Let $k=8$ and $q_1=12$, $q_2=2, q_3=5$. We note that $k\mid q_1q_2q_3$, but $k\nmid q_1$ and $k\nmid q_2$ and $k\nmid q_3$.
Also
$$ 8=k_1k_2k_3=4*2*1, $$
where $k_1\mid q_1$, $k_2\mid q_2$, and $k_3\mid q_3$.
I want to prove that if $k\mid q_1...q_n$, then we can find $k_i$ such that $k=k_1...k_n$ and for every $i$ holds $k_i\mid q_i$, like in the example of the beginning.
My idea is to descompose every $q_i$ in prime numbers by the Fundamental theorem of aritmethic. Then
$$ q_1...q_n=p_1^{a_1}...p_s^{a_s}. $$
So if $k\mid q_1...q_n$, then
$$ k=p_1^{b_1}..p_s^{b_s} $$
and
$$ 0\leq b_i\leq a_i. $$
Now I know every $q_i$ is a combination of powers of $p_i$ so if want to make a $k_i$ such that $k_i\mid q_i$, the $k_i$ should be a combination of this $p_i$ and its exponent should be less than $b_i$, let's call it $c_{i,j}$. Then $\sum_{j=1} c_{i,j}=b_i$.
(Do I make myself clear?)
In the example $$q_1\cdot q_2\cdot q_3=(2^2\cdot 3)\cdot(2)\cdot(5)=2^3\cdot3\cdot5$$ then $$8=2^{b_1}\cdot3^{b_2}\cdot5^{b_3}$$ Obviously by the FTA $b_1=3$, $b_2=0$, $b_3=0$
Note $q_1$ and $q_2$ has powers of $2$ then we need $c_{1,1}$ and $c_{1,2}$
$c_{1,1}=2$ and $c_{1,2}=1$ and it verify $c_{1,1}+c_{1,2}=2+1=3=b_1$
Then
$k_1=2^{c_{1,1}}\cdot 3^{b_2}=2^2\cdot3^0=4$
$k_2=2^{c_{1,2}}=2^1=2$
$k_3=5^{b_3}=5^0=1$
I don't know how to make more clear my proof or if it is incorrect or just I don't proof anything. Can you help me please?