Question

5. Баба Яга в своей избушке на курьих ножках завела сказочных животных. Все они, кроме двух - Говорящие Коты; все, кроме двух – Мудрые Совы; остальные – Усатые Тараканы. Сколько обитателей в избушке у Бабы Яги?13. Дима провел социальный опрос и выяснил про жителей своего подъезда, что 25 из них играют в шахматы, 30 были в Архангельске, 28 летали на самолете. Среди летавших на самолете 18 играют в шахматы и 17 были в Архангельске. 16 жителей играют в шахматы и были в Архангельске, притом среди них 15 еще и летали на самолете. Сколько человек живет в подъезде? 19. Сколько существует целых чисел от 1 до 33000, которые не делятся ни на 3, ни на 5, но делятся на 11?

Answer 1

У бабы Яги 14 животных

Answer 2

Question 5: Баба Яга в своей избушке на курьих ножках завела сказочных животных. Все они, кроме двух - Говорящие Коты; все, кроме двух – Мудрые Совы; остальные – Усатые Тараканы. Сколько обитателей в избушке у Бабы Яги?

To determine the number of inhabitants in Baba Yaga's hut, we need to analyze the given information.

According to the statement, all the animals in the hut can be categorized into three groups: Говорящие Коты (Talking Cats), Мудрые Совы (Wise Owls), and Усатые Тараканы (Mustached Cockroaches).

Let's denote the number of inhabitants in each group as follows: - Number of Говорящие Коты: x - Number of Мудрые Совы: y - Number of Усатые Тараканы: z

From the given information, we can deduce the following equations: 1. All the animals in the hut, except for two, are Говорящие Коты. This can be represented as: x + 2 = y + z. 2. All the animals in the hut, except for two, are Мудрые Совы. This can be represented as: y + 2 = x + z. 3. The remaining animals in the hut are Усатые Тараканы. This can be represented as: z = x + y.

To solve these equations, we can substitute the values of x and y from equations 1 and 2 into equation 3:

(x + 2) = (y + z) - 2 (x + 2) = (x + y) - 2 2 = y - x

Substituting this value back into equation 1: x + 2 = (y + z) x + 2 = (x + y) + z 2 = z

Now we have the values of x = -2, y = 0, and z = 2. However, since we are dealing with the number of inhabitants, negative values are not possible. Therefore, we can conclude that there are 0 inhabitants in Baba Yaga's hut.

Question 13: Дима провел социальный опрос и выяснил про жителей своего подъезда, что 25 из них играют в шахматы, 30 были в Архангельске, 28 летали на самолете. Среди летавших на самолете 18 играют в шахматы и 17 были в Архангельске. 16 жителей играют в шахматы и были в Архангельске, притом среди них 15 еще и летали на самолете. Сколько человек живет в подъезде?

To determine the number of people living in the building, we need to analyze the given information.

Let's denote the number of people who play chess as A, the number of people who have been to Arkhangelsk as B, and the number of people who have flown on an airplane as C.

From the given information, we can deduce the following equations: 1. 25 people play chess: A = 25. 2. 30 people have been to Arkhangelsk: B = 30. 3. 28 people have flown on an airplane: C = 28. 4. 18 people who have flown on an airplane also play chess: A ∩ C = 18. 5. 17 people who have been to Arkhangelsk have also flown on an airplane: B ∩ C = 17. 6. 16 people play chess and have been to Arkhangelsk: A ∩ B = 16. 7. 15 people who play chess and have been to Arkhangelsk have also flown on an airplane: A ∩ B ∩ C = 15.

To solve these equations, we can use the principle of inclusion-exclusion.

The principle of inclusion-exclusion states that the number of elements in the union of multiple sets can be calculated by summing the sizes of the individual sets, subtracting the sizes of their intersections, adding back the sizes of the intersections of three sets, and so on.

Using this principle, we can calculate the number of people living in the building as follows:

A ∪ B ∪ C = A + B + C - (A ∩ B) - (A ∩ C) - (B ∩ C) + (A ∩ B ∩ C)

Substituting the given values into the equation:

A ∪ B ∪ C = 25 + 30 + 28 - 16 - 18 - 17 + 15 A ∪ B ∪ C = 67

Therefore, there are 67 people living in the building.

Question 19: Сколько существует целых чисел от 1 до 33000, которые не делятся ни на 3, ни на 5, но делятся на 11?

To find the number of integers between 1 and 33000 that are divisible by 11 but not divisible by 3 or 5, we can use the principle of inclusion-exclusion.

Let's denote the set of integers divisible by 11 as A, the set of integers divisible by 3 as B, and the set of integers divisible by 5 as C.

To find the number of integers that satisfy the given conditions, we need to calculate the size of the set A ∩ (B' ∩ C'), where B' represents the complement of B (integers not divisible by 3) and C' represents the complement of C (integers not divisible by 5).

Using the principle of inclusion-exclusion, we can calculate the size of the set A ∩ (B' ∩ C') as follows:

|A ∩ (B' ∩ C')| = |A| - |A ∩ B| - |A ∩ C| + |A ∩ B ∩ C|

Substituting the given values into the equation:

|A ∩ (B' ∩ C')| = |A| - |A ∩ B| - |A ∩ C| + |A ∩ B ∩ C| |A ∩ (B' ∩ C')| = floor(33000/11) - floor(33000/33) - floor(33000/55) + floor(33000/165) |A ∩ (B' ∩ C')| = 3000 - 1000 - 600 + 200 |A ∩ (B' ∩ C')| = 1600

Therefore, there are 1600 integers between 1 and 33000 that are divisible by 11 but not divisible by 3 or 5.