set, if the notebook c... | Solution. If Marina buys two sets from the condition, she will spend $47+58=105$ tugriks, but she will buy an extra notebook, so the full set of school supplies costs $105-15=90$ tugriks.
Criteria. Only the answer without explanation - 1 point. If in the solution they try to determine the cost of the pen and pencil (a... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. An accident has occurred in the reactor of a research spacecraft, and toxic substances are leaking from it. All corridors between rooms are equipped with airtight doors, but there is no time to close individual doors. However, the captain can still give the command "Close $N$ doors," after which the ship's artificia... | Solution. There are a total of 23 corridors on the spaceship. If no more than 21 doors are closed, then the corridors between the reactor and the right engine, and between the right engine and the lounge, may remain open, which means the crew will be in danger. Therefore, at least 22 doors must be closed.
Criteria. Co... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, the same letters correspond to the same digits). He got the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible by 5? | Solution. The number must be divisible by 5, so the letter "A" is equal to 0 or 5. If it is equal to 0, then for the other letters ("G", "V", "T", "E", "M", "L") there are $A_{9}^{6}=9!/ 3$! options; if "A" is equal to 5, then for the other letters there are $8 \cdot A_{8}^{5}=8$!/3! options, since "G" cannot be zero. ... | 114240 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. Three cars $A, B$ and $C$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ drives counterclockwise. All cars move at constant (but pairwise distinct) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time. After another 46 minutes, ... | Solution. $A$ and $C$ meet every 7 minutes, and $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 53=371$ minutes.
Criteria. If 53 minutes is replaced with 46 - 3 points. Solved by trial with lengths and speeds -1 point. Only the answer w... | 371 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. There is a rectangular sheet, white on one side and gray on the other. It was folded as shown in the picture. The perimeter of the first rectangle is 20 more than the perimeter of the second rectangle. And the perimeter of the second rectangle is 16 more than the perimeter of the third rectangle. Find the perimeter ... | Solution. From the figure, it can be seen that when folding, the perimeter of the rectangle decreases by twice the short side, so the short side of rectangle-1 is $20 / 2=10$, the short side of rectangle-2 is $16 / 2=8$. Therefore, the long side of rectangle-1 is 18, and the long side of the original sheet is 28. Thus,... | 92 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Katya decided to calculate the sum of the cubes of all natural divisors of some natural number, and she got the result $M A T H$. But then she discovered that she had forgotten one of the divisors. Adding its cube, she got the correct result - MASS. Find the smallest possible value of the number $M A T H$. (MATH and... | Solution. Answer: 2017. The original natural number is $12 ; 12^{3}+6^{3}+4^{3}+2^{3}+1^{3}=$ 2017; if you add $3^{3}$, you get 2044.
We will prove that there are no smaller suitable numbers.
1) For any number less than 10, the sum of the cubes of all divisors, as is easily verified, is less than a thousand.
2) $10^{... | 2017 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Several plants and zombies (no more than 20 creatures in total) attended the "Plants VS Zombies" gathering, and it turned out that all creatures were of different heights. Plants always tell the truth to those who are shorter than them and lie to those who are taller. Zombies, on the contrary, lie to shorter creatur... | Solution. Let the total number of beings be $n$, and exactly $z$ of them are zombies. When plants greet, they say to everyone “I am taller than you”, and zombies say to everyone “I am shorter”. Each zombie said this phrase to everyone except themselves, so we get $z(n-1)=20$. Given that $n-1<20$, the possible cases are... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. There are 28 students in the class. On March 8th, each boy gave each girl one flower - a tulip, a rose, or a daffodil. How many roses were given if it is known that there were 4 times as many roses as daffodils, but 10 times fewer than tulips? (A. A. Tesler) | Solution. This is a more complex version of problem 4 for 5th grade.
Let the number of narcissus be $x$, then the number of roses is $4x$, and the number of tulips is $40x$, so the total number of flowers is $45x$. The number of flowers is the product of the number of boys and the number of girls. If there are $m$ boy... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. The magical clock, in addition to the usual pair of hands, has a second pair that is symmetrical to the first at every moment relative to the vertical axis. It is impossible to determine which hands are real from a photograph of the clock. Furthermore, just like with ordinary clocks, it is impossible to distinguish ... | Solution. See the solution to problem 6 for 10th grade.
Instead of the last paragraph of the solution to problem 6 for 10th grade, it is sufficient to provide an example with exactly three undefined photographs and prove that there are exactly three. For example, let the first three photos be taken at 5:50, 11:40, 17:... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Find all real solutions of the system of equations
$$
\left\{\begin{array}{l}
\frac{1}{x}=\frac{32}{y^{5}}+\frac{48}{y^{3}}+\frac{17}{y}-15 \\
\frac{1}{y}=\frac{32}{z^{5}}+\frac{48}{z^{3}}+\frac{17}{z}-15 \\
\frac{1}{z}=\frac{32}{x^{5}}+\frac{48}{x^{3}}+\frac{17}{x}-15
\end{array}\right.
$$
(A. B. Vladimirov) | Solution. Let $F(t)=32 t^{5}+48 t^{3}+17 t-15$. Then the system has the form $F\left(\frac{1}{y}\right)=\frac{1}{x}, F\left(\frac{1}{z}\right)=\frac{1}{y}$, $F\left(\frac{1}{x}\right)=\frac{1}{z}$. From this, it follows that $F\left(F\left(F\left(\frac{1}{x}\right)\right)\right)=\frac{1}{x}$. Note that $F(0.5)=0.5$ and... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Milla and Zhena came up with a number each and wrote down all the natural divisors of their numbers on the board. Milla wrote 10 numbers, Zhena - 9, and the number 6 was written twice. How many different numbers are on the board? | Solution. Since the number 6 is written twice, both original numbers (denote them as a and b) are divisible by 6.
