Question

Four consecutive positive prime numbers have a sum that is divisible by three. What is the smallest possible value of this sum?

Four consecutive positive prime numbers have a sum that is divisible by three. What is the smallest possible value of this sum?

Answer 1

Step-by-step explanation:

The smallest possible value of the sum of four consecutive positive prime numbers that is divisible by three is 18.

To find the solution, we need to understand a few key points:

1. Prime numbers are positive integers greater than 1 that have no divisors other than 1 and themselves. Examples of prime numbers are 2, 3, 5, 7, 11, and so on.

2. Consecutive numbers are numbers that follow each other in order without any gaps. For example, 1, 2, 3, 4, and 5 are consecutive numbers.

3. Divisibility by three means that a number can be divided evenly by 3 without leaving a remainder. For example, 9 is divisible by 3 because 9 ÷ 3 = 3 with no remainder.

Now, let's find the four consecutive prime numbers whose sum is divisible by 3:

Starting with the first prime number, which is 2, we need to find three more consecutive prime numbers. The next prime number after 2 is 3. So far, the sum of the two prime numbers is 2 + 3 = 5.

To find the next prime number, we need to check if 5 + 1 = 6 is a prime number. However, 6 is not a prime number because it is divisible by 2 and 3. Therefore, we need to find the next prime number after 5, which is 7.

Now, let's add the prime number 7 to our sum: 5 + 7 = 12.

To find the last prime number, we need to check if 12 + 1 = 13 is a prime number. Luckily, 13 is indeed a prime number.

Finally, let's add 13 to our sum: 12 + 13 = 25.

The sum of the four consecutive prime numbers, 2, 3, 7, and 13, is 25, which is not divisible by 3.

Since our initial attempt did not yield a sum divisible by 3, let's try again with a different set of consecutive prime numbers.

Starting with the first prime number, which is 2, the next prime number after 2 is 3. So far, the sum of the two prime numbers is 2 + 3 = 5.

To find the next prime number, we need to check if 5 + 1 = 6 is a prime number. However, 6 is not a prime number because it is divisible by 2 and 3. Therefore, we need to find the next prime number after 5, which is 7.

Now, let's add the prime number 7 to our sum: 5 + 7 = 12.

To find the last prime number, we need to check if 12 + 1 = 13 is a prime number. Luckily, 13 is indeed a prime number.

Finally, let's add 13 to our sum: 12 + 13 = 25.

Again, the sum of the four consecutive prime numbers, 2, 3, 7, and 13, is 25, which is not divisible by 3.

It seems that there is no set of four consecutive prime numbers whose sum is divisible by 3 with the numbers we have tried so far. Therefore, the smallest possible value of this sum is not found among the given four consecutive prime numbers.

Answer 2

The smallest sum of four consecutive positive prime numbers that is divisible by three is 36, formed by the primes 5, 7, 11, and 13.

Step-by-step explanation:

The question involves finding four consecutive prime numbers whose sum is divisible by three. We know that except for the number 2, all prime numbers are odd. Since the sum of any three odd numbers is odd and adding one more odd (the fourth prime) would make it even, it's impossible for four consecutive odd primes to have a sum divisible by three unless one of them is 2. The smallest consecutive primes starting from 2 are 2, 3, 5, and 7.

Adding these together, 2 + 3 + 5 + 7, we get a sum of 17, which is not divisible by three. However, the sum of the next possible sequence, 3, 5, 7, and 11, provides us with 26, which is also not divisible by three. We thus look at sequences starting with 5 and find that the next set of consecutive primes are 5, 7, 11, and 13. Their sum is 36, which is divisible by three.

Therefore, the smallest possible value of the sum of four consecutive prime numbers that is divisible by three is 36.