Find $x,y$ given $\gcd(x,y)$ and ${\rm lcm}(x,y)$

$\begin{align}{\bf Hint}\ \ &\gcd(X,Y) = d,\ \ \ {\rm lcm}(X,Y) = m \ \ \ \text{yields by cancelling $\,d\neq 0$}\\[.3em] \iff\ &\gcd(x,\,y)\ \ = 1,\qquad\ \ \, x\cdot y\ =\, m/d,\ \ {\rm for}\ \ x = X/d,\,\ y = Y/d \end{align}$

$\begin{align}{\rm e.g.}\ \ \ &\gcd(X,Y) = 15,\ \, {\rm lcm}(X,Y) = 150\\[.3em] \iff\ &\gcd(x,\,y)\ \ =\ 1,\qquad\ \ \, x\cdot y\,\ =\,\ 10 \end{align}$

This method quickly and simply solves all of them. Cancelling to reduce to the coprime case is a common way to simplify homogeneous divisibility problems.

Here is the relevant fact:

Let $d=gcd(a,b)$ and $m=lcm(a,b)$. Then $v_p(d)=\min(v_p(a),v_p(b))$ and $v_p(m)=\max(v_p(a),v_p(b))$.

Here, $v_p(n)$ is the exponent of the prime $p$ in the factorization of $n$.

If you have two numbers with prime factorizations

$$x = p_1^{a_1}p_2^{a_2}p_3^{a_3}\cdots p_n^{a_n}$$ $$y = p_1^{b_1}p_2^{b_2}p_3^{b_3}\cdots p_n^{b_n}$$

then

$$GCD(x,y) = p_1^{min(a_1, b_1)}p_2^{min(a_2, b_2)}p_3^{min(a_3, b_3)}\cdots p_n^{min(a_n, b_n)}$$

and

$$LCM(x,y) = p_1^{max(a_1, b_1)}p_2^{max(a_2, b_2)}p_3^{max(a_3, b_3)}\cdots p_n^{max(a_n, b_n)}$$

where $min(a,b)$ and $max(a,b)$ are the minimum and maximum of $a$ and $b$, respectively.

Does this help?