Finding unknowns from HCF and LCM The Highest common factor and Lowest common multiple of two numbers $A$ and $B$ are $12$ and $168$ respectively . Find the possible values of $A$ and $B$ with exception of $12$ and $168$.
Workings -
HCF $(A,B) = 12 = 2^2 \times 3$
LCM $(A,B) = 168 = 2^3 \times 3 \times 7$
$A \times B$ = LCM $\times$ HCF = $12 \times 168$
$A \times B = (2^3 \times 3) \times (2^3 \times 3 \times 7)$ 
I'm stuck from here onwards on how to find the possible values. Thanks for any help.
 A: (1) $12$ has to be part of both numbers - since it is the GCD.
(2) In order to get the LCM as $168$ we got to distribute the remaining $14$ among these two numbers, being careful that the distribution has no common factors.
(3) The factors of $14$ are $1,2,7,14$.  So we take the first number as $12.1$, $12.2$, $12.7$, $12.14$ and the second number as $12.14$,$12.7$,$12.2$ and $12.1$ respectively
(4) Given that we exclude the extremities, we have only one value up to symmetry.  
Finally, we have $\color{blue}{(A,B) \in \{(24,84), (84,24)\}}$
Some Notes
This problem becomes a little interesting if the LCM given was say $336$. Now you have to be more careful in distributing the $28$. It cannot be split as $2$ and $2$, otherwise it will affect the GCD. So you have to take the $4$ factor completely. 
lab bhattacharjee has of of course a more formal approach.
A: WLOG let $\dfrac Aa=\dfrac Bb=12\implies(a,b)=1$
and lcm$(A,B)=12$lcm$(a,b)=12ab\implies ab=14$
which can also be derived as follow: 
As $A\cdot B=$HCF$(A,B)\cdot$LCM$(A,B)$
$$12a\cdot12b=12\cdot168\iff ab=?$$
As $(a,b)=1$  the possible set of values of $(a,b)$ are $$\{(1,14);(2,7);(7,2);(14,1)\}$$
