Question

Что больше: 22017 + 32017 или 52017? На доске 8×8 в 33 клетках есть мины. За ход можно узнать количество мин в любом прямоугольнике, каком захочешь. Как за 5 вопросов найти пару соседних клеток с минами? В стране есть 100 городов и совсем нет железных дорог. Каждый год в стране строится отрезок железной дороги между двумя какими-то городами (не проходящий через другие города). Докажите, что через 98 лет всё ещё найдутся два города, не связанные друг с другом железными дорогами. а) Сколько раз в сутки часовая и минутная стрелки совпадают? б) Сколько раз в сутки часовая, минутная и секундная стрелки совпадают? (Считайте, что полночь бывает один раз в сутки.) Докажите, что число 12k −25 — составное при любом целом k>1. На столе лежат несколько отрезков, которые образуют четыре квадрата разных (то есть попарно различных) размеров. Докажите, что можно переложить их так, чтобы получилось шесть квадратов разных размеров.

Answer 1

1) больше 22.017+32.017>52.017 потому что 22.017+32.017=54.034

2) 100-98=2(г) ответ: через 98 лет еще найдутся 2 города, не связанные друг с другом железными дорогами

3) а) сколько раз в сутки часовая и минутная стрелки совпадают?
простите что не все ответы, я просто не в 5 классе учусь, а в 4 классе

Answer 2

Comparing Numbers: 22017, 32017, and 52017

To determine which number is greater between 22017, 32017, and 52017, we can simply compare the values.

- 22017 is less than both 32017 and 52017. - 32017 is greater than 22017 but less than 52017. - 52017 is greater than both 22017 and 32017.

Therefore, 52017 is the greatest number among the three. [[1]]

Finding Mines on a 8x8 Board

On an 8x8 board with 33 cells containing mines, you want to find a pair of neighboring cells that contain mines. Here's a strategy to find such a pair in 5 questions:

1. Start by asking about the top-left cell (1,1). If it contains a mine, move to step 2. Otherwise, move to step 3. 2. Ask about the cell to the right (1,2). If it contains a mine, you have found a pair of neighboring cells with mines. If not, move to step 3. 3. Ask about the cell below (2,1). If it contains a mine, you have found a pair of neighboring cells with mines. If not, move to step 4. 4. Ask about the cell to the right and below (2,2). If it contains a mine, you have found a pair of neighboring cells with mines. If not, move to step 5. 5. Ask about the cell to the right and below (3,3). If it contains a mine, you have found a pair of neighboring cells with mines. If not, there are no neighboring cells with mines.

By following this strategy, you can find a pair of neighboring cells with mines in 5 questions or less. [[2]]

Proof of the Existence of Two Cities Unconnected by Railroads

In a country with 100 cities and no existing railroads, a new segment of railroad is built each year between two randomly chosen cities, without passing through any other cities. We need to prove that after 98 years, there will still be two cities unconnected by railroads.

To prove this, we can use the concept of a complete graph. A complete graph is a graph in which every pair of distinct vertices is connected by an edge. In this case, the complete graph represents all possible connections between the cities.

Initially, with no railroads, the complete graph has 100 vertices and no edges. Each year, a new edge (railroad segment) is added between two cities, which means one edge is added to the complete graph. After 98 years, there will be 98 edges in the complete graph.

According to the formula for the number of edges in a complete graph, the number of edges is given by n(n-1)/2, where n is the number of vertices. Plugging in n = 100, we get 100(100-1)/2 = 4950 edges.

Since after 98 years, there will be only 98 edges, there will still be 4950 - 98 = 4852 unconnected pairs of cities. Therefore, after 98 years, there will still be at least two cities unconnected by railroads. [[3]]

Coinciding Clock Hands

a) The hour and minute hands of a clock coincide 22 times in a 24-hour period. This happens when the minute hand is at the 12 o'clock position and the hour hand is at one of the positions where it coincides with the minute hand. Therefore, the hour and minute hands coincide 22 times in a day. [[4]]

b) To find the number of times the hour, minute, and second hands coincide in a day, we need to consider the relative speeds of the hands. The minute hand moves 12 times faster than the hour hand, and the second hand moves 60 times faster than the minute hand.

The three hands coincide when the hour, minute, and second hands align perfectly. This happens once every 12 hours when the minute and hour hands coincide, and once every 720 minutes when the second hand aligns with the minute and hour hands.

Therefore, the hour, minute, and second hands coincide 22 + 2 = 24 times in a day. [[5]]

Proof that 12k - 25 is Composite for any Integer k > 1

To prove that the number 12k - 25 is composite for any integer k > 1, we can use algebraic manipulation.

Let's assume k > 1 and consider the expression 12k - 25. We can factor out a common factor of 12 from the expression:

12k - 25 = 12(k - 2) + 24 - 25 = 12(k - 2) - 1

Now, we can rewrite the expression as:

12(k - 2) - 1 = 3(4(k - 2)) - 1

Since k > 1, the term (k - 2) is greater than or equal to 0. Therefore, 4(k - 2) is greater than or equal to 0. Let's denote 4(k - 2) as m.

Now, we have:

12(k - 2) - 1 = 3m - 1

Since m is a non-negative integer, 3m is always divisible by 3. Therefore, 3m - 1 is not divisible by 3.

Since 3m - 1 is not divisible by 3, it cannot be a prime number. Therefore, the expression 12k - 25 is composite for any integer k > 1. [[6]]

Rearranging Squares on a Table

To prove that it is possible to rearrange the squares on a table to form six squares of different sizes, we can use the pigeonhole principle.

Let's assume that it is not possible to rearrange the squares to form six squares of different sizes. This means that no matter how we arrange the squares, there will always be at least two squares of the same size.

Now, let's consider the possible sizes of the squares. Since there are four squares, there are four possible sizes: A, B, C, and D.

According to the pigeonhole principle, if we have more pigeons (squares) than pigeonholes (possible sizes), there must be at least two pigeons (squares) in the same pigeonhole (size).

In this case, we have four squares and four possible sizes. Therefore, there must be at least two squares of the same size.

This contradicts our assumption that it is not possible to rearrange the squares to form six squares of different sizes. Therefore, it must be possible to rearrange the squares on the table to form six squares of different sizes. [[7]]

I hope this helps! Let me know if you have any further questions.