Divisible by 19? 
If $b$ has to be a whole number, I don't understand why $19$ HAS to be a factor of:


*

*$2$ (impossible) 

*$a$ (also impossible because $a$ is a digit $(0-9)$)

*$5 \times 10^{m – 1} – 1$ (possible)


Because as I understand it, 19 could also be a factor of either:


*

*$2a$ (impossible since $a$ is a digit $(0-9)$ so max $18$)

*$2 \times (5 \times 10^{m – 1} – 1)$ 

*$a \times (5 \times 10^{m – 1} – 1)$

*$2a \times (5 \times 10^{m – 1} – 1)$


Maybe I'm totally wrong but it's like saying $(8\times 2) / 16$ is not a whole number because $16$ isn't a factor of $8$ or $2$ (But it is a factor of $8\times 2$).
Thanks in advance
 A: $19$ is a prime number, which means that if $19$ is a factor of $ab$ it is a factor of $a$ or a factor of $b$.
So if $19$ is a factor of $a(bc)$ it is a factor of $a$ or a factor of $bc$. And if it is a factor of $bc$ it is a factor of $b$ or a factor of $c$.
A: Prime $\,p\mid a_1\cdots a_n\,\Rightarrow\, p\mid a_i\,$ for some $i,\,$ i.e. if a prime divides a product then it must divide one of the factors of the product. This is an immediate consequence of the Fundamental Theorem of Arithmetic, i.e. for every natural $> 1$ there $\rm\color{#0a0}{exists}$  a factorization into a product of primes that is $\rm\color{#c00}{unique}$ (up to order).
Indeed, $\,p b = a_1\cdots a_n\,$ thus by $\rm\color{#0a0}{existence}$ we can replace $b$ and the $a_i$ by their prime factorizations. Thus by $\rm\color{#c00}{uniqueness}$, since $p$ occurs in the LHS prime factorization, it must also occur in the RHS, so it occurs in the prime factorization of some $a_i,\,$ hence $\,p\mid a_i.$
Remark $ $ Conversely, the above Prime Divisor Property implies uniqueness of prime factorizations, hence it is equivalent to the uniqueness of prime factorizations.