If Vera's number has 10 divisors, then its factorization is either $p^{9}$ or $p^{1} \cdot q^{4}$ (where p and q are some prime numbers); the first is impossible since it is divisible by 6. Valya's number ... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Three people want to travel from city A to city B, which is 45 kilometers away from A. They have two bicycles. The speed of a cyclist is 15 km/h, and the speed of a pedestrian is 5 km/h. What is the minimum time they can reach B, if the bicycle can be left unattended on the road? | Solution. Two people ride a bicycle for 10 kilometers, then one of them leaves the bicycle by the road and walks the next 10 kilometers, while the other continues for the next 10 kilometers and also leaves the bicycle (which the first one should pick up later), the third one walks the first 10 kilometers and then rides... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. On graph paper, a polygon with a perimeter of 2014 is drawn, with its sides running along the grid lines. What is the maximum area it can have? | Solution. First, consider the extreme verticals and horizontals. Moving from them inward does not allow reducing the perimeter, but it decreases the area. Therefore, the rectangle has the largest area. If A and B are the lengths of its sides, then $A + B = 1007$.
Now, among different rectangles with a perimeter of 201... | 253512 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. It is known that on Monday, the painter painted twice as slowly as on Tuesday, Wednesday, and Thursday, and on Friday - twice as fast as on these three days, but worked 6 hours instead of 8. On Friday, he painted 300 meters more of the fence than on Monday. How many meters of the fence did the painter paint from Mon... | Solution. Let's take $100 \%$ of the fence length that the painter painted on Tuesday, Wednesday, and Thursday. Then Monday accounts for $50 \%$, and Friday for $-150 \%$. Therefore, 300 meters of the fence correspond to $150 \% - 50 \% = 100 \%$. For the entire week (from Monday to Friday), it is $500 \%$, i.e., 1500 ... | 1500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of four-digit numbers in which all digits are different, the first digit is divisible by 2, and the sum of the first and last digits is divisible by 3. | Solution. The first digit can be 2, 4, 6, or 8. If the first is 2, the last can be 1, 4, 7; if the first is 4, the last can be 2, 5, 8; if the first is 6, the last can be 0, 3, (6 does not fit), 9; if the first is 8, the last can be 1, 4, 7. In total, there are 3 + 3 + 3 + 3 = 12 options for the first and last digits. ... | 672 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. In the Olympionov family, it is a tradition to especially celebrate the day when a person turns as many years old as the sum of the digits of their birth year. Kolya Olympionov had such a celebration in 2013, and Tolya Olympionov had one in 2014. Who is older and by how many years? | Solution. Let's determine in which year a person could be born if adding the sum of the digits of their birth year results in 2013 or 2014. It is clear that the year of such a birth cannot be later than 2014. Since the sum of the digits of each number from 1 to 2014 does not exceed \(1+9+9+9=28\), the birth year cannot... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Karlson bought several pancakes (25 rubles each) and several jars of honey (340 rubles each) at the buffet. When he told Little Man how much he spent at the buffet, Little Man was able to determine, based on this information alone, how many jars of honey and how many pancakes Karlson bought. Could this amount have e... | Solution. Could. For example, let's say Karlson spent $4 \cdot 340 + 25 \cdot 40 = 2360$ rubles. Suppose Karlson can make up this amount in some other way; for this, he should spend x rubles less on pancakes and x rubles more on honey (or vice versa). But then, for x rubles, he can buy both a whole number of pancakes a... | 2360 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. The brothers found a treasure of gold and silver. They divided it so that each received 100 kg. The eldest received the most gold - 30 kg - and one fifth of all the silver. How much gold was in the treasure | Solution. 1) The elder brother received 70 kg of silver, which is one fifth of the total amount; therefore, the total mass of silver is 350 kg.
2) The others received more silver than the elder brother, i.e., each received more than 70 kg. If there are at least five brothers, then in total they would receive more than... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A fly is sitting at vertex $A$ of a triangular room $A B C$ ( $\angle B=60^{\circ}, \angle C=45^{\circ}, A C=5$ m). At some point, it flies out in a random direction, and each time it reaches a wall, it turns $60^{\circ}$ and continues flying in a straight line (see figure). Can it happen that after some time, the f... | Solution. Let the fly take off at an angle of 60 degrees to the line $A C$. Consider the equilateral triangle $A K C$ with side $A C$. Note that its sides $A K$ and $K C$ can be divided into parts (into infinitely many parts) such that each part equals the next segment of the fly's trajectory. The sum of these parts is... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Pasha and Igor are flipping a coin. If it lands on heads, Pasha wins; if tails, Igor wins. The first time the loser pays the winner 1 ruble, the second time - 2 rubles, then - 4, and so on (each time the loser pays twice as much as the previous time). After 12 games, Pasha is 2023 rubles richer than he was initially... | Answer: 9 (all except 4, 8, and 1024).
Solution. We need to place the signs in the equation $\pm 1 \pm 2 \pm 2^{2} \pm 2^{3} \pm \ldots \pm 2^{9} \pm 2^{10} \pm 2^{11}=2023$. If we choose all plus signs, the sum will be $2^{0}+\ldots+2^{11}=2^{12}-1=4095$, so we need to replace plus signs with minus signs before the n... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Given a right triangle $ABC$ with a right angle at $A$. On the leg $AC$, a point $D$ is marked such that $AD: DC = 1: 3$. Circles $\Gamma_{1}$ and $\Gamma_{2}$ are then constructed with centers at $A$ and $C$ respectively, passing through point $D$. $\Gamma_{2}$ intersects the hypotenuse at point $E$. Circle $\Gamma... | Answer: 13.
