Given a set of primes S, show that no prime in S divides the sum of staggered products of these primes.

Question

Definition: Given a set of primes $S = \{p_1, p_2\, \cdots, p_n\}$ with $|S| \ge 2$, form the sum of staggered products using the following formula: $$ \sum_{i} \prod_{j \ne i} p_j $$

Example: If $S = \{2, 3, 5 \}$, then $\sum_{i} \prod_{j \ne i} p_j = 3 \cdot 5 + 2 \cdot 5 + 2 \cdot 3 = 31$

Question:

Is it true that no prime in $S$ divides the sum of staggered products? i.e. Does $p_k | \sum_{i} \prod_{j \ne i} p_j$ for any $k \in \{1, \cdots, n \}$ ?

Comments: I've done some numerical experiments and it seems to be true, but I'm not sure how to go about proving it. I've considered rewriting the sum of staggered products as $\sum_{i} \prod_{j} \frac{p_j}{p_i}$, so I could maybe work in modulo $p_k$ and maybe show that $\sum_{i} \frac{1}{p_i} \not\equiv 0$, but I'm not sure how to actually do this. Any suggestions would be helpful.

[This isn't a homework question, but I'd still like it if full answers weren't provided :)]

Answer 1

Correct. Consider a simpler statement: if $p\mid a$ and $p\mid(a+b)$, then $p\mid b$.

Each prime in the set $S$ is a factor of each term of the sum except one term. If $p_k$ divided the sum of staggered products, it would also have to divide each term in the sum.

Answer 2

Consider your example: each $p$ in $S$ divides all of the sum terms except for one. Can you see why the sum term that $p$ doesn't divide "spoils" the divisibility the sum?