Past exam of the options I course of the University of Oxford / 2004 / C / C1 / a / ii / Solution

Write $a={\prod }_p{p}^{r_p}$ and $b={\prod }_p{p}^{s_p}$. Divisibility is exactly comparison of prime exponents. Thus the greatest common divisor uses $\min (r_p,s_p)$ and the least common multiple uses $\max (r_p,s_p)$. Since their sum is $r_p+s_p$ for each prime,