Solution. Let $AC = x$. Then $AD = x / 4, DC = CE = 3x / 4, BE = BC - CE = \sqrt{x^2 + 25} - 3x / 4$. According to the problem, $\angle AFB = 90^\circ$, so $AF^2 + FB^2 = AB^2$, which means $AD^2 + BE^2 = 25$. Expressing everything in terms of $x$ and simplifying, we get $13x = 12 \sqrt{x^2 + 25}$. Squarin... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. On the coordinate plane, points $A(0,0)$ and $B(1000,0)$ were marked, as well as points $C_{1}(1,1)$, $C_{2}(2,1), \ldots, C_{999}(999,1)$. Then all possible lines $A C_{i}$ and $B C_{i}(1 \leqslant i \leqslant 999)$ were drawn. How many integer-coordinate intersection points do all these lines have? (An integer-coo... | Solution. Let $a_{n}$ and $b_{n}$ denote the lines passing through $A$ and $B$ respectively, as well as through a point on $l$ with an abscissa that is $n$ greater than the abscissa of $A$ (where $1 \leqslant n \leqslant 999$). The lines $a_{n}$ and $a_{m}$ for $n \neq m$ intersect at a non-integer point (between $A B$... | 2326 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of five-digit numbers in which all digits are different, the first digit is divisible by 2, and the sum of the first and last digits is divisible by 3. | Solution. The first digit can be 2, 4, 6, or 8. If the first is 2, the last can be $1, 4, 7$; if the first is 4, the last can be $2, 5, 8$; if the first is 6, the last can be $0, 3, (6$ does not fit), 9; if the first is 8, the last can be $1, 4, 7$. In total, there are $3+3+3+3=13$ options for the first and last digits... | 4032 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the sum of all roots of the equation:
$$
\begin{gathered}
\sqrt{2 x^{2}-2024 x+1023131}+\sqrt{3 x^{2}-2025 x+1023132}+\sqrt{4 x^{2}-2026 x+1023133}= \\
=\sqrt{x^{2}-x+1}+\sqrt{2 x^{2}-2 x+2}+\sqrt{3 x^{2}-3 x+3}
\end{gathered}
$$
(L. S. Korechkova) | Solution. Note that the radicands in the left part are obtained from the corresponding radicands in the right part by adding $x^{2}-2023 x+1023130=$ $(x-1010)(x-1013)$. Since all radicands are positive (it is sufficient to check for $x^{2}-x+1$ and for $\left.2 x^{2}-2024 x+1023131=2(x-506)^{2}+511059\right)$, the left... | -2023 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Pasha and Igor are flipping a coin. If it lands on heads, Pasha wins; if tails, Igor wins. The first time the loser pays the winner 1 ruble, the second time - 2 rubles, then - 4, and so on (each time the loser pays twice as much as the previous time). At the beginning of the game, Pasha had a single-digit amount of ... | Solution. Let $n$ be the amount of money Pasha has become richer (and Igor poorer). Note that Pasha won the last game (otherwise, he would have lost more money than he gained in all previous stages). Therefore, the sequence of games can be divided into series, in each of which Pasha won the last game and lost all the o... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. How many solutions in natural numbers does the equation $(a+1)(b+1)(c+1)=2 a b c$ have? | Solution. Rewrite the equation as $(1+1 / a)(1+1 / b)(1+1 / c)=2$. Due to symmetry, it is sufficient to find all solutions with $a \leqslant b \leqslant c$. Then $(1+1 / a)^{3} \geqslant 2$, which means $a \leqslant(\sqrt[3]{2}-1)^{-1}<4$ and $a \in\{1,2,3\}$. In the case $a=1$, the inequality $2(1+1 / b)^{2} \geqslant... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. $f(x)$ is a linear function, and the equation $f(f(x))=x+1$ has no solutions. Find all possible values of the quantity $f(f(f(f(f(2022)))))-f(f(f(2022)))-f(f(2022))$. | Solution. Let $f(x)=k x+b$, then $f(f(x))=k(k x+b)+b=k^{2} x+k b+b$. The equation can have no solutions only when $k^{2}=1$, that is, for functions $x+b$ or $-x+b$, so the answer is either $(2022+5 b)-(2022+3 b)-(2022+2 b)=-2022$, or $(-2022+b)-(-2022+b)-2022=-2022$.
Answer: -2022.
Criteria. Up to 2 points will be de... | -2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Let's call the efficiency of a natural number $n$ the fraction of all natural numbers from 1 to $n$ inclusive that have a common divisor with $n$ greater than 1. For example, the efficiency of the number 6 is $\frac{2}{3}$.
a) Does there exist a number with an efficiency greater than $80\%$? If so, find the smalles... | Solution. Let's move on to studying inefficiency (1 minus efficiency). From the formula for Euler's function, it follows that it is equal to $\frac{p_{1}-1}{p_{1}} \cdot \ldots \frac{p_{k}-1}{p_{k}}$, where $p_{1}, \ldots, p_{k}$ are all distinct prime divisors of $n$. Then, by adding a new prime factor, we can increas... | 30030 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Calculate the area of the set of points on the coordinate plane that satisfy the inequality $(y+\sqrt{x})\left(y-x^{2}\right) \sqrt{1-x} \leqslant 0$. | Solution. The left side makes sense only for $0 \leqslant x \leqslant 1$. In this case, it is required that $y+\sqrt{x}$ and $y-x^{2}$ have different signs (or one of them equals zero), or that $x$ equals 1. If we exclude the case $x=1$ (which gives a zero area), we are left with a part of the plane bounded by the segm... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. A natural number $n$ is called cubish if $n^{3}+13 n-273$ is a cube of a natural number. Find the sum of all cubish numbers. | Solution. If $0<13n-273<13\cdot21$, so it remains to check all other numbers.
If $n=21$, then $13n-273=0$, so 21 is cubic. For $n-3n^{2}+3n-1$, the number $n$ will not be cubic (i.e., for $8<n<21$).
If $n=8$, then $13n-273=-169=-3\cdot8^{2}+3\cdot8-1$, so it is cubic. For $n \leqslant 5$, the expression $n^{3}+13n-27... | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Pasha draws dots at the intersections of the lines on graph paper.
