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Problem 9. (12 points)
Ivan, a full-time student, started working at his long-desired job on March 1 of the current year. In the first two months, during the probationary period, Ivan's salary was 20000 rubles. After the probationary period, the salary increased to 25000 rubles per month. In December, for exceeding th... | Answer: 32500.
## Comment
Solution: Personal Income Tax from salary $=(20000 \times 2+25000 \times 8+10000) \times 13\% = 32500$ rubles. The scholarship is not subject to Personal Income Tax. | 32500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (8 points)
Lena receives 50,000 rubles per month and spends 45,000 rubles per month. She gets her salary on the 6th of each month. Lena has: a deposit in the bank $\mathrm{X}$ with a $1\%$ monthly interest rate with monthly capitalization and the possibility of topping up, but no withdrawals allowed, with... | Solution:
1) Of all the bank's products, the deposit brings Lena the highest income. She can use $50-45=5$ thousand rubles for savings, depositing them immediately after receiving her salary. Over a year, this will bring Lena $5000 * \frac{1.01 * (1.01^{12}-1)}{1.01-1} - 5000 * 12 \approx 4000$.
2) The remaining 45 th... | 4860 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6. (8 points)
Vasily is planning to graduate from the institute in a year. Only 270 out of 300 third-year students successfully pass their exams and complete their bachelor's degree. If Vasily ends up among the 30 expelled students, he will have to work with a monthly salary of 25,000 rubles. It is also know... | Solution:
Four years after graduating from school, Fedor will earn $25000 + 3000 * 4 = 37000$ rubles (2 points)
The expected salary of Vasily is the expected value of the salary Vasily can earn under all possible scenarios (2 points). It will be 270/300*(1/5*60 000 + 1/10*80 000 + 1/20*25 000 + (1 - 1/5 - 1/10 - 1/20... | 45025 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In the store "Third is Not Excessive," there is a promotion: if a customer presents three items at the cash register, the cheapest of them is free. Ivan wants to buy 11 items costing $100, 200, 300, \ldots, 1100$ rubles. For what minimum amount of money can he buy these items? | Answer: 4800. It is clear that items should be listed in descending order of price, then the cost of the purchase will be $1100+1000+800+700+500+400+200+100=4800$ rubles. | 4800 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. A supermarket discount card that gives a $3 \%$ discount costs 100 rubles. Masha bought 4 cakes for 500 rubles each and fruits for 1600 rubles for her birthday. The seller at the cash register offered her to buy the discount card before the purchase. Should Masha agree?
1) no, they offer these cards to everyone
2) y... | Answer: 2. The cost of Masha's purchase is $4 * 500 + 1600 = 3600$. If Masha buys a discount card, she will spend $100 + 3600 * 0.97 = 3592$. | 3592 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. Vanya decided to give Masha a bouquet of an odd number of flowers for her birthday, consisting of yellow and red tulips, so that the number of flowers of one color differs from the number of flowers of the other by exactly one. Yellow tulips cost 50 rubles each, and red ones cost 31 rubles. What is the largest numbe... | Answer: 15.
A bouquet with one more red tulip than yellow ones is cheaper than a bouquet with the same total number of flowers but one more yellow tulip. Therefore, Vanya should buy a bouquet with one more red tulip. The remaining flowers can be paired into red and yellow tulips, with each pair costing 81 rubles. Let'... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. (8 points)
Irina Mikhailovna opened a foreign currency deposit in the "Western" Bank for an amount of $23,904 for a term of 1 year. The interest rate on the deposit was 5% per annum. Exactly 3 months later, the Bank of Russia revoked the license of the "Western" Bank. The official exchange rate on the date ... | Answer: $b$.
