On finding triplets that satisfy a certain GCD and LCM property.

$10=2^15^1$ and $100=2^25^2$. Let $a=2^{a_2}5^{a_5}$, and similarly for $b$ and $c$. Because the gcd expresses the minimum exponent for each prime across its arguments, we have $$\min(a_2,b_2,c_2)=1\qquad \min(a_5,b_5,c_5)=1$$ Similarly, the lcm expresses the maximum prime exponents, and we have $$\max(a_2,b_2,c_2)=2\qquad \max(a_5,b_5,c_5)=2$$ One of $a_2,b_2,c_2$ has to be $1$ and another $2$. The third may be either $1$ or $2$. This gives six possibilities for the triple $(a_2,b_2,c_2)$ (as can be directly enumerated easily), and by symmetry there are six possibilities for $(a_5,b_5,c_5)$ independent of the other set of variables. Thus there are $6×6=36$ triples $(a,b,c)$ satisfying the conditions.

All three are divisible by $10$, so we divide each by $10$ and find the triples with GCD = $1$ and LCM = $10$.

There can only be factors $2$ and $5$, and both must be there. So, the numbers can be $1$, $2$, $5$ or $10$, and either $2$ and $5$ or just $10$ must be present. They cannot all three be divisible by $2$, or all three divisible by $5$. If we take the numbers in sorted order:

$$ (1,1,10);\ (1,2,5);\ (1,2,10);\ (1,5,10);\ (1,10,10);\ (2,2,5);\ (2,5,5);\ (2,5,10) $$

That’s $8$ solutions, and you can arrange the numbers in different order. Multiply by $10$ to solve the original problem.