If we have 3 consecutive odd numbers, then at least 1 is divisible by 3. Thus, no $ p \geq 4$ can satisfy the condition. Hence largest possible number is $p=3$, and it indeed does work.

A slightly more interesting variant: Quote: A "really awesome prime" $p$ is a prime number, such that for all primes $q<p$ also $p+2q$ is prime. Find the largest really awesome prime.

How can it be p7? :thonk: $\textbf{Sol :}$ Clearly $p=3$ works Let $p>3 \implies p,p+2,p+4,p+6 \in prime$. As $p>3, p\equiv 1,2 \pmod 3$ For, $$p\equiv 1 \pmod 3, p+2 \equiv 0 \pmod 3, p+2>3$$For, $$p\equiv 2 \pmod 3, p+4 \equiv 0 \pmod 3, p+4>3$$Hence $p\le 3 \implies p=3\blacksquare$

test20 wrote: A slightly more interesting variant: Quote: A "really awesome prime" $p$ is a prime number, such that for all primes $q<p$ also $p+2q$ is prime. Find the largest really awesome prime. $\textbf {Sol :}$ Clearly $p=7$ works $p\equiv 2 \pmod 3$ doesn't work as then $p+4>3 \in prime$ can't happen. Hence, $p=11,17$ doesn't work. $p=13,19$ doesn't work as it gives $(13+2\times7, 19+2\times3) \in prime$ Let $p > 19$ Then consider $X=(p,p+4,p+6,p+22,p+38) \in prime$ Taking $mod 5$ we are done, hence $p=7\blacksquare$