## Comment:
23904 USD $\times 58.15$ RUB $\times(1+5\% / 4)=1407393$ RUB. Since the Deposit Insurance Agency compensates deposits up to 1400000 RUB, this amount will be paid to Irina Mikhailovna. | 1400000 | Algebra | MCQ | Yes | Yes | olympiads | false |
Problem 13. (8 points)
Dar'ya received a New Year's bonus of 60 thousand rubles, which she decided to save for a summer vacation. To prevent the money from losing value, the girl chose between two options for saving the money - to deposit the money at an interest rate of $8.6 \%$ annually for 6 months or to buy dollar... | Answer: the loss incurred from the second option for placing funds is (rounded to the nearest whole number) $\underline{3300 \text{ RUB}}$ | 3300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. (8 points)
Anna Ivanovna bought a car from her neighbor last November for 300,000 rubles with an engine power of 250 hp, and in May she purchased a used rowing catamaran for 6 rubles. The transport tax rate is set at 75 rubles per 1 hp. How much transport tax should Anna Ivanovna pay? (Provide the answer as... | Solution: transport tax $=250 \times 75 \times 2 / 12=3125$ rubles. A rowing catamaran is not a taxable object. | 3125 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. (8 points)
The earned salary of the citizen was 23,000 rubles per month from January to June inclusive, and 25,000 rubles from July to December. In August, the citizen, participating in a poetry competition, won a prize and was awarded an e-book worth 10,000 rubles. What amount of personal income tax needs ... | Solution: Personal Income Tax from salary $=(23000 \times 6+25000 \times 6) \times 13\%=37440$ rubles.
Personal Income Tax from winnings $=(10000-4000) \times 35\%=2100$ rubles.
Total Personal Income Tax = 37440 rubles +2100 rubles$=39540$ rubles. | 39540 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 14. (1 point)
Calculate the amount of personal income tax (NDFL) paid by Sergey for the past year, if he is a Russian resident and during this period had a stable income of 30000 rub./month and a one-time vacation bonus of 20000 rub. In the past year, Sergey sold a car he inherited two years ago for 250000 rub.... | Answer: 10400.
## Comment:
Solution: tax base $=30000 \times 12+20000+250000=630000$ rubles. The amount of the tax deduction $=250000+300000=550000$ rubles. The amount of personal income tax $=13 \% \times(630000-$ $550000)=10400$ rubles. | 10400 | Other | math-word-problem | Yes | Yes | olympiads | false |
Problem 20. (6 points)
Ivan Sergeyevich decided to raise quails. In a year, he sold 100 kg of poultry meat at a price of 500 rubles per kg, and also 20000 eggs at a price of 50 rubles per dozen. The expenses for the year amounted to 100000 rubles. What profit did Ivan Sergeyevich receive for this year? (Provide the an... | Answer: 50000.
Comment:
Solution: revenue $=100 \times 500 + 50 \times 20000 / 10 = 150000$ rubles. Profit $=$ revenue costs $=150000-100000=50000$ rubles. | 50000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 21. (8 points)
Dmitry's parents decided to buy him a laptop for his birthday. They calculated that they could save the required amount in two ways. In the first case, they need to save one-tenth of their salary for six months. In the second case, they need to save half of their salary for one month, and then d... | Answer: 25000.
## Comment:
Solution:
Mom's salary is $x$, then dad's salary is $1.3x$. We set up the equation:
$(x + 1.3x) / 10 \times 6 = (x + 1.3x) / 2 \times (1 + 0.03 \times 10) - 2875$
$1.38x = 1.495x - 2875$
$x = 25000$ | 25000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. When insuring property, the insurance amount cannot exceed its actual value (insurance value) at the time of concluding the insurance contract.
Insurance tariff - the rate of the insurance premium or the insurance premium (insurance premium) expressed in rubles, payable per unit of the insurance amount, which is us... | # Solution:
In accordance with the instruction, the base rate is $0.2\%$ of the insurance amount, apply a reducing factor for the absence of a change in ownership over the past 3 years $(0.8)$ and an increasing factor for the absence of certificates from the PND and ND $(1.3)$.
In total: $0.2 * 0.8 * 1.3=0.208\%$
Th... | 31200 | Other | math-word-problem | Yes | Yes | olympiads | false |
3. In July, Volodya and Dima decided to start their own business producing non-carbonated bottled mineral water called "Dream," investing 1,500,000 rubles, and used these funds to purchase equipment for 500,000 rubles. The technical passport for this equipment indicates that the maximum production capacity is 100,000 b... | # Answer:
a) The norm for 1 bottle of water = initial cost / maximum quantity: $500000 / 100000 = 5$ rubles;
- depreciation in July $5 \times 200 = 1000$ rubles;
- depreciation in March $15000 \times 5 = 75000$ rubles;
- depreciation in September $12300 \times 5 = 61500$ rubles, Total depreciation 137500 (6 points).
