Higher Secondary P7

If there exists a prime number $p$ such that $p+2q$ is prime for all positive integer $q$ smaller than $p$, then $p$ is called an "awesome prime". Find the largest "awesome prime" and prove that it is indeed the largest such prime.
To find the largest "awesome prime" \( p \), we need to ensure that \( p + 2q \) is prime for all positive integers \( q \) smaller than \( p \). Let's analyze the given solution step-by-step.

1. **Verification for \( p = 7 \)**:
   - For \( p = 7 \), we need to check if \( 7 + 2q \) is prime for all \( q \) such that \( 1 \leq q < 7 \).
   - Check:
     - \( q = 1 \): \( 7 + 2 \times 1 = 9 \) (not prime)
     - \( q = 2 \): \( 7 + 2 \times 2 = 11 \) (prime)
     - \( q = 3 \): \( 7 + 2 \times 3 = 13 \) (prime)
     - \( q = 4 \): \( 7 + 2 \times 4 = 15 \) (not prime)
     - \( q = 5 \): \( 7 + 2 \times 5 = 17 \) (prime)
     - \( q = 6 \): \( 7 + 2 \times 6 = 19 \) (prime)
   - Since \( 9 \) and \( 15 \) are not prime, \( p = 7 \) does not satisfy the condition for all \( q \).

2. **Verification for \( p = 11 \)**:
   - For \( p = 11 \), we need to check if \( 11 + 2q \) is prime for all \( q \) such that \( 1 \leq q < 11 \).
   - Check:
     - \( q = 1 \): \( 11 + 2 \times 1 = 13 \) (prime)
     - \( q = 2 \): \( 11 + 2 \times 2 = 15 \) (not prime)
     - \( q = 3 \): \( 11 + 2 \times 3 = 17 \) (prime)
     - \( q = 4 \): \( 11 + 2 \times 4 = 19 \) (prime)
     - \( q = 5 \): \( 11 + 2 \times 5 = 21 \) (not prime)
     - \( q = 6 \): \( 11 + 2 \times 6 = 23 \) (prime)
     - \( q = 7 \): \( 11 + 2 \times 7 = 25 \) (not prime)
     - \( q = 8 \): \( 11 + 2 \times 8 = 27 \) (not prime)
     - \( q = 9 \): \( 11 + 2 \times 9 = 29 \) (prime)
     - \( q = 10 \): \( 11 + 2 \times 10 = 31 \) (prime)
   - Since \( 15 \), \( 21 \), \( 25 \), and \( 27 \) are not prime, \( p = 11 \) does not satisfy the condition for all \( q \).

3. **Verification for \( p = 13 \)**:
   - For \( p = 13 \), we need to check if \( 13 + 2q \) is prime for all \( q \) such that \( 1 \leq q < 13 \).
   - Check:
     - \( q = 1 \): \( 13 + 2 \times 1 = 15 \) (not prime)
     - \( q = 2 \): \( 13 + 2 \times 2 = 17 \) (prime)
     - \( q = 3 \): \( 13 + 2 \times 3 = 19 \) (prime)
     - \( q = 4 \): \( 13 + 2 \times 4 = 21 \) (not prime)
     - \( q = 5 \): \( 13 + 2 \times 5 = 23 \) (prime)
     - \( q = 6 \): \( 13 + 2 \times 6 = 25 \) (not prime)
     - \( q = 7 \): \( 13 + 2 \times 7 = 27 \) (not prime)
     - \( q = 8 \): \( 13 + 2 \times 8 = 29 \) (prime)
     - \( q = 9 \): \( 13 + 2 \times 9 = 31 \) (prime)
     - \( q = 10 \): \( 13 + 2 \times 10 = 33 \) (not prime)
     - \( q = 11 \): \( 13 + 2 \times 11 = 35 \) (not prime)
     - \( q = 12 \): \( 13 + 2 \times 12 = 37 \) (prime)
   - Since \( 15 \), \( 21 \), \( 25 \), \( 27 \), \( 33 \), and \( 35 \) are not prime, \( p = 13 \) does not satisfy the condition for all \( q \).

4. **Verification for \( p = 19 \)**:
   - For \( p = 19 \), we need to check if \( 19 + 2q \) is prime for all \( q \) such that \( 1 \leq q < 19 \).
   - Check:
     - \( q = 1 \): \( 19 + 2 \times 1 = 21 \) (not prime)
     - \( q = 2 \): \( 19 + 2 \times 2 = 23 \) (prime)
     - \( q = 3 \): \( 19 + 2 \times 3 = 25 \) (not prime)
     - \( q = 4 \): \( 19 + 2 \times 4 = 27 \) (not prime)
     - \( q = 5 \): \( 19 + 2 \times 5 = 29 \) (prime)
     - \( q = 6 \): \( 19 + 2 \times 6 = 31 \) (prime)
     - \( q = 7 \): \( 19 + 2 \times 7 = 33 \) (not prime)
     - \( q = 8 \): \( 19 + 2 \times 8 = 35 \) (not prime)
     - \( q = 9 \): \( 19 + 2 \times 9 = 37 \) (prime)
     - \( q = 10 \): \( 19 + 2 \times 10 = 39 \) (not prime)
     - \( q = 11 \): \( 19 + 2 \times 11 = 41 \) (prime)
     - \( q = 12 \): \( 19 + 2 \times 12 = 43 \) (prime)
     - \( q = 13 \): \( 19 + 2 \times 13 = 45 \) (not prime)
     - \( q = 14 \): \( 19 + 2 \times 14 = 47 \) (prime)
     - \( q = 15 \): \( 19 + 2 \times 15 = 49 \) (not prime)
     - \( q = 16 \): \( 19 + 2 \times 16 = 51 \) (not prime)
     - \( q = 17 \): \( 19 + 2 \times 17 = 53 \) (prime)
     - \( q = 18 \): \( 19 + 2 \times 18 = 55 \) (not prime)
   - Since \( 21 \), \( 25 \), \( 27 \), \( 33 \), \( 35 \), \( 39 \), \( 45 \), \( 49 \), \( 51 \), and \( 55 \) are not prime, \( p = 19 \) does not satisfy the condition for all \( q \).

5. **Verification for \( p = 5 \)**:
   - For \( p = 5 \), we need to check if \( 5 + 2q \) is prime for all \( q \) such that \( 1 \leq q < 5 \).
   - Check:
     - \( q = 1 \): \( 5 + 2 \times 1 = 7 \) (prime)
     - \( q = 2 \): \( 5 + 2 \times 2 = 9 \) (not prime)
     - \( q = 3 \): \( 5 + 2 \times 3 = 11 \) (prime)
     - \( q = 4 \): \( 5 + 2 \times 4 = 13 \) (prime)
   - Since \( 9 \) is not prime, \( p = 5 \) does not satisfy the condition for all \( q \).

6. **Verification for \( p = 3 \)**:
   - For \( p = 3 \), we need to check if \( 3 + 2q \) is prime for all \( q \) such that \( 1 \leq q < 3 \).
   - Check:
     - \( q = 1 \): \( 3 + 2 \times 1 = 5 \) (prime)
     - \( q = 2 \): \( 3 + 2 \times 2 = 7 \) (prime)
   - Since both \( 5 \) and \( 7 \) are prime, \( p = 3 \) satisfies the condition for all \( q \).

Therefore, the largest "awesome prime" is \( p = 3 \).

The final answer is \( \boxed{3} \).