Find the sum of primes p such that p²+11 has exactly six different positive divisors (including 1 and the number itself).
To find the sum of primes p that satisfy the given conditions, we need to break down the problem into steps.
Step 1: Identify the prime numbers
First, let's identify the prime numbers that we need to test. We start by checking the smallest prime number, which is 2, and continue checking subsequent odd numbers. If a number is prime, we move to the next step.
Step 2: Calculate p²+11
For each prime number p, we calculate p²+11. This will give us a new number, which we'll call n.
Step 3: Find the divisors of n
To find the divisors of n, we need to iterate from 1 up to the square root of n and check if each number divides n without a remainder. If it does, we count it as a divisor.
Step 4: Count the number of divisors
Once we have found all the divisors of n, we count how many there are. If there are exactly six divisors, we move to the next step.
Step 5: Sum up the primes
If a number satisfies the condition of having exactly six divisors, we add the corresponding prime number p to the running total sum.
Step 6: Repeat steps 2-5
We repeat steps 2-5 for all prime numbers until we have checked all possible primes.
Step 7: Return the sum
After checking all primes, we return the sum of the found primes.
Let's implement these steps in code to solve the problem efficiently.
To find the sum of primes p such that p²+11 has exactly six different positive divisors, including 1 and the number itself, we need to follow these steps:
Step 1: Understand the problem.
We are looking for primes (represented by p) that satisfy a certain condition. The condition is that when we square the prime number, add 11 to it, and count the number of positive divisors, the count should be exactly six.
Step 2: Break down the problem.
To solve this problem, we need to consider two key aspects:
- Prime numbers: We need to iterate through prime numbers and check if they meet the specified condition.
- Count of divisors: To count the divisors of a number, we need to understand how divisors are related to prime factorization.
Step 3: Determine the prime numbers to consider.
To find the prime numbers that satisfy the condition, we can use a prime number generation algorithm, such as the Sieve of Eratosthenes. This algorithm will generate primes up to a certain limit. We can choose a large enough limit based on the range of numbers we expect to encounter.
Step 4: Test the condition for each prime.
For each prime number generated, we square it, add 11, and count the number of divisors. We need to ensure that the count of divisors is exactly six.
Step 5: Sum the primes that satisfy the condition.
For each prime that meets the condition, add it to a running sum.
Step 6: Output the final sum.
After checking all primes and summing the ones that satisfy the condition, output the final sum.
Note that implementing the Prime number generation algorithm and Divisor counting algorithm are further steps that need to be explained. Let me know if you would like to proceed with the explanations for those steps as well.
if the prime factors of p²+1, excluding 1 and itself, are of the form {a²,b | a≠b}, then the possible factors are:
{1,a,b,a², ab, a²b} total exactly 6 factors.
The first few such sums are:
1²+11=12=2²*3 [but 1 is not a prime]
3²+11=20=2²*5 [cannot be the sum of primes]
8²+11=75=3*5² factors are {1,3,5,15,25,75}, and 8=3+5, both primes.
9²+11=92=2²*23; factors are {1,2,4,23,46,92}, and 9=2+2+5
14²+11=207=3²*23; factors are {1,3,9,23,69,207}, and 14=2+5+7
15²+11=236=2²*59; factors are {1,2,4,59,118,236}, and 15=3+5+7
...
I guess we can go on and on.