Solve for $A,B$: $\mathrm{LCM}(A,B)=168$, $\mathrm{HCF(A,B)}=12$

Question

The highest common factor and the lowest common multiple of two numbers $A$ and $B$ are $12$ and $168$ respectively. Find the possible values of $A$ and $B$ with the exception of $12$ and $168$.

Help me with this equation please. I kind of get it but I don't understand the concept behind it... I do not want to memorize.

Answer 1

HINT:

1. Use the property that for two positive integers A and B,

   Product of the two numbers = L.C.M. $\times$ H.C.F. i.e. $A \times B = \mathrm{L.C.M.} \times \mathrm{H.C.F.}$

   If you are not aware of this property, then you can check the proof of this formula from here.

2. Use the prime factorisation of $A \times B$ to obtain the different values of $A$ and $B$.

Answer 2

You can also use the fact that the prime decomposition of hcf is obtained using the primes common in A and B, elevated to their highest powers apparing in both A and B, and the prime decomposition of lcm is obtained using all possible primes in A or B, elevated at the highest power in either A or B. So wlog, $A=2^{m_1}.3.7^{n_1}$ and $B=2^{m_2}.3.7^{n_2}$, with $n_1=1$ and $n_2=0$ or the converse, and $m_i\geq 2$ and one of the $m_i=3$, but not the other.

Answer 3

You did not say that $A,B$ are positive. We have $A=x L$ and $B=y L$ where $hcf (A,B)=L=12$ and $hcf(x,y)=1.$ We have $168=lcm (A,B)=|x y L|=12 |x y|$ because the convention is that $lcm$ means least common positive multiple.(Also see Appendix, below.) So $$|A B|=|12 x|\cdot |12 y|= |x y L^2|=|x y L| L=(168)(12)=2016.$$ Therefore $|x y|=2016/12^2=14.$ Since $hcf (x,y)=1,$ this means $\{x,y\}$ is one of $$\{1,14\},\{-1,-14\},\{2,7\},\{-2,-7\}.$$So $\{A,B\},$ which is $\{12 x, 12 y\},$ is one of $$\{12,168\},\{-12,-168\},\{24,84\},\{-24,-84\}.$$ The set $\{24,84\}$ appears to be the one you are interested in. APPENDIX: $gcd$ (for greatest common divisor) is the usual term for $hcf.$ Also, for any positive integers $p,q$ we have $ p=x l$ and $q=y l$ where $l=gcd (p,q)$ and $gcd (x,y)=1$; Now $x z l$ is a multiple of $p$ whenever $z$ is a positive integer, but $$q| x z l \implies (x z l)/q=x z l/y l=x z/y \in N \implies z/y\in N$$...(because $gcd (x,y)=1$)... so $q|x z l\implies z\geq y.$ This gives $lcm (p,q)\geq x y l.$ But $x y l$ is a common multiple of $p,q$. Therefore $lcm (p,q)=x y l.$ And so we have $$p q=(x l) (y l)=(x y l)l= lcm (p,q)\cdot gcd (p,q).$$