A prime number problem. If $n$ is a positive integer and $(p_1,p_2,p_3,p_4,\ldots, p_n)$ are distinct positive primes, show that the integer $(p_1\cdot p_2\cdot p_3\cdot p_4\cdots p_n)+1$ is divisible by none of these primes. How do I figure out that none of the primes divide that new integer?
 A: Suppose $7$ is one of the primes, so $p_1\cdots p_n$ is a multiple of $7$.
The next multiple of $7$ after $p_1\cdots p_n$ is $(p_1\cdots p_n)+7$.  So $(p_1\cdots p_n)+1$ is not a multiple of $7$.
More formally, suppose $7$ divides $(p_1\cdots p_n)+1$.  Then for some integers $j$, $k$,
$$
\begin{align}
(p_1 \cdots p_n) & = 7j \tag 1 \\
(p_1 \cdots p_n) + 1 & = 7k \tag 2 \\[10pt]
\text{Then subtracting $(1)$ from $(2)$, we get: } 1 & = 7k - 7j = 7(k-j) \\
\end{align}
$$
So $1 = 7(k-j)$.
A: We will need the following theorem:
Theorem. For any two natural numbers $x, y$ ($ y \geq 1 $), there are unique natural numbers $q, r$ where $ 0 \leq r < y $ such that $x = qy + r $.
Proof. If $ x < y $ then we must have $ q = 0 $ and $ r = x $, so in this case the $q, r$ exist and are unique. Assume that there is a natural number $ x \geq y $ such that no such $q, r$ exist, then without loss of generality we may assume that $ x $ is the smallest such number. However, then $x - y$ cannot be written in the desired form either, as if we had $x - y = qy + r $ then this would imply $x = (q+1)y + r$. This contradicts the minimality of $ x $, as $ x - y $ is a smaller natural number with the same property. Therefore, such $q, r$ must exist for all values of $x$.
To prove uniqueness, assume that we had $x = q_1 y + r_1 = q_2 y + r_2$ and $r_1 > r_2$, then $r_1 - r_2 = y(q_2 - q_1)$. On the other hand, $r_1 - r_2 \leq r_1 < y$, so that we have $y(q_2 - q_1) < y$ and therefore $q_1 = q_2$. QED.
Corollary. $x = qn + 1$ ($q, n \in \mathbb{N}$) is not divisible by $ q $ for any $q \geq 2$.
Proof. If it were, we would have $x = qm$ for some $ m $ and $x = qn + 1$ simultaneously, contradicting the uniqueness proved in the above theorem.
The statement in the question follows immediately from the corollary.