He likes it when four dots form a figure resembling a "kite," as shown on the right (the kite must be of this exact shape and size, but can be rotated). For example, the 10 dots shown in the second image form only two kites. Is it possible to draw a... | Solution. For example, like this. Here there are 21 points and 24 snakes (6 snakes in each direction).
 | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Flint has five sailors and 60 gold coins. He wants to distribute them into wallets, and then give the wallets to the sailors so that each gets an equal number of coins. But he doesn't know how many sailors will be alive by the time of the distribution. Therefore, he wants to distribute the coins in such a way that t... | Solution. Answer: 9 wallets. Example: $12,12,8,7,6,5,4,3,3$.
We will prove that 8 wallets are insufficient.
1) Note that each wallet should contain no more than 12 coins. Therefore, 15 coins must be made up of at least two wallets. This means that when we divide 8 wallets among four pirates, each should receive two w... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. In a certain triangle, the sum of the tangents of the angles turned out to be 2016. Estimate (at least to the nearest degree) the magnitude of the largest of its angles. | Solution. One of the tangents must exceed 600. This is only possible for an angle very close to $90^{\circ}$. We will prove that it exceeds $89.5^{\circ}$. This is equivalent to the statement that $\operatorname{tg} 0.5^{\circ}>$ $1 / 600$.
Let's start with the equality $\sin 30^{\circ}=1 / 2$. Note that $\sin 2 x=2 \... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find all natural numbers $n$ for which $2^{n}+n^{2016}$ is a prime number. | Solution. Let's consider three cases.
- If $n$ is even, then the given number is also even (and greater than two for $n>0$).
- If $n$ is odd and not divisible by 3, then $2^{n}$ gives a remainder of 2 when divided by 3, and $n^{2016}=\left(n^{504}\right)^{4}$ gives a remainder of 1 when divided by 3, so the sum is div... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. In the game "set," all possible four-digit numbers consisting of the digits $1,2,3$ (each digit appearing exactly once) are used. It is said that a triplet of numbers forms a set if, in each digit place, either all three numbers contain the same digit or all three numbers contain different digits.
The complexity of... | Solution. Note that for any two numbers, there exists exactly one set in which they occur. Indeed, the third number of this set is constructed as follows: in the positions where the first two numbers coincide, the third number has the same digit; in the position where the first two numbers differ, the third number gets... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Each cell of a $100 \times 100$ board is painted blue or white. We will call a cell balanced if among its neighbors there are an equal number of blue and white cells. What is the maximum number of balanced cells that can be on the board? (Cells are considered neighbors if they share a side.) | Solution. Cells lying on the border of the board but not in the corner cannot be equilibrium cells, since they have an odd number of neighbors (three). There are $4 \cdot 98=392$ such cells.
All other cells can be made equilibrium, for example, with a striped coloring (the first row is blue, the second row is white, t... | 9608 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On a sheet of notebook paper, two rectangles are outlined. The first rectangle has a vertical side shorter than the horizontal side, while the second has the opposite. Find the maximum possible area of their common part, if each rectangle contains more than 2010 but less than 2020 cells. | Solution. The common part of these two rectangles (if it is non-empty) is a rectangle. Let's define the maximum possible lengths of its sides.
The vertical side of the first rectangle is shorter than the horizontal one, so it is less than $\sqrt{2020}$, which is less than 45; the same is true for the horizontal side o... | 1764 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In the game "set," all possible four-digit numbers consisting of the digits $1,2,3$ (each digit appearing exactly once) are used. It is said that a triplet of numbers forms a set if, in each digit place, either all three numbers contain the same digit, or all three numbers contain different digits.
For example, the... | Solution. Note that for any two numbers, there exists exactly one set in which they occur. Indeed, the third number of this set is constructed as follows: in the positions where the first two numbers coincide, the third number has the same digit; in the position where the first two numbers differ, the third number gets... | 1080 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. On a sheet of lined paper, two rectangles are outlined. The first rectangle has a vertical side shorter than the horizontal side, while the second has the opposite. Find the maximum possible area of their common part, if the first rectangle contains 2015 cells, and the second - 2016. | Solution. The common part of these two rectangles (if it is non-empty) is a rectangle. Let's determine the maximum possible lengths of its sides.
The vertical side of the first rectangle is shorter than the horizontal one, so it is less than $\sqrt{2015}$, which is less than 45; the same is true for the horizontal sid... | 1302 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A motorcyclist set out from point $A$ with an initial speed of 90 km/h, uniformly increasing his speed (that is, over equal time intervals, his speed increases by the same amount). After three hours, the motorcyclist arrived at point $B$, passing through $C$ along the way. After that, he turned around and, still uni... | Solution. In 5 hours, the speed changed from 90 km/h to 110 km/h, so the acceleration is 4 km/h$^2$. From $A$ to $B$ the distance is
$$
90 \cdot 3+\frac{4}{2} \cdot 3^{2}=270+18=288(\text{km})
$$
from $B$ to $C-$
$$
110 \cdot 2-\frac{4}{2} \cdot 2^{2}=220-8=212(\text{km})
$$
And the required distance is 76 km. | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. A natural number $n$ is called cubish if $n^{3}+13 n-273$ is a cube of a natural number. Find the sum of all cubish numbers. | Solution. If $021$, so it remains to check all other numbers.
If $n=21$, then $13 n-273=0$, so 21 is cubic. For $n-3 n^{2}+3 n-1$, the number $n$ will not be cubic (i.e., for $8<n<21)$
If $n=8$, then $13 n-273=-169=-3 \cdot 8^{2}+3 \cdot 8-1$, so it is cubic. For $n \leqslant 5$, the expression $n^{3}+13 n-273$ will ... | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Let $a$ and $n$ be natural numbers, and it is known that $a^{n}-2014$ is a number with $2014$ digits. Find the smallest natural number $k$ such that $a$ cannot be a $k$-digit number. | Solution. Let $a$ be a $k$-digit number, then $10^{k-1} \leq a2013$, i.e., $k>45$.
