Question

Let $p_1,p_2,\dots,p_n$ be distinct pisitive primes. Show that $(p_1p_2\dots p_n)+1$ is divisible by none of these primes.

Answer 1

Assume that there exists a prime say $p_i$, where $i\leq n$ such that $p_i$ divides $p_1p_2\dots p_n+1$. Then clearly $p_i|p_1p_2\dots p_n$ and $p_i|p_1p_2\dots p_n+1$ implies that $p_i|1=(p_1\dots p_n+1)-(p_1\dots p_n)$.

Which is imposible as $p_i\geq2$. Hence none of the $p_i$'s divides $p_1p_2\dots p_n+1$.