If $n_{1}|a,....n_{k}|a$ and $gcd(n_i,n_{j})=1$ for all $i \neq j$, then the product of all the $n_i$ divides a?

Question

Hello I know that if $a|bc$ and $gcd(a,b)=1$ then $a|c$

but is this the same as

if $n_{1}|a,....n_{k}|a$

and $gcd(n_i,n_{j})=1$ for all $i \neq j$

then the product of all the $n_i$ divides a?

I know the result is true as I saw it in the proof of the Chinese remainder thereom, but is the proof for this similar? What is it?

thanks

Answer 1

If $gcd(n_1,n_{2})=1$ and $n_{1}|a,n_{2}|a$

there is $k$ such that $a=n_1k$ since $n_2|a=n_1 k$ and $gcd(n_1,n_{2})=1$ thus $n_2|k$ so there is $k_2$ such that $n_2k_2=k$, hence $a=n_1n_2k_2$ means $n_1n_2|a$

Now by using induction you will have the result.