Show that $S=\frac{1}{a_1}+\cdots+\frac{1}{a_n}$ is not an integer. So I have a bunch of integers $a_1,...,a_n$ and a prime number $p$ that divides only one of the numbers in this sequence, say $a_k$
I want to show, that $S=\dfrac{1}{a_1}+\cdots+\dfrac{1}{a_n}$ is not an integer.
Well we can express the sum in another way, by putting them under a commond denominator, then the sum is
$$\sum_{i=1}^n\dfrac{\prod_{j=1,j \neq i}^n a_j}{\prod_{j=1}^n a_j}$$
All of the members of this sum are divisible by $p$ expect the $k$'th one. But is this helpful in any way? 
 A: $$
S = \frac{\sum_{l=1}^n \prod_{j\in\{1,..,n\}, j\neq l} a_j}{\prod_{j=1}^n a_j} = \frac{\sum_{l\in\{1,..,n\}, l\neq k} \prod_{j\in\{1,..,n\}, j\neq l} a_j + \prod_{j\in\{1,..,n\}, j\neq k} a_j}{\prod_{j=1}^n a_j} 
$$
Each term in the left sum is divisible by p exactly once and the denominator is divisible by p.  So there exist integers $r$ and $s$ such that
$$ S = \frac{r\cdot p + \prod_{j\in\{1,..,n\}, j\neq k} a_j}{s\cdot p}$$
The right term is not divisible by p, so the numerator is not divisible by p.  Hence S is not an integer.
A: Assume by contradiction the sum is an integer, lets say $m$.
Then
$$\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_{k-1}}+\frac{1}{a_{k+1}}+...+\frac{1}{a_n} =m-\frac{1}{a_k} \,.$$
Now, when you do the addition on the LHS, $p$ doesn't divide the denominator, so it won't divide it even after reducing the fraction.
 On the RHS, $\frac{ma_{k}-1}{a_k}$ is already reduced and $p$ divides the denominator...
Thus, you get two reduced fractions which are equal but have different denominators, which is the contradiction.
A: The pedantic way: The $p$-adic absolute value on $\mathbf Q$ satisfies the ultrametric inequality $|x+y|_p \leq \max\{|x|_p, |y|_p\}$. Let $S$ be your sum. Then
$$|S-1/a_k| \leq \max_{i \neq k}\{|1/a_i|_p\} \leq  1$$
because the $a_i$'s are $p$-adic units for $i \neq k$. 
On the other hand, $$p\leq  |1/a_k|_p \leq \max\{|S-1/a_k|_p, |S|_p\}\leq \max\{1, |S|_p\}$$
implies $|S|_p\geq p$, so $S$ is not an integer.
