Problem 1641

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 $\boxed{3}$.