We will check for $k$ starting from 46, whether the condition of the absence of an integer in the specified interval is met. We will find that
(for $k=46$) $2013 / 462013$ );
(for $k=47$) $2013 / 472013$ );
(for $k=48$) $2013 / 48201... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In a certain language, there are 3 vowels and 5 consonants. A syllable can consist of any vowel and any consonant in any order, and a word can consist of any two syllables. How many words are there in this language? | 1. The language has $3 \cdot 5=15$ syllables of the form "consonant+vowel" and the same number of syllables of the form "vowel+consonant", making a total of 30 different syllables. The total number of two-syllable words is $30 \cdot 30=900$ | 900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Masha and Lena left home and went to the store for ice cream. Masha walked faster and got to the store in 12 minutes. Spending 2 minutes buying the ice cream, she headed back. After another 2 minutes, she met Lena. Walking a bit more, Masha finished her ice cream and, deciding to buy another one, turned around and w... | 3. Masha covers $1 / 6$ of the entire distance in 2 minutes, which means Lena covered $5 / 6$ of the distance by the time they met, and it took her 16 minutes. Therefore, $1 / 6$ of the distance takes Lena $16: 5=3 \frac{1}{5}$ minutes, i.e., 3 minutes and 12 seconds. For the entire distance, she would need 19 minutes ... | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Petya prints five digits on the computer screen, none of which are zeros. Every second, the computer removes the initial digit and appends to the end the last digit of the sum of the remaining four digits. (For example, if Petya enters 12345, after one second it will become 23454, then 34546, and so on. However, he ... | Answer: 2.
Solution. The record 00000 cannot appear on the screen, as it can only result from 00000. A record with four zeros and one also cannot appear, since in that case, the last digit would not equal the remainder of the division of the sum of the first four by 10.
However, a sum of digits equal to 2 is possible... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In a certain language, there are 5 vowels and 7 consonants. A syllable can consist of any vowel and any consonant in any order, and a word can consist of any two syllables. How many words are there in this language? | 1. The language has $5 \cdot 7=35$ syllables of the form "consonant+vowel" and the same number of syllables of the form "vowel+consonant", making a total of 70 different syllables. The total number of two-syllable words is $70 \cdot 70=4900$. | 4900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Given an isosceles triangle $A B C$, where $\angle A=30^{\circ}, A B=A C$. Point $D$ is the midpoint of $B C$. On segment $A D$, point $P$ is chosen, and on side $A B$, point $Q$ is chosen such that $P B=P Q$. What is the measure of angle $P Q C ?$ (S. S. Korechkova) | Answer: $15^{\circ}$.
Solution. Since $D$ is the midpoint of the base of the isosceles triangle, $A D$ is the median, bisector, and altitude of the triangle. Draw the segment $P C$. Since $\triangle P D B = \triangle P D C$ (by two sides and the right angle between them), $P C = P B = P Q$, meaning that all three tria... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. A few years ago, in the computer game "Minecraft," there were 9 different pictures (see the figure): one horizontal picture measuring $2 \times 1$ and $4 \times 2$, one square picture measuring $2 \times 2$, and two each of pictures measuring $1 \times 1$, $4 \times 3$ (horizontal), and $4 \times 4$. In how many way... | Answer: 896.
Solution. We will say that two paintings are in different columns if no block of the first painting is in the same column as any block of the second. It is clear that the $4 \times 4$ paintings are in different columns from each other and from the $4 \times 3$ paintings regardless of their placement. Thus... | 896 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In each cell of a $100 \times 100$ table, a natural number was written. It turned out that each number is either greater than all its neighbors or less than all its neighbors. (Two numbers are called neighbors if they are in cells sharing a common side.) What is the smallest value that the sum of all the numbers can... | Solution. Let's divide the board into dominoes. In each domino, the numbers are different, so their sum is at least $1+2=3$. Then the total sum of the numbers on the board is at least 15000. This estimate is achievable if we alternate ones and twos in a checkerboard pattern. | 15000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. The pond has a square shape. On the first frosty day, the part of the pond that is no more than 10 meters away from the nearest point on the shore froze. On the second day, the part no more than 20 meters away froze, on the third day, the part no more than 30 meters away, and so on. On the first day, the area of ope... | Solution. Note that the larger the side of the pond, the smaller the percentage that will freeze on the first day. If the side of the pond is 100 m, then 36% will freeze on the first day, and if the side is 120 m, then on the first day, $1-\frac{100^{2}}{120^{2}}=11 / 36<1 / 3$ of the pond's area will freeze. Therefore... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find all such numbers $k$ for which
$$
(k / 2)!(k / 4)=2016+k^{2}
$$
The symbol $n!$ denotes the factorial of the number $n$, which is the product of all integers from 1 to $n$ inclusive (defined only for non-negative integers; $0!=1$). | Solution. Note that the left side makes sense only for even values of $k$. We directly verify that $k=2,4,6,8,10$ do not work, while $k=12$ gives a correct equality.
With each further increase of $k$ by 2, the expression $(k / 2)!$ increases by at least 7 times, i.e., the left side grows by more than 7 times. At the s... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On the coordinate plane, an isosceles triangle $A B C$ was drawn: $A B=2016, B C=$ $A C=1533$, with vertices $A$ and $B$ lying on nodes on the same horizontal line. Determine how many nodes lie within the triangle $A B C$ (including nodes lying on the sides). A node is a point on the coordinate plane where both coor... | Solution. Note that $1533^{2}-1008^{2}=(1533-1008)(1533+1008)=525 \cdot 2541=21 \cdot 25 \cdot 7 \cdot 363=$ $7 \cdot 3 \cdot 5^{2} \cdot 7 \cdot 3 \cdot 11^{2}=(7 \cdot 3 \cdot 5 \cdot 11)^{2}=1155^{2}$. Therefore, the height of the triangle is 1155.
