HCF $(x,y) = 16$ and LCM $(x,y) = 48000$. Then the possible number of pairs $(x,y)$ Let $S$ be the set of all ordered pairs $(x,y)$ of positive integers, with HCF $(x,y) = 16$ and LCM $(x,y) = 48000$. The
number of elements in $S$ is
My Attempt : $48000= 2^7. 3 . 5^3$  As the L.c.m contains $2^7$ as a factor and G.c.d contains $2^4$ as a factor , we can assure that one element contains $2^7$ and the other one contains $2^4$. And $3$ and $5^3$ will be divided between two numbers in a manner so that the L.C.M and G.C.D remain as given.
So the possible pairs are $(2^7 .3 . 5^3 , 2^4)$ , $(2^7 .3 , 5^3.2^4)$ , $(2^7 , 3.5^3.2^4)$ , $(2^7.5^3 , 3.2^4)$.
So I think the number of elements in the set $S$ is $4$. Have I gone wrong anywhere? Can anyone please help me ?
 A: Let $m$ and $n$ be positive integers such that $m\mid n$.  Then the number of solutions $(x,y)$ of positive integers $x,y$ such that
$$\gcd(x,y)=m$$
and
$$\operatorname{lcm}(x,y)=n$$
is
$$2^{\omega(n/m)},$$
where $\omega$ is the prime $\omega$-function, namely, $\omega(d)$ is the number of distinct prime divisors of $d$ (e.g., $\omega(1)=0$, $\omega(9)=1$, $\omega(12)=2$).  For a proof, each prime $p$ dividing $n/m$ has two choices, namely, 


*

*$p^{s_p} \parallel x$ and $p^{t_p}\parallel y$, or

*$p^{t_p}\parallel x$ and $p^{s_p}\parallel y$.


Here, $p^k \parallel d$ if $p^k\mid d$ but $p^{k+1}\nmid d$, and $s_p$ and $t_p$ are such that $p^{s_p}\parallel m$ and $p^{t_p}\parallel n$.
In particular, for $m=16$ and $n=48000$, we have
$$\frac{n}{m}=3000=2^3\cdot 3\cdot 5^3$$
so that $\omega(n/m)=3$.  Hence, the answer is $2^3=8$.  
If the number of unordered pairs $\{x,y\}$ is to be calculated or if you require that $x\leq y$, then the answer is
$$\left\lceil 2^{\omega\left(\frac{n}{m}\right)-1}\right\rceil\,.$$
(Normally, you wouldn't need the ceiling function.  It is needed only in the special case where $m=n$.)  In particular, for $m=16$ and $n=48000$, the number of solutions with $x\leq y$ is $\frac{2^3}{2}=4$.
A: If $$\gcd(x,y)=16=2^4$$ and $$\operatorname{lcm}(x,y)=48000=2^7\cdot3\cdot5^3$$ then $$xy=2^{11}\cdot3\cdot5^3$$
So we can let 
$$x = 2^\alpha\cdot3^\beta\cdot5^\gamma \qquad \text{and} \qquad
  y = 2^{11-\alpha}\cdot3^{1-\beta}\cdot5^{3-\gamma}$$
We need
$$\begin{align}
   \min(\alpha, 11-\alpha) &= 4 \\
   \max(\alpha, 11-\alpha) &= 7 \\
   \min(\beta, 1-\beta) &= 0 \\
   \max(\beta, 1-\beta) &= 1 \\
   \min(\gamma, 3-\gamma) &= 0 \\
   \max(\gamma, 3-\gamma) &= 3
\end{align}$$
So $$\begin{align}
   \alpha &\in \{4,7\} \\
   \beta &\in \{0,1\} \\
   \gamma &\in \{0,3\}
\end{align}$$
So there are $2\cdot2\cdot2 = 8$ possible $(x,y)$ pairs.
A: WLOG let $\dfrac xX=\dfrac yY=16$ so that $(X,Y)=1$
We have $48000\cdot16=xy=16^2XY\iff XY=3000=3\cdot2^3\cdot5^3$
So, the possible values of $X$ can be 
take none of the factors $$1$$
take one of the factors $$3;2^3;5^3$$
take two of the factors $$2^3\cdot5^3;3\cdot5^3;2^3\cdot3$$
take all three of the factors $$3\cdot2^3\cdot5^3$$
