Let $7(p_1\cdot p_2\dotsb p_s)+3=q_1\cdot q_2 \dotsb q_t$ where the $p_1,p_2,\dots,p_s$ and $q_1,q_2,\dots,q_t$ are primes and where none of the $p_i$ is $3$. Argue that none of the $q_j$ are 3 and that each $q_j$ is different from each $p_i$.
I'm not sure how to go about solving this. Therefore I will start off by taking the first 10 primes (excluding the number $3$ for some odd unknown reason to me) and making sure that when $7(\text{product of primes})+3=\text{a prime}$. If this is not true, then I would think this suggests that what was stated is false.
$7(2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 29\cdot31)+3=467974476973 $
However, $467974476973 $ is not a prime, because $14519|467974476973 $. Therefore, I clearly don't know what I'm talking about.
Perhaps, my strategy excluding the $3$ was incorrect.
$7(2\cdot 3\cdot5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 29)+3=45287852613 $
Again, I am wrong, because $3|45287852613$, obviously.
In other words, I need some type of assistance regarding how to solve what is being asked.