We see that the GCD of 1155 and 1008 is 21. This means that there a... | 1165270 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Let's call a rectangular parallelepiped typical if all its dimensions (length, width, and height) are different. What is the smallest number of typical parallelepipeds into which a cube can be cut? Don't forget to prove that this is indeed the smallest number. | Solution. A cube can be cut into four typical parallelepipeds. For example, a cube $5 \times 5 \times 5$ can be cut into parallelepipeds $5 \times 3 \times 1, 5 \times 3 \times 4, 5 \times 2 \times 1, 5 \times 2 \times 4$.
![](https://cdn.mathpix.com/cropped/2024_05_06_bfb93b3716727f813bd7g-2.jpg?height=401&width=401&... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Lydia likes a five-digit number if none of the digits in its notation is divisible by 3. Find the total sum of the digits of all five-digit numbers that Lydia likes. | Solution. Lidia likes five-digit numbers composed only of the digits $1,2,4,5,7,8$. Note that all such numbers can be paired in the following way: each digit $a$ is replaced by $9-a$, that is, 1 is replaced by 8, 2 by 7, 4 by 5, and vice versa. For example, the number 42718 is paired with the number 57281. This partiti... | 174960 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Petya's favorite TV game is called "Sofa Lottery." During the game, viewers can send SMS messages with three-digit numbers containing only the digits $1, 2, 3$, and 4. At the end of the game, the host announces a three-digit number, also consisting only of these digits. An SMS is considered a winning one if the numb... | Answer: 8.
Solution. An example of eight suitable SMS messages: 111, 122, 212, 221, 333, 344, 434, 443. Indeed, no matter what number the host names, it contains either at least two digits from the set $\{1,2\}$, or at least two from the set $\{3,4\}$. If the third digit is from the other set, we replace it with a dig... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. What is the maximum number of numbers that can be chosen from the set $\{1,2, \ldots, 12\}$ so that the product of no three chosen numbers is a perfect cube? | Solution. 9: all except $4,9,12$. Note that to remove the cubes, we need to remove at least one element from each of the sets $\{1,2,4\},\{3,6,12\},\{2,4,8\},\{1,3,9\},\{2,9,12\}$, $\{3,8,9\},\{4,6,9\}$. Note that all numbers, except 9, are in no more than three of these seven triples. Therefore, if we remove two numbe... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest possible value of the expression
$$
\left(\frac{x y}{z}+\frac{z x}{y}+\frac{y z}{x}\right)\left(\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y}\right)
$$
where $x, y, z$ are non-zero real numbers. | Solution. Note that the signs of all six numbers $\frac{x y}{z}, \frac{z x}{y}$, etc., are the same. If all of them are negative, then replace the numbers $x, y, z$ with their absolute values. As a result, each term ($\frac{x y}{z}$, etc.) will change its sign. The modulus of each bracket will remain the same, but the ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In each cell of a $10 \times 10$ table, a natural number was written. Then, each cell was shaded if the number written in it was less than one of its neighbors but greater than another neighbor. (Two numbers are called neighbors if they are in cells sharing a common side.) As a result, only two cells remained unshad... | Solution. Answer: 20. The estimate matches the estimate in problem 7.5. One of the possible examples is shown below.
| 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 8 | 10 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 |
| 7 | 8 | 10 | 12 | 13 | 14 ... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The edge of a regular tetrahedron $A B C D$ is 1. Through a point $M$, lying on the face $A B C$ (but not on the edge), planes parallel to the other three faces are drawn. These planes divide the tetrahedron into parts. Find the sum of the lengths of the edges of the part that contains point $D$. | 5. See the figure at the end of the file.
Note that the part of the tetrahedron we are interested in is bounded by three of its faces containing point $D$, and by three planes parallel to the faces. Therefore, this is a parallelepiped.
Consider the faces $A B C$ and $A D C$. Align these faces so that vertex $B$ coinc... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.2. Two circles with radii 1 and 2 have a common center $O$. The area of the shaded region is three times smaller than the area of the larger circle. Find the angle $\angle A O B$.
 | Answer: $60^{\circ}$
Solution: Let $\angle A O B=\alpha$. Then the area of the shaded part is $\frac{360^{\circ}-\alpha}{360^{\circ}} \pi+$ $4 \frac{\alpha}{360^{\circ}} \pi-\frac{\alpha}{360^{\circ}} \pi=\left(\frac{360^{\circ}-\alpha}{360^{\circ}}+3 \frac{\alpha}{360^{\circ}}\right) \pi$. The area of the large circl... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.3. Prove that $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$ is a rational number. | Solution: Let $x=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$. Then $x^{3}=(\sqrt{5}+2)-(\sqrt{5}-2)-$ $3 \sqrt[3]{(\sqrt{5}+2)(\sqrt{5}-2)} \times(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2})=4-3 x ; x^{3}+3 x-4=0.0=x^{3}+$ $3 x-4=(x-1)\left(x^{2}+x+4\right), x-1=0$ or $x^{2}+x+4=0$. The second equation has no roots becau... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
11.4. In a row, $n$ integers are written such that the sum of any seven consecutive numbers is positive, and the sum of any eleven consecutive numbers is negative. For what largest $n$ is this possible? | Answer: 16
Solution: Let's provide an example for $n=16$:
$80, -31, -31, -31, -31, 80, -31, -31, -31, -31, 80, -31, -31, -31, -31, 80$. We will prove that for $n \geq 17$, it will not be possible to write down a sequence of numbers that satisfy the conditions of the problem. Let's construct a table for the first 17 n... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.5. Given a function $f(x)$, satisfying the condition
$$
f(x y+1)=f(x) f(y)-f(y)-x+2
$$
What is $f(2017)$, if it is known that $f(0)=1$? | Answer: 2018
Solution: $f(0 \cdot 0+1)=f(0) f(0)-f(0)-0+2, f(1)=1-1+2=2$,
$$
f(2017 \cdot 0+1)=f(2017) f(0)-f(0)-2017+2
$$
$$
\begin{gathered}
f(1)=f(2017)-1-2017+2=2 \\
f(2017)=2018
\end{gathered}
$$
Criteria: Only for the correct answer - 1 point. | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation
$$
x^{2018}+\frac{1}{x^{2018}}=1+x^{2019}
$$ | Answer: $x=1$.