... | 372500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Newlyweds Alexander and Natalia successfully got jobs at an advertising company in April. With their earnings, they want to buy new phones next month: phone "A" for Alexander, which costs 57,000 rubles, and phone "B" for Natalia, costing 37,000 rubles. Will they be able to do this, given the following data?
- Alexa... | Answer:
a) Total expenses: $17000+15000+12000+20000+30000+30000=$ $=124000$ (4 points);
b) Net income: $(125000+61000) \times 13 \% = 24180.186000 - 24180=$ $=161820$ (6 points);
c) Remaining funds: $161820-124000=37820$. The phone can only be bought for Natalia, and to buy a phone for Alexander, it is necessary to ... | 37820 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let's say you have two bank cards for making purchases: a debit card and a credit card. Today, you decided to buy airline tickets for 20,000 rubles. If you pay for the purchase with a credit card (the credit limit allows it), you will have to return the money to the bank in $\mathrm{N}$ days to avoid going beyond th... | # Solution:
When paying by credit card, the amount of 20,000 rubles will be on your debit card for $\mathrm{N}$ days, which will earn you $\frac{6 \mathrm{~N}}{100 \cdot 12 \cdot 30} \cdot 20000$ rubles in interest on the remaining funds.
You will also receive $20000 \times 0.005 = 100$ rubles in cashback.
When payi... | 31 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Two friends, Arthur and Timur, with the support of their parents, decided to open several massage salons in Moscow. For this purpose, a business plan was drawn up, the economic indicators of which are presented below.
- Form of ownership - LLC
- Number of employees - no more than 50 people
- Planned revenues for th... | # Solution:
Reference information: Chapter 26.2, Part 2 of the Tax Code of the Russian Federation.
- Criteria applicable under the simplified tax system (STS) - Article 346.13;
- Taxable object - Article 346.14;
- Determination of income - Article 346.15;
- Determination of expenses - Article 346.16;
- Recognition of... | 222000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let's say you have two bank cards for making purchases: a debit card and a credit card. Today, at the beginning of the month, you decided to buy airline tickets for 12,000 rubles. If you pay for the purchase with a credit card (the credit limit allows it), you will have to return the money to the bank in $\mathrm{N}... | # Solution:
When paying by credit card, the amount of 12,000 rubles will be on your debit card for $\mathrm{N}$ days, which will earn you $\frac{6 \mathrm{~N}}{100 \cdot 12 \cdot 30} \cdot 12000$ rubles in interest on the remaining funds. Additionally, you will receive $12000 \times 0.01 = 120$ rubles in cashback.
Wh... | 59 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Twin brothers, Anton Sergeyevich, a civil aviation pilot by profession, and Mikhail Sergeyevich, a neurologist, born on 05.06.1977, decided to go on vacation together and purchase a life and health insurance policy for 2,000,000 rubles. Anton Sergeyevich and Mikhail Sergeyevich had the same height - 187 cm and weigh... | # Solution:
Let's calculate the age. Both men are 40 years old. The base rate is $0.32\%$. By occupation class: Mikhail (doctor) $0.32\% \times 1.02 = 0.3264\%$.
Anton (pilot) $0.32\% \times 1.5 = 0.48\%$.
Calculate the BMI $= 98 / 1.87^2 = 98 / 3.4969 = $ (approximately) 28.025.
Find the increasing coefficient fro... | 3072 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Anna and Ekaterina have opened a cosmetic salon in New Moscow. The enterprise applies the general taxation system. Ekaterina attended a seminar on taxation and learned about the Simplified System of Taxation (USNO). To avoid changing the document flow and control over financial and economic operations, the friends d... | # Solution:
Reference information: Chapter 26.2, Chapter 25, Part 2 of the Tax Code of the Russian Federation.
- Criteria applicable under the simplified tax system (STS) - Article 346.13;
- Taxable object - Article 346.14;
- Determination of income - Article 346.15;
- Determination of expenses - Article 346.16;
- Re... | 172800 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. (8 points)
The earned salary of the citizen was 23,000 rubles per month from January to June inclusive, and 25,000 rubles from July to December. In August, the citizen, participating in a poetry contest, won a prize and was awarded an e-book worth 10,000 rubles. What amount of personal income tax needs to b... | Solution: Personal Income Tax from salary $=(23000 \times 6+25000 \times 6) \times 13\%=37440$ rubles.
