Question

Сколькими способами можно покрасить все натуральные числа от 1 1 до 100 100 в красный и синий цвета так, чтобы никакая сумма двух различных одноцветных чисел не равнялась степени двойки?

Answer 1

Ответ:

Пошаговое объяснение:

3

Answer 2

Number of Ways to Paint Natural Numbers from 1 to 100

To determine the number of ways to paint all natural numbers from 1 to 100 in red and blue such that no sum of two different numbers of the same color equals a power of two, we can analyze the problem step by step.

Let's consider the possible sums of two different numbers of the same color:

1. Red Numbers: If we have two red numbers, their sum can be any number from 3 to 200. Since we want to avoid sums that are powers of two, we need to exclude all powers of two from this range. The powers of two between 3 and 200 are 4, 8, 16, 32, 64, 128. So, we need to exclude these six numbers from the range. Therefore, the possible sums of two different red numbers are from 3 to 200 excluding 4, 8, 16, 32, 64, and 128.

2. Blue Numbers: Similarly, if we have two blue numbers, their sum can also be any number from 3 to 200 excluding the powers of two mentioned above.

Now, let's calculate the number of ways to paint the numbers from 1 to 100.

1. We have two choices for the first number: red or blue. 2. For each subsequent number, we have two choices: it can be the same color as the previous number or a different color. 3. We need to ensure that the sum of two different numbers of the same color does not equal a power of two.

Using these rules, we can calculate the number of ways to paint the numbers from 1 to 100.

Calculation

To calculate the number of ways, we can use a recursive approach or dynamic programming. Let's use dynamic programming for simplicity.

We can define a 2D array `dp`, where `dp[i][j]` represents the number of ways to paint the numbers from 1 to `i` such that the last number is `j` (0 for red, 1 for blue).

The base case is `dp[1][0] = 1` and `dp[1][1] = 1` since there is only one number, and we can paint it either red or blue.

For each subsequent number `i` (2 to 100), we can calculate `dp[i][0]` and `dp[i][1]` as follows:

- `dp[i][0]` is the sum of `dp[i-1][1]` and all `dp[i-1][0]` values excluding the sums that equal a power of two. - `dp[i][1]` is the sum of `dp[i-1][0]`` and all `dp[i-1]]` values excluding the sums that equal a power of two.

Finally, the total number of ways to paint the numbers from 1 to 100 is `dp[100][0] + dp[100]]`.

Implementation

Here's a Python implementation of the dynamic programming approach:

dp = [[0, 0] for _ in range(101)]
dp[1][0] = 1
dp[1][1] = 1

for i in range(2, 101):
    for j in range(2):
        for k in range(2):
            if (j + k) not in [4, 8, 16, 32, 64, 128]:
                dp[i][j] += dp[i-1][k]

total_ways = dp[100][0] + dp[100][1]

The `total_ways` variable will contain the number of ways to paint the numbers from 1 to 100.

Answer

The number of ways to paint all natural numbers from 1 to 100 in red and blue such that no sum of two different numbers of the same color equals a power of two is {{total_ways}}.

Please note that the calculation may take some time, so it's recommended to use a computer program to obtain the exact result.