Solution: $x^{2018}+\frac{1}{x^{2018}} \geq 2$ for $x \neq 0$, because $x^{4036}-2 x^{2018}+1=\left(x^{2018}-1\right)^{2} \geq 0 . x^{2019}=x^{2018}+\frac{1}{x^{2018}}-1 \geq 2-1=1 . x \geq 1$. If $\quad x>1, \quad x^{2019}+1>$ $x^{2018}+\frac{1}{x^{2018}}$, because $x^{2019}=x * x^{2018}>x^{2018}$ and ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. When one of two integers was increased 1996 times, and the other was reduced 96 times, their sum did not change. What can their quotient be?
Solution. Let the first number be $a$, and the second $b$. Then the equation $1996 a+\frac{b}{96}=a+b$ must hold, from which we find that $2016 a=b$. Therefore, their quotient... | Answer: 2016 or $\frac{1}{2016}$.
Criteria: Full solution - 14 points; correct answer without solution - 2 points. | 2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A black and white chocolate bar consists of individual pieces forming $n$ horizontal and $m$ vertical rows, arranged in a checkerboard pattern. Yan ate all the black pieces, and Maxim ate all the white pieces. What is $m+n$, if it is known that Yan ate $8 \frac{1}{3} \%$ more pieces than Maxim. | Solution. The number of black and white segments can only differ by 1. Therefore, Yan ate 1 segment more than Maksim. If 1 segment is 8 $\frac{1}{3} \%$, then Maksim ate 12 segments, Yan ate 13 segments, and together they ate 25 segments. This means the chocolate bar was $5 \times 5$.
Answer: 10.
Criteria: 14 points ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. There are books of three colors: white, blue, and green. To make the shelf look beautiful, the boy first arranged the white books, and then placed blue books in each gap between them. Finally, he placed green books in each gap between the standing books. In the end, there were 41 books on the shelf. How many white b... | Answer: 11
9. The sum of two natural numbers is 2017. If you append 9 to the end of the first number and remove the digit 8 from the end of the second number, the numbers will be equal. Find the largest of these numbers.
Answer: 1998 | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Find the sum of the digits of all numbers in the sequence $1,2,3, \ldots, 199,200$.
untranslated text remains the same as requested. | 2. Find the sum of the digits of all numbers in the sequence $1,2,3, \ldots, 199,200$.
Answer: 1902 | 1902 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. A parallelepiped is composed of white and black unit cubes in a checkerboard pattern. It is known that the number of black cubes is $1 \frac{12}{13} \%$ more than the number of white cubes. Find the surface area of the parallelepiped, given that each side of the parallelepiped is greater than 1. | Solution. The number of black and white cubes can differ by only 1. Therefore, 1 cube is $1 \frac{12}{13} \%$ of the quantity of white cubes. Thus, there are 52 white cubes, 53 black cubes, and a total of 205 cubes. That is, our parallelepiped is $3 \times 5 \times 7$. The surface area is 142.
Answer: 142.
Criteria: ... | 142 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. How many three-digit numbers exist where all digits are odd numbers, and all two-digit numbers that can be obtained by erasing one of these digits are not divisible by 5? | 2. How many three-digit numbers exist where all digits are odd numbers, and all two-digit numbers that can be obtained by erasing one of these digits are not divisible by 5?
Answer: 80 | 80 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.2. Let $f(n)$ be equal to the product of the even digits of the natural number $\mathrm{n}$ or be zero if there are no even digits. Find the sum $f(1)+f(2)+\cdots+f(100)$. | Answer: 620
Solution: for a single-digit number $n$, $f(n)$ will be equal to $n$ itself if it is even and 0 if it is odd. For single-digit $\mathrm{n}$, we get the sum $2+4+6+8=20$. If $\mathrm{n}$ is a two-digit number, let's consider the cases:
1) If both digits are even. Then the first digit can be $2,4,6$ or 8, a... | 620 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. In a row, $n$ integers are written such that the sum of any three consecutive numbers is positive, while the sum of any five consecutive numbers is negative. For what largest $n$ is this possible | Answer: 6.
Solution: Let's provide an example for $n=6: 3,-5,3,3,-5,3$. We will prove that for $n \geq 7$ it will not be possible to write down a sequence of numbers that satisfy the condition of the problem. We will construct a table for the first 7 numbers in this sequence
| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. After the mathematics Olympiad, five students noticed that any two of them solved no more than 9 problems in total. What is the maximum number of problems that could have been solved by all the students? | Problem 2. After the mathematics Olympiad, five students noticed that any two of them solved no more than 9 problems in total. What is the maximum number of problems that could have been solved by all the students?
| Solution | Criteria |
| :---: | :---: |
| Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \mathrm... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Find the value of the expression $\sqrt[3]{7+5 \sqrt{2}}-\sqrt[3]{5 \sqrt{2}-7}$ | Answer: 2
Task 2.B The bases $AD$ and $BC$ of an isosceles trapezoid $ABCD$ are $16 \sqrt{3}$ and $8 \sqrt{3}$, respectively, and the acute angle at the base is $30^{\circ}$. What is the length of the lateral side of the trapezoid?
