All pair of $m,n$ satisfying $lcm(m,n)=600$ 
Find the number of pairs of positive integers $(m,n)$, with $m \le n$, such that
  the ‘least common multiple’ (LCM) of $m$ and $n$ equals $600$.

My tries:
It's very clear that $n\le600$, always.
Case when $n=600=2^3\cdot 3\cdot 5^2$, and let $m=2^{k_1}\cdot 3^{k_2}\cdot 5^{k_3}$, all possible values of $k_1=3+1=4,\ k_2=1+1=2,\ k_3=2+1=3$. So number of $m$ which satisfy above will be $4\cdot 2 \cdot 3=24$
Help me analyzing when $n<600$.
 A: Forget about the condition $m\leq n$ for the moment. Since $600=2^3\cdot 3^1\cdot 5^2$ we have
$$m=2^{\alpha_2}3^{\alpha_3}5^{\alpha_5},\quad n=2^{\beta_2}3^{\beta_3}5^{\beta_5}$$
with $\alpha_i$, $\beta_i\geq0$ and
$$\max\{\alpha_2,\beta_2\}=3,\quad \max\{\alpha_3,\beta_3\}=1,\quad \max\{\alpha_5,\beta_5\}=2\ .$$
It follows that
$$\eqalign{(\alpha_2,\beta_2)&\in\{(0,3),(1,3),(2,3),(3,3),(3,2),(3,1),(3,0)\}\>,\cr
(\alpha_3,\beta_3)&\in\{(0,1),(1,1),(1,0)\}\>,\cr
(\alpha_5,\beta_5)&\in\{(0,2),(1,2),(2,2),(2,1),(2,0)\}\cr}$$
are admissible, allowing for $7\cdot3\cdot5=105$ combinations. Exactly one of them has $m=n$, namely $m=n=600$, and in all other $104$ cases $m\ne n$. Since we want $m\leq n$ we have to throw out half of these cases, leaving $52+1=53$ different solutions of the problem.
A: As you have very nicely written, you have the factorisation
$$600=2^3\cdot  3\cdot 5^2.$$
Now recall that the lcm of two numbers, given their prime factorisation, is the product of their prime factors to the highest power. For example 
$$lcm(2\cdot 3\cdot 5^7,2^2\cdot 5^6)=2^2\cdot3\cdot5^7$$
So you have to find all the possible numbers of the form $2^p3^q5^r$, $p\leq 3,q\leq 1,r\leq 2\ $  so that at least one of them contains $2^3$, $3^5$ and $5^2$ as factors.
