To Prove the relation between HCF and LCM of three numbers
if $p,q,r$ are three positive integers prove that
$$LCM(p,q,r)=\frac{pqr \times HCF(p,q,r)}{HCF(p,q) \times HCF(q,r) \times HCF(r,p)}$$
I tried in this way;
Let $HCF(p,q)=x$ hence $p=xm$ and $q=xn$ where $m$ and $n$ are relatively prime.
similarly let $HCF(q,r)=y$ hence $q=ym_1$ and $r=yn_1$ where $m_1$ and $n_1$ are Relatively prime.
Alo let $HCF(r,p)=z$ hence $r=zm_2$ and $p=zn_2$
we have $$p=xm=zn_2$$
$$q=xn=ym_1$$and
$$r=yn_1=zm_2$$
can i have any hint to proceed?
$\endgroup$ 32 Answers
$\begingroup$I decided to write my comment as an answer. Rather than start with naming $HCF(p,q)$, $HCF(q,r)$ and $HCF(r,p)$, start with $HCF(p,q,r)$. So let's call $HCF(p,q,r) = h$.
Next, write $HCF(p,q) = xh$, $HCF(q,r) = yh$ and $HCF(r,p) = zh$. It should be clear why we can assume the factor $h$ appears in all three, but you also know that $x,y,z$ are relatively prime. (Why?)
Thus, you can write $p = p'xzh$ for some $p'$, and similarly $q = q'xyh$ and $r = r'yzh$ (again, why?). What do you get when you plug those into your equation?
$\endgroup$ 2 $\begingroup$Let $P, Q$ and $R$ be the sets of prime factors of the numbers $p,q$ and $r$ respectively. Let the universal set be $P\cup Q\cup R$. Let there be product function $v(S)$ which gives the product of the elements of set S.
Then, celarly,
$$pqr=v(P\cap Q' \cap R')*v(Q\cap P' \cap R')*v(R\cap P'\cap Q')*v^2(P\cap Q\cap R')*v^2(Q\cap R\cap P')*v^2(R\cap P\cap Q')*v^3(P\cap Q\cap R)$$
Now,
$$LCM(p,q,r)=v(P\cap Q' \cap R')*v(Q\cap P' \cap R')*v(R\cap P'\cap Q')*v(P\cap Q\cap R')*v(Q\cap R\cap P')*v(R\cap P\cap Q')*v(P\cap Q\cap R)$$
$$HCF(p,q)=v(P\cap Q\cap R')*v(P\cap Q\cap R)$$
$$HCF(q,r)=v(Q\cap R\cap P')*v(P\cap Q\cap R)$$
$$HCF(r,p)=v(R\cap P\cap Q')*v(P\cap Q\cap R)$$
$$HCF(p,q,r)=v(P\cap Q\cap R)$$
These expressions satisfy your equation.
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