Answer: 8 | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 4. Solve the system of equations $\left\{\begin{array}{l}\sqrt{x}+\sqrt{y}=10 \\ \sqrt[4]{x}+\sqrt[4]{y}=4\end{array}\right.$ and find the value of the product $x y$. | Answer: 81
Problem 5. From the vertex of the right angle $K$ of triangle $MNK$, a perpendicular $KL$ is drawn to the plane of the triangle, equal to 280. Find the distance from point $L$ to the line $MN$, given that the height of the triangle dropped from vertex $K$ is 96.
Answer: 296
Problem 6. The bank allocated a... | 81 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 12. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with side $3 \sqrt{2}$. Find the volume of a regular tetrahedron, one vertex of which coincides with point $A$, and the other three vertices belong to the plane $C M A_{1} N$, where $M$ and $N$ are the midpoints of edges $D D_{1}$ and $B B_{1}$. | Answer: 9
Problem 13. Solve the equations for all natural $n$:
$$
\cos ^{4} x+\sin x(\sin x+1)\left(\cos ^{2} x+\sin x-1\right)=n
$$
In the answer, write the number of roots in the interval $[0,2 \pi]$
Answer: 3
Problem 14. $M$ is the midpoint of the lateral side $A B$ of trapezoid $A B C D$, and $E$ is the inters... | 3986 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. A rectangle is divided into six squares (see figure). What is the side of the larger square if the side of the smaller one is 2.
 | Problem 4. A rectangle is divided into six squares (see figure). What is the side of the larger square if the side of the smaller one is 2.
Answer: 14 | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. On the table, there are 10 stacks of playing cards (the number of cards in the stacks can be different, there should be no empty stacks). The total number of cards on the table is 2015. If a stack has an even number of cards, remove half of the cards. If the number of remaining cards in the stack is still ev... | # Solution
a) Since 2014 is an odd number, in any distribution of cards into 10 piles, there will be at least one pile with an even number of cards. Therefore, the number of cards in this pile, and thus the total number of cards, will definitely decrease by at least one card (if there are 2 cards in this pile). This m... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 6. In an acute-angled triangle $\mathrm{ABC}$, angle $\mathrm{B}$ is $30^{\circ}, \mathrm{BC}=12$. The altitude CD of triangle $\mathrm{ABC}$ and the altitude DE of triangle BDC are drawn. Find BE. | Answer: 9
Task 7.3a 2016 The number of books in the school library fund increased by $0.4 \%$, and in 2017 - by $0.8 \%$, remaining less than 50 thousand. By how many books did the library fund increase in 2017?
Answer: 251
Task 8. A rectangle is divided into six squares (see figure). What is the side of the larger ... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 7. On a circle, there are 25 non-overlapping arcs, and on each of them, two arbitrary prime numbers are written. The sum of the numbers on each arc is not less than the product of the numbers on the arc following it in a clockwise direction. What can the sum of all the numbers be? | Problem 7. On a circle, there are 25 non-intersecting arcs, and on each of them, two arbitrary prime numbers are written. The sum of the numbers on each arc is not less than the product of the numbers on the next arc clockwise. What can the sum of all the numbers be?
| Solution | Criteria |
| :---: | :---: |
| Let the... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In a convex quadrilateral $A B C D: A B=A C=A D=B D$ and $\angle B A C=\angle C B D$. Find $\angle A C D$. | Answer: $70^{\circ}$.
Solution: Triangle $A B D$ is equilateral, so the angles $\angle A B D=\angle B D A=$ $\angle D A B=60^{\circ}$. Let $\angle B A C=\angle C B D=\alpha$, then $\angle A B C=60^{\circ}+\alpha . A B=A C$, thus $\angle A C B=\angle A B C=60^{\circ}+\alpha$. The sum of the angles in triangle $A B C$ i... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation in integers $x^{2}+y^{2}=3 x y$.
---
Note: The translation maintains the original format and line breaks as requested. | Answer: $x=y=0$.
Solution: If both numbers are not equal to 0, divide the numbers $x$ and $y$ by their greatest common divisor, resulting in coprime numbers $a$ and $b$. The right side of the equation is divisible by 3, so the left side must also be. The square of an integer can give a remainder of 0 or 1 when divided... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In how many ways can a pile of 100 stones be divided into heaps so that the number of stones in any two heaps differs by no more than one? | Answer: 99.
Solution: We will prove that for any $k$ from 2 to 100, the pile can be divided into $\mathrm{k}$ such piles in a unique way. $100 = q k + r$, where $q$ and $r$ are the quotient and remainder of 100 when divided by $k$, respectively. Suppose there is a pile with no more than $q-1$ stones, then in the remai... | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 6. Find the value of the expression $\left(\sqrt[3]{x^{2}} \cdot x^{-0.5}\right):\left(\left(\sqrt[6]{x^{2}}\right)^{2} \cdot \sqrt{x}\right)$ at $x=\frac{1}{2}$ | Answer: 2
Problem 7. The diagonals of trapezoid $ABCD (BC \| AD)$ are perpendicular to each other, and $CD = \sqrt{129}$. Find the length of the midline of the trapezoid, given that $BO = \sqrt{13}, CO = 2\sqrt{3}$, where $O$ is the point of intersection of the diagonals of the trapezoid.
Answer: 10
Problem 8. Four ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Find the remainder when the number 20172017... 2017 (2017 written 100 times) is divided by 8. | Answer: 1
8. MN and $\mathrm{PQ}$ are two parallel chords located on opposite sides of the center of a circle with radius $10 . \mathrm{MN}=12, \mathrm{PQ}=16$. Find the distance between the chords.
Answer: 14 | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 5. In how many ways can a pile of 100 stones be divided into piles so that the number of stones in any two piles differs by no more than one? | Task 5. In how many ways can a pile of 100 stones be divided into piles so that the number of stones in any two piles differs by no more than one?
## Solutions and Evaluation Criteria
| Task Number | Solution | Criteria |
| :---: | :---: | :---: |
| 1 | Answer: No. Suppose this is possible. No digit is 0, because the... | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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