Personal Income Tax from winnings $=(10000-4000) \times 35\%=2100$ rubles.
Total Personal Income Tax = 37440 rubles +2100 rubles$=39540$ rubles. | 39540 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 4.
Every day after lunch, 7 half-eaten pieces of white bread are left on the desks of the second-grade students. If these pieces are put together, they make up half a loaf of bread. How many loaves of bread will the second-grade students save in 20 days if they do not leave these pieces? How much money will the s... | Solution: 1. 0.5 (1/2) * 20 = 10 (loaves); 10 * 35 = 350 rubles; 2. 0.5 (1/2) * 60 = 30 (loaves); 30 * 35 = 1050 rubles.
Themes for extracurricular activities: "Young Economist," "Bread is the Head of Everything," "Saving and Frugality in Our School Canteen," "Journey to the School of the Frugal," "Frugality - the Mai... | 350 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Once, a god sent a little cheese to two ravens. The first raven received 100 g, from which part was taken by a fox. The piece of the second raven turned out to be twice as large as that of the first, but she managed to eat only half as much as the first raven. The portion of cheese that the fox got from the second r... | Solution. Let the first crow eat $x$ grams of cheese. Then the fox received $100-x$ grams of cheese from the first crow. The second crow ate $\frac{x}{2}$ grams of cheese. From the second crow, the fox received $200-\frac{x}{2}$ grams of cheese. This was three times more, so: $200-\frac{x}{2}=3(100-x)$. Solution: $x=40... | 240 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. 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. Since $A$ with $C$ and $B$ with $D$ meet every 7 minutes, their approach speeds are equal: $V_{A}+V_{C}=V_{B}+V_{D}$. In other words, the speeds of separation of $A, B$ and $C, D$ are equal: $V_{A}-V_{B}=V_{D}-V_{C}$. Therefore, since $A$ and $B$ meet for the first time at the 53rd minute, $C$ and $D$ will al... | 53 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. 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 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Marina needs to buy a notebook, a pen, a ruler, and a pencil to participate in the Olympiad. If she buys a notebook, a pencil, and a ruler, she will spend 47 tugriks. If she buys a notebook, a ruler, and a pen, she will spend 58 tugriks. If she buys only a pen and a pencil, she will spend 15 tugriks. How much money ... | Solution. If Marina buys all three sets from the condition at once, she will spend $47+58+$ $15=120$ tugriks, and she will buy each item twice, so the full set of school supplies costs $120 / 2=60$ tugriks.
Criteria. Only the answer without explanation - 1 point. If in the solution they try to determine the cost of th... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. 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 |
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). The result was the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible ... | Solution. The number must be divisible by 25, so “$\lambda$A” equals 25, 50, or 75 (00 cannot be, as the letters are different). If “LA” equals 50, then for the other letters (“G”, “V”, “T”, “E”, “M”) there are $A_{8}^{5}$ options; otherwise, for the other letters there are $7 \cdot A_{7}^{4}$ options. In total, $8!/ 6... | 18480 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. 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. Since $A$ with $C$ and $B$ with $D$ meet every 7 minutes, their approach speeds are equal: $V_{A}+V_{C}=V_{B}+V_{D}$. In other words, the speeds of separation of $A, B$ and $C, D$ are equal: $V_{A}-V_{B}=V_{D}-V_{C}$. Therefore, since $A$ and $B$ meet for the first time at the 53rd minute, $C$ and $D$ will al... | 53 | Other | math-word-problem | Yes | Yes | olympiads | false |
5. A few years ago, in the computer game "Minecraft," there were 11 different pictures (see the figure): one horizontal with dimensions $2 \times 1$, and two each with dimensions $1 \times 1$, $1 \times 2$ (vertical), $2 \times 2$, $4 \times 3$ (horizontal), and $4 \times 4$. In how many ways can all 11 pictures be pla... | Answer: 16896.
Solution. We will say that two pictures are in different columns if no block of the first picture is in the same column as any block of the second. It is clear that the $4 \times 4$ pictures are in different columns from each other and from the $4 \times 3$ pictures in any arrangement. Thus, the $4 \tim... | 16896 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. In the Thrice-Tenth Kingdom, there are 17 islands, each inhabited by 119 people. The inhabitants of the kingdom are divided into two castes: knights, who always tell the truth, and liars, who always lie. During the census, each person was first asked: "Excluding you, are there an equal number of knights and liars on... | Answer: 1013.
