Prove that for any number N there is String of N consecutive numbers none of which is prime.
To prove that for any number N there exists a string of N consecutive numbers, none of which is prime, we can use the concept of prime numbers and divisibility.
First, let's define a prime number. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. In other words, it cannot be divided evenly by any number other than 1 and itself.
To tackle this problem, we can construct a string of N consecutive numbers and show that at least one of them is not prime.
Let's consider the string of numbers starting from (N+1)! + 2 up to (N+1)! + (N+1). Here, (N+1)! represents the factorial of N+1, which is the product of all positive integers from 1 to N+1.
Now, we will explain how to show that none of the numbers in this string are prime:
1. Divisibility by 2: The number (N+1)! is divisible by 2 because it is the product of all positive integers up to N+1, and as a result, it is even. Adding any number to an even number will still result in an even number. Therefore, (N+1)! + 2 is divisible by 2.
2. Divisibility by 3: Similarly, (N+1)! is divisible by 3 because it contains the factor 3. Again, adding any number to a number divisible by 3 will also be divisible by 3.
By following this pattern, for each prime number less than or equal to N+1, (N+1)! + p will be divisible by p, where p is the prime number. Therefore, at least one number in the string (N+1)! + 2, (N+1)! + 3, ..., (N+1)! + (N+1) will be divisible by a prime number, making none of them prime.
Hence, we have proven that for any number N, there exists a string of N consecutive numbers, none of which is prime, by constructing the string (N+1)! + 2, (N+1)! + 3, ..., (N+1)! + (N+1).