Here's the code for the algorithm:
Foo(n) lcm = 1 for i = 2 to n lcm = lcm*i/Euclid(lcm,i) return lcm
The running time of
Euclid$(a, b)$ is given as $O(\log(\min(a, b)))$
So the running time of the for loop will be $O(n)$, so would this be the final running time? or do I have to take the $O(\log(\min(a, b)))$ into account as well?