## Solution.
1) Consider the first question. A "yes" answer would be given by either a knight on an island with exactly 60 knights, or a liar if the number of knights is different. A "no" answer would be given by either a liar on an island with 59 knights, or a knight if the number of knights is differe... | 1013 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Find all real solutions to the system of equations
$$
\left\{\begin{array}{l}
\sqrt{x-997}+\sqrt{y-932}+\sqrt{z-796}=100 \\
\sqrt{x-1237}+\sqrt{y-1121}+\sqrt{3045-z}=90 \\
\sqrt{x-1621}+\sqrt{2805-y}+\sqrt{z-997}=80 \\
\sqrt{2102-x}+\sqrt{y-1237}+\sqrt{z-932}=70
\end{array}\right.
$$
(L. S. Korechkova, A. A. Tessl... | Answer: $x=y=z=2021$.
Solution. First, we prove that the solution is unique if it exists. Let $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ be two different solutions and, without loss of generality, $x_{1} \leqslant x_{2}$. Then there are four possible cases: $y_{1} \leqslant y_{2}$ and $z... | 2021 | 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. This value is achieved if the unshaded cells are in opposite corners and contain the numbers 1 and 19.
Evaluation.
1) The cells that contain the minimum and maximum numbers are definitely not shaded. This means that the minimum and maximum each appear exactly once, and they are in the unshaded c... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. How many five-digit numbers are divisible by their last digit?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 2. The total number of five-digit numbers is $-99999-9999=90000$, and among them, there are an equal number of numbers ending in $0,1, \ldots, 9$, that is, 9000 numbers of each type.
Let $n_{i}$, where $i=0,1, \ldots, 9$, be the number of numbers ending in $i$ that are divisible by $i$. Then
$n_{0}=0$ (a number canno... | 42036 | Number Theory | proof | Yes | Yes | olympiads | false |
1. In one move, you can either add one of its digits to the number or subtract one of its digits from the number (for example, from the number 142 you can get $142+2=144, 142-4=138$ and several other numbers).
a) Can you get the number 2021 from the number 2020 in several moves?
b) Can you get the number 2021 from th... | Solution. a) Yes, for example, like this: $20 \mathbf{2 0} \rightarrow 20 \mathbf{1 8} \rightarrow \mathbf{2 0 1 9} \rightarrow 2021$.
b) Yes. For example, by adding the first digit (one), we can reach the number 2000; by adding the first digit (two), we can reach 2020; then see part a.
Criteria. Part a) 3 points, b)... | 2021 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In a rectangular grid 303 cells long and 202 cells wide, two diagonals were drawn and all cells through which they passed were painted. How many cells were painted?
(O. A. Pyayve, A. A. Tseler)
Notice that each diagonal intersects 101 such rectangles (passing through their vertices), and in each of them, it passes through 4 cells. Thus, the two diagonals, it seems, pass thro... | 806 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. 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 3 times fewer than tulips?
(A. A. Tesler) | Solution. Let the number of narcissus be $x$, then the number of roses is $4x$, and the number of tulips is $12x$, so the total number of flowers is $17x$. The number of flowers is the product of the number of boys and the number of girls. Since 17 is a prime number, one of these quantities must be divisible by 17, mea... | 44 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Once, a god sent a little cheese to two ravens. The first raven received 100 g, from which a part was taken by a fox. The piece of the second raven turned out to be twice as large as that of the first, but she managed to eat only half as much as the first raven. The portion of cheese that the fox got from the second... | Solution. Let the first crow eat $x$ grams of cheese. Then the fox got $100-x$ grams of cheese from the first crow. The second crow ate ${ }_{2}^{x}$ grams of cheese. From the second crow, the fox received $200-\frac{x}{2}$ grams of cheese. This was three times more, so: $200-\frac{x}{2}=3(100-x)$. Solution: $x=40$. Th... | 240 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. 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. Since $A$ with $C$ and $B$ with $D$ meet every 7 minutes, their approach speeds are equal: $V_{A}+V_{C}=V_{B}+V_{D}$. In other words, the speeds of separation of $A, B$ and $C, D$ are equal: $V_{A}-V_{B}=V_{D}-V_{C}$. Therefore, since $A$ and $B$ meet for the first time at the 53rd minute, $C$ and $D$ will al... | 53 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. 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). The result was the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible ... | Solution. The letter A must equal 0. The remaining 6 letters are non-zero digits with a sum that is a multiple of 3. Note that each remainder when divided by 3 appears three times. By enumeration, we find all sets of remainders whose sum is a multiple of three: 000111, 000222, 111222, 001122. We count the 6-element sub... | 21600 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. How many ways are there to cut a $10 \times 10$ square into several rectangles along the grid lines such that the sum of their perimeters is 398? Ways that can be matched by rotation or flipping are considered different. | Solution: 180 ways.
If the entire square is cut into 100 unit squares, the sum of the perimeters will be $4 \times 100=400$. Therefore, we need to reduce this sum by 2, which is achieved by keeping one internal partition intact (in other words, the square is cut into 98 squares and 1 domino). There are a total of 180 ... | 180 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A rectangle $11 \times 12$ is cut into several strips $1 \times 6$ and $1 \times 7$. What is the minimum total number of strips? | Solution. Answer: 20. The example is shown in the figure.
Evaluation: we will paint every seventh diagonal so that 20 cells are shaded (see figure). Each strip contains no more than one cell, so there are no fewer than 20 strips.
. The result was the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible ... | Solution. For a number to be divisible by 8, "АЛА" must be divisible by 8, with "А" -
the expression in parentheses is clearly divisible by 8, so it is sufficient to require that ("А" + 2 * "... | 67200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. 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 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Here is a problem from S. A. Rachinsky's problem book (late 19th century): "How many boards 6 arshins long and 6 vershoks wide are needed to cover the floor of a square room with a side of 12 arshins?" The answer to the problem is: 64 boards. Determine from these data how many vershoks are in an arshin. | Solution. The area of the room is $12 \cdot 12=144$ square arshins. Therefore, the area of each board is $144 / 64=2.25$ square arshins. Since the length of the board is 6 arshins, its width should be $2.25 / 6=3 / 8=6 / 16$ arshins. Thus, 6 vershoks make up $6 / 16$ arshins, meaning 1 vershok is $1 / 16$ arshin.
Anot... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. On a glade, two firs, each 30 meters tall, grow 20 meters apart from each other. The branches of the firs grow very densely, and among them are some that are directed straight towards each other, and the length of each branch is half the distance from it to the top. A spider can crawl up or down the trunk (strictly ... | Solution. From the diagram, it can be seen that the branches of the firs intersect at a height of no more than 10 meters from the ground. Indeed, at this height, the distance to the treetop is 20 meters, so the length of each branch is $20 / 2 = 10$ meters, and the total length of the branches of the two firs is equal ... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Nikita has a magic jar. If you put $n$ candies in the jar and close it for an hour, the number of candies inside will increase by the sum of the digits of $n$. For example, if there were 137 candies, it would become $137+1+3+7=148$. What is the maximum number of candies Nikita can get in 20 hours and 16 minutes, if ... | Solution. We need to strive to have as many candies as possible at the end of each hour. However, this does not mean that we should always put all the candies in the jar. The greatest sum of digits (i.e., the greatest increase in the number of candies) is achieved with a number where all digits (except the first) are n... | 267 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. How many numbers from 1 to 999 without the digit "0" are written in the Roman numeral system exactly one symbol longer than 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 tabl... | Solution. Note that, regardless of the digit place and other digits of the number, a decimal digit $a$ is written as:
- 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 $a=8$.
Thus, in suitable numbers, only the... | 68 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In a spring math camp, between 50 and 70 children arrived. On Pi Day (March 14), they decided to give each other squares if they were just acquaintances, and circles if they were friends. Andrey calculated that each boy received 3 circles and 8 squares, while each girl received 2 squares and 9 circles. And Katya not... | Answer: 60.
Solution. Let the number of boys be $m$, and the number of girls be $-d$. Then $3 m + 9 d = 8 m + 2 d$ (the number of circles equals the number of squares). Transforming, we get $5 m = 7 d$, which means the number of boys and girls are in the ratio $7: 5$. Therefore, the total number of children is divisib... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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. 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